DBMS LAB MANUAL

INTRODUCTION TO SQL-PLSQL

2011-11-16T21:14:00.000-08:00

Introduction to SQL-PLSQL

How to Write and execute sql, pl/sql commands/programs:

1). Open your oracle application by the following navigation

Start->all programs->Oracle Database 10g Express Edition

->Run SQL Command Line

2). You will be asked for user name, password.

You have to enter user name, pass word.

3). Upon successful login you will get SQL prompt (SQL>).

In two ways you can write your programs:

a) directly at SQL prompt (or)

b) in sql editor.

If you type your programs at sql prompt then screen will look like follow:

SQL> SELECT ename,empno,

2 sal from

3 emp;

where 2 and 3 are the line numbers and rest is the command

/program……

to execute above program/command you have to press ‘/’ then enter.

Here editing the program is somewhat difficult; if you want to edit the previous command then you have to open sql editor (by default it displays the sql buffer contents). By giving ‘ed’ at sql prompt.(this is what I mentioned as a second method to type/enter the program).

in the sql editor you can do all the formatting/editing/file operations directly by selecting menu options provided by it.

To execute the program which saved; do the following

SQL> @ programname.sql (or)

SQL> Run programname.sql

Then press ‘\’ key and enter.

To save the day`s session ;do the following

SQL>commit;

This how we can write, edit and execute the sql command and

programs.

Always you have to save your programs in your own logins.

Background Theory

Oracle workgroup or server is the largest selling RDBMS product.it is

estimated that the combined sales of both these oracle database product

account for aroud 80% of the RDBMSsystems sold worldwide.

These products are constantly undergoing change and evolving. The

natural language of this RDBMS product is ANSI SQL,PL/SQL a superset

of ANSI SQL.oracle 8i and 9i also under stand SQLJ.

Oracle corp has also incorporated a full-fledged java virtual machine

into its database engine.since both executable share the same memory

space the JVM can communicate With the database engine with ease

and has direct access to oracle tables and their data.

SQL is structure query language.SQL contains different data types those

are

1. char(size)

2. varchar(size)

3. varchar2(size)

4. date

5. number(p,s) //** P-PRECISION S-SCALE **//

6. number(size)

7. raw(size)

8. raw/long raw(size)

Different types of commands in SQL:

A).DDL commands: - To create a database objects

B).DML commands: - To manipulate data of a database objects

C).DQL command: - To retrieve the data from a database.

D).DCL/DTL commands: - To control the data of a database…

DDL commands:

1. The Create Table Command: - it defines each column of the table

uniquely. Each column has minimum of three attributes, a name , data

type and size.

Syntax:

Create table <table name> (<col1> <datatype>(<size>),<col2>

<datatype><size>));

Ex:

create table emp(empno number(4) primary key, ename char(10));

2. Modifying the structure of tables.

a)add new columns

Syntax:

Alter table <tablename> add(<new col><datatype(size),<new

col>datatype(size));

Ex:

alter table emp add(sal number(7,2));

3. Dropping a column from a table.

Syntax:

Alter table <tablename> drop column <col>;

Ex:

alter table emp drop column sal;

4. Modifying existing columns.

Syntax:

Alter table <tablename> modify(<col><newdatatype>(<newsize>));

Ex:

alter table emp modify(ename varchar2(15));

5. Renaming the tables

Syntax:

Rename <oldtable> to <new table>;

Ex:

rename emp to emp1;

6. truncating the tables.

Syntax:

Truncate table <tablename>;

Ex:

trunc table emp1;

7. Destroying tables.

Syntax:

Drop table <tablename>;

Ex:

drop table emp;

DML commands:

8. Inserting Data into Tables: - once a table is created the most

natural thing to do is load this table with data to be manipulated later.

Syntax 1:

insert into <tablename> (<col1>,<col2>…..<col n>) values(<val 1>,

<val 2>…….<val n>);

Syntax 2:

insert into <tablename> values(&<col1>,&<col2>……,&<col n>);

Syntax 3:

insert into <tablename> values(<val 1>,<val 2>…….,<val n>);

Ex 1:

Insert into skc (sname,rollno,class,dob,fee_paid)

values(‘sri’,’104B’,’cse’,’27-feb-05’,10000.00);

Ex 2:

insert into skc values(&sname,&roll no,&class);

enter sname:’sri’

enter roll no:’104B’

enter class:’cse’

1 row created.

Ex 3:

insert into skc values(‘sri’,’104B’,cse’,’27-feb-05’,10000.00);

9. Delete operations.

a) remove all rows

Syntax:

delete from <tablename>;

b) removal of a specified row/s

Syntax:

delete from <tablename> where <condition>;

10. Updating the contents of a table.

a) updating all rows

Syntax:

Update <tablename> set <col>=<exp>,<col>=<exp>;

b) updating seleted records.

Syntax:

Update <tablename> set <col>=<exp>,<col>=<exp>

where <condition>;

11. Types of data constrains.

a) not null constraint at column level.

Syntax:

<col><datatype>(size)not null

b) unique constraint

Syntax:

Unique constraint at column level.

<col><datatype>(size)unique;

c) unique constraint at table level:

Syntax:

Create table

tablename(col=format,col=format,unique(<col1>,<col2>);

d) primary key constraint at column level

Syntax:

<col><datatype>(size)primary key;

e) primary key constraint at table level.

Syntax:

Create table tablename(col=format,col=format

primary key(col1>,<col2>);

f) foreign key constraint at column level.

Syntax:

<col><datatype>(size>) references <tablename>[<col>];

g) foreign key constraint at table level

Syntax:

foreign key(<col>[,<col>]) references

<tablename>[(<col>,<col>)

h) check constraint

check constraint constraint at column level.

Syntax: <col><datatype>(size) check(<logical expression>)

i) check constraint constraint at table level.

Syntax: check(<logical expression>)

DQL Commands:

12. Viewing data in the tables: - once data has been inserted into a

table, the next most logical operation would be to view what has been

inserted.

a) all rows and all columns

Syntax:

Select <col> to <col n> from tablename;

Select * from tablename;

13. Filtering table data: - while viewing data from a table, it is rare

that all the data from table will be required each time. Hence, sql must

give us a method of filtering out data that is not required data.

a) Selected columns and all rows:

Syntax:

select <col1>,<col2> from <tablename>;

b) selected rows and all columns:

Syntax:

select * from <tablename> where <condition>;

c) selected columns and selected rows

Syntax:

select <col1>,<col2> from <tablename> where<condition>;

14. Sorting data in a table.

Syntax:

Select * from <tablename> order by <col1>,<col2> <[sortorder]>;

DCL commands:

Oracle provides extensive feature in order to safeguard information

stored in its tables from unauthoraised viewing and damage.The rights

that allow the user of some or all oracle resources on the server are

called privileges.

a) Grant privileges using the GRANT statement

The grant statement provides various types of access to database

objects such as tables,views and sequences and so on.

Syntax:

GRANT <object privileges>

ON <objectname>

TO<username>

[WITH GRANT OPTION];

b) Reoke permissions using the REVOKE statement:

The REVOKE statement is used to deny the Grant given on an object.

Syntax:

REVOKE<object privilege>

FROM<user name>;

WEEK-1

CREATING,ALTERING AND DROPPING TABLES AND INSERTING ROWS INTO A TABLE (USE CONSTRAINTS WHILE CREATING TABLES) EXAMPLES USING

SELECT COMMAND .

EXAMPLE 1:

CREATING A STUDENT RELATION TABLE WITH ALL DATATYPES:

SQL> create table student252(

sid number(5),

sname varchar(20),

sbranch char(5),

dob date,

spercent number(3,2));

Table created.

RELATIONAL SCHEMA FOR STUDENT RELATION :

SQL> desc student252;

Name Null? Type

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

SID NUMBER(5)

SNAME VARCHAR2(20)

SBRANCH CHAR(5)

DOB DATE

SPERCENT NUMBER(5,2)

INSERT THE RECORDS INTO STUDENT RELATION:

METHOD 1:

SQL>Insert into

Student252(sid,sname,sbranch,dob,spercent) values(104,‘sri’,,’cse’,’27-

feb-05’,70);

1 row created.

METHOD 2:

SQL>Insert into

Student252 values(104,‘sri’,,’cse’,’27-feb-05’,70);

1 row created.

METHOD 3:

SQL>Insert into

Student252(sid,sname,sbranch,dob,spercent)

values(&sid, &sname,&sbranch,&dob,&spercent);

1 row created.

METHOD 4:

SQL>Insert into

Student252(sid,sname,sbranch,dob,spercent)

values(&sid, ‘&sname’,’&sbranch’,’&dob’,&spercent);

1 row created.

QUERY THE TABLE VALUES:

ALL ROWS AND ALL COLUMNS:

SQL> select * from student252;

SID SNAME SBRANCH DOB SPERCENT

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

130 ravi it 30-1-95 60

131 teja cse 21-07-87 55

129 kiran mech 12-05-92 60

104 sri cse 30-07-90 70

133 sajith eee 12-06-89 55

137 ram ece 07-07-85 40

WEEK 2 (cont…1)

1) Creation, altering and dropping tables and inserting rows into a table (use constraints while creating tables) examples using SELECT command.

MODIFYING THE STRUCTURE OF TABLE

ADDING A NEW COLUMN

SQL> ALTER TABLE Emp252

ADD (age number(3),

phno number(10));

Table altered.

MODIFYING EXISTING COLUMN

SQL> ALTER TABLE Emp252

MODIFY (phno varchar(20));

Table altered.

DROPING A COLUMN

SQL> ALTER TABLE Emp252

DROP COLUMN phno;

Table altered.

QUERY FOR THE TABLE VALUES

SQL> SELECT *

FROM Emp252;

ENO ENAME ESAL DEPTNO AGE

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

30 ravi 51000 3

31 teja 31000 2

29 kiran 31200 1

45 allen 41000 3

33 sajith 51000 4

46 geetha 11000 4

90 veena 16000 3

85 pragna 61000 1

84 harsha 91000 3

40 sanjeev 1500 13

10 rows selected.

UPDATING ENTIRE COLUMN

SQL> UPDATE Emp252

SET age=18;

10 rows updated.

QUERY THE TABLE VALUES

SQL> SELECT *

FROM Emp252;

ENO ENAME ESAL DEPTNO AGE

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

30 ravi 51000 3 18

31 teja 31000 2 18

29 kiran 31200 1 18

45 allen 41000 3 18

33 sajith 51000 4 18

46 geetha 11000 4 18

90 veena 16000 3 18

85 pragna 61000 1 18

84 harsha 91000 3 18

40 sanjeev 1500 13 18

10 rows selected.

RENAMING THE TABLE:

SQL> RENAME Emp252

TO Emp1252;

Table renamed.

SELECTING THE TABLE VALUES

SQL> SELECT *

FROM Emp1252;

Example 3

CREATING A DEPARTMENT RELATION TABLE

CREATING A DEPARTMENT TABLE

SQL> CREATE TABLE Dept252(

dname VARCHAR(10),

dno CHAR(5),

dloc VARCHAR(25));

Table created.

DESCRIBE A STUDENT TABLE

SQL> desc Dept252;

Name Null? Type

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

DNAME VARCHAR2(10)

DNO CHAR(5)

DLOC VARCHAR2(25)

DROPING THE TABLE

SQL> DROP TABLE Dept252;

Table dropped.

WEEK 3 (cont…1)

1) Creation, altering and dropping tables and inserting rows into a table (use constraints while creating tables) examples using SELECT command.

CREATING A TABLE WITH KEY CONSTRAINTS

Example 1

CREATING A TABLE WITH ‘UNIQUE ‘, ‘NOT NULL’, ‘CHECK’ AND ‘DEFAULT’ CONSTRAINT:

SQL> CREATE TABLE emp252

(eid NUMBER(5) UNIQUE,

ename VARCHAR(10) DEFAULT(‘UNKNOWN’),

age NUMBER(3) NOT NULL,

esal NUMBER(7) CHECK(esal > 1000));

Table created.

INSERTING RECORDS INTO TABLE:

SQL> INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal);

Enter value for eid: 1

Enter value for ename: 'ravi'

Enter value for age: 18

Enter value for esal: 10000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (1, 'ravi', 18, 10000)

1 row created.

SQL> /

Enter value for eid: 2

Enter value for ename: 'teja'

Enter value for age: 18

Enter value for esal: 20000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (2, 'teja', 18, 20000)

1 row created.

SQL> /

Enter value for eid: 3

Enter value for ename: 'kiran'

Enter value for age: 19

Enter value for esal: 25000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (3, 'kiran', 19, 25000)

1 row created.

SQL> /

Enter value for eid: 4

Enter value for ename: 'srinivas'

Enter value for age: 19

Enter value for esal: 30000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (4, 'srinivas', 19, 30000)

1 row created.

SQL> /

Enter value for eid: 1

Enter value for ename: 'alan'

Enter value for age: 19

Enter value for esal: 29000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (1, 'alan', 19, 29000)

INSERT INTO emp252 VALUES (1, 'alan', 19, 29000)

[SHOWING AN ERROR WHILE VIOLATING UNIQUE KEY CONSTRAINT]

ERROR at line 1:

ORA-00001: unique constraint (SYSTEM.SYS_C003875) violated

SQL> /

Enter value for eid: 7

Enter value for ename: 'dravid'

Enter value for age: null

Enter value for esal: 100000

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (7, 'dravid', null, 100000)

INSERT INTO emp252 VALUES (7, 'dravid', null, 100000)

[SHOWING AN ERROR AS NOT NULL KEY CONSTRAINT IS VIOLATED] *

ERROR at line 1:

ORA-01400: cannot insert NULL into ("SYSTEM"."EMP230"."AGE")

SQL> /

Enter value for eid: 8

Enter value for ename: 'sachin'

Enter value for age: 35

Enter value for esal: 100

old 1: INSERT INTO emp252 VALUES (&eid, &ename, &age, &esal)

new 1: INSERT INTO emp252 VALUES (8, 'sachin', 35, 100)

INSERT INTO emp252 VALUES (8, 'sachin', 35, 100)

[NOT ALLOWING AS IT VOILATES CHECK CONSTRAINT FOR esal > 1000 VALUE]

ERROR at line 1:

ORA-02290: check constraint (SYSTEM.SYS_C003874) violated

Example 2

CREATING A TABLE WITH ‘PRIMARY KEY’ CONSTRAINT:

SQL> CREATE TABLE mdept252

(dno NUMBER(5),

dname CHAR(10),

dloc VARCHAR(10),

PRIMARY KEY (dno));

Table created.

SQL> desc mdept252;

Name Null? Type

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

DNO NOT NULL NUMBER(5)

DNAME CHAR(10)

DLOC VARCHAR2(10)

INSERTING RECORDS INTO MASTER DEPARTMENT TABLE:

SQL> INSERT INTO mdept252 VALUES (&dno, &dname, &dloc);

Enter value for dno: 1

Enter value for dname: 'ravi'

Enter value for dloc: 'hyd'

old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)

new 1: INSERT INTO mdept252 VALUES (1, 'ravi', 'hyd')

1 row created.

SQL> /

Enter value for dno: 1

Enter value for dname: 'teja'

Enter value for dloc: 'sec'

old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)

new 1: INSERT INTO mdept252 VALUES (1, 'teja', 'sec')

INSERT INTO mdept252 VALUES (1, 'teja', 'sec')

ERROR at line 1:

ORA-00001: unique constraint (SYSTEM.SYS_C003876) violated

SQL> /

Enter value for dno: null

Enter value for dname: 'sajithulhuq'

Enter value for dloc: 'kmm'

old 1: INSERT INTO mdept252 VALUES (&dno, &dname, &dloc)

new 1: INSERT INTO mdept252 VALUES (null, 'sajithulhuq', 'kmm')

INSERT INTO mdept252 VALUES (null, 'sajithulhuq', 'kmm')

ERROR at line 1:

ORA-01400: cannot insert NULL into ("SYSTEM"."MDEPT230"."DNO")

ADDING A PRIMARY KEY TO AN EXISTING TABLE:

SQL> ALTER TABLE student252 ADD PRIMARY KEY (sid);

Table altered.

SQL> ALTER TABLE emp252 ADD PRIMARY KEY (eid);

ALTER TABLE emp252 ADD PRIMARY KEY (eid)

[ GIVING AN ERROR AS ONE TABLE CAN HAVE A SINGLE PRIMARY KEY AT COLUMN LAVEL]

ERROR at line 1:

ORA-02261: such unique or primary key already exists in the table

Example 3

CREATING A TABLE WITH ‘FORIEGN KEY’ CONSTRAINT:

SQL> CREATE TABLE detailemp252

(eid NUMBER(5) REFERENCES mdept230 (dno),

ename VARCHAR(10),

esal NUMBER(7));

Table created.

INSERING RECORDS INTO DETAIL EMPLOYEE TABLE:

SQL> INSERT INTO detailemp252 VALUES (2, 'ravi', 50000);

INSERT INTO detailemp252 VALUES (2, 'ravi', 50000)

ERROR at line 1:

ORA-02291: integrity constraint (SYSTEM.SYS_C003877) violated - parent key not

found

SQL> INSERT INTO detailemp252 VALUES (1, 'teja', 60000);

1 row created.

SQL> DELETE FROM mdept252 where dno=1;

DELETE FROM mdept252 where dno=1

ERROR at line 1:

ORA-02292: integrity constraint (SYSTEM.SYS_C003877) violated - child record

found

SQL> SELECT * FROM detailemp252;

EID ENAME ESAL

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

1 teja 60000

SQL> SELECT * FROM mdept252;

DNO DNAME DLOC

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

1 ravi hyd

Exercise

CREATING A CUSTOMER TABLE USING CONSTRAINTS :

SQL> CREATE TABLE cust252

(cnum NUMBER(5),

cname VARCHAR(10),

state VARCHAR(10) DEFAULT ('ap'),

phno NUMBER(5),

CONSTRAINT cnum_pkkey PRIMARY KEY (cnum));

Table created.

SQL> INSERT INTO cust252 VALUES (&cnum, &cname, &state, &phno);

Enter value for cnum: 1

Enter value for cname: 'ravi'

Enter value for state: 'bihar'

Enter value for phno: 001

old 2: (&cnum, &cname, &state, &phno)

new 2: (1, 'ravi', 'bihar', 001)

1 row created.

SQL> /

Enter value for cnum: 2

Enter value for cname: 'teja'

Enter value for state: 'up'

Enter value for phno: 007

old 2: (&cnum, &cname, &state, &phno)

new 2: (2, 'teja', 'up', 007)

1 row created.

SQL> /

Enter value for cnum: 2

Enter value for cname: 'yama'

Enter value for state: 'ap'

Enter value for phno: 006

old 2: (&cnum, &cname, &state, &phno)

new 2: (2, 'yama', 'ap', 006)

INSERT INTO cust252 VALUES

ERROR at line 1:

ORA-00001: unique constraint (SYSTEM.CNUM_PKKEY) violated

SQL> /

Enter value for cnum: 4

Enter value for cname: 'huu'

Enter value for state: 'ap'

Enter value for phno: 101

old 2: (&cnum, &cname, &state, &phno)

new 2: (4, 'huu', 'ap', 101)

1 row created.

SQL> SELECT * FROM cust252;

CNUM CNAME STATE PHNO

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

1 ravi bihar 1

2 teja up 7

4 huu ap 101

CREATING AN ITEM TABLE USING CONSTRAINTS:

SQL> CREATE TABLE itm252

(ino NUMBER(3),

iname VARCHAR(10),

iprice NUMBER(4,3),

qtyonhand VARCHAR(5),

CONSTRAINT itm252_ino_pkkey PRIMARY KEY (ino),

CONSTRAINT itm230_qtyoh_chk CHECK (qtyonhand>1));

Table created.

SQL> INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand);

Enter value for ino: 1

Enter value for iname: 'rubber'

Enter value for iprice: 3.50

Enter value for qtyonhand: 3

old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)

new 1: INSERT INTO itm252 VALUES (1, 'rubber', 3.50, 3)

1 row created.

SQL> /

Enter value for ino: 1

Enter value for iname: 'pencil'

Enter value for iprice: 1.00

Enter value for qtyonhand: 3

old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)

new 1: INSERT INTO itm252 VALUES (1, 'pencil', 1.00, 3)

INSERT INTO itm252 VALUES (1, 'pencil', 1.00, 3)

ERROR at line 1:

ORA-00001: unique constraint (SYSTEM.ITM230_INO_PKKEY) violated

SQL> /

Enter value for ino: 2

Enter value for iname: 'powder'

Enter value for iprice: 3.00

Enter value for qtyonhand: 0

old 1: INSERT INTO itm252 VALUES (&ino, &iname, &iprice, &qtyonhand)

new 1: INSERT INTO itm252 VALUES (2, 'powder', 3.00, 0)

INSERT INTO itm252 VALUES (2, 'powder', 3.00, 0)

ERROR at line 1:

ORA-02290: check constraint (SYSTEM.ITM230_QTYOH_CHK) violated

1* CREATE TABLE invoice252(ivnno NUMBER(5), itemno NUMBER(5), qty NOT NULL, CONSTRAINT invoice252_ivnno_pkkey PRIMARY KEY(ivnno), CONSTRAINT FOREIGN KEY(itemno) REFERENCES cust252)

SQL> desc cust252;

Name Null? Type

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

CNUM NOT NULL NUMBER(5)

CNAME VARCHAR2(10)

STATE VARCHAR2(10)

PHNO NUMBER(5)

CREATING A INVOICE TABLE USING CONSTRAINTS:

SQL> CREATE TABLE invoice252

(ivnno NUMBER(5),

itemno NUMBER(5),

qty NUMBER(5) NOT NULL,

CONSTRAINT invoice252_ivnno_pkkey PRIMARY KEY (ivnno),

CONSTRAINT fk_inv252 FOREIGN KEY (itemno) REFERENCES cust252 (cnum))

Table created.

SQL> CREATE TABLE invitm252

(invno NUMBER(5),

itmno NUMBER(5),

qty NUMBER(5) NOT NULL,

CONSTRAINT invitm252_invno_itmno_pkkey PRIMARY KEY (invno, itmno));

Table created.

WEEK 4

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints.

Example: select the rollno and name of the student who secured 4^th rank in the class

TABLE DEFINITIONS

SQL> CREATE TABLE Customer (

cust_no NUMBER(4) PRIMARY KEY,

last_name VARCHAR2(20),

first_name VARCJHAR2(20) NOT NULL,

address1 VARCHAR2(20),

address2 VARCHAR2(20),

city VARCHAR2(3),

state VARCHAR2(20),

pin VARCHAR2(6),

birth_date DATE,

status VARCHAR2(1),

CHECH (status IN (‘V’, ‘I’, ‘A’))

);

Table created.

Insert the following data:

1 row created.

CUST NO	LAST NAME	FIRST NAME	ADDRESS1	ADDRESS2	CITY	STATE	PIN	BIRTH DATE	STATUS
1001	UDUPI	RAJ	UPENDRABAUG	NEAR KALPANA	UDPP	KARNARATA	576101	12-DEC-62	A
1002	KUMAR	RAJ							A
1003	BAHADUR	RAJ	SHANTHI VILLA	NEAR MALLIKA	UDP	KARNATAKA	576101	1-AUG-70	V
1004	SIMON	FELIX	M-J-56	ALTOBETIM	PJM	GOA	403002	12-FEB-71	A
1005	KUTTY	RAJAN	A1 TRADERS	NEAR RLY STATION	KNR	KERALA	67001	9-JUN-71	A
1006	PAI	SHILPA	12/4B	POLICE QUARTERS	MNG	KARNATAKA	574154	11-DEC-70	I
1007	JAIN	RAKSHIT	BOSCO	R.K PLAZA	BNG	KARNATAKA	576201	1-JAN-71	A

QUERIES

1) To list all the fields from the table Customer.

SEELCT *

FROM Customer;

2) To list the first name, last name.

SELECT first_name, last_name

FROM Customer;

3) To list the first name and last name of persons in Karnataka.

SELECT first_name, last_name

FROM Customer

WHERE state = ‘KARNATAKA’;

4) To list all the columns for invalid persons.

SELECT *

FROM Customer

WHERE status = ‘I’;

5) To list the names of active customers.

SELECT first_name, last_name

FROM Customer

WHERE status = ‘A’;

6) To list the name and address using concatenation.

SELECT first_name || ‘ ‘ || last_name, address1 || ‘,’ || address2 || ‘,’ || city || ‘,’ || state || ‘-‘ || pin

FROM Customer;

7) To select records where the pin code has not been entered.

SELECT *

FROM Customer

WHERE pin IS NULL;

8) To select the single occurrence of any value from the table.

SELECT DISTINCT state

FROM Customer;

9) To select rows of valid customers from Karnataka.

SELECT *

FROM Customer

WHERE state = ‘KARNATAKA’

AND status = ‘V’;

10) To select rows of customers from Karnataka or Kerala.

SELECT *

FROM Customer

WHERE state = KARNATAKA’

OR state = ‘KERALA’;

11) To sort the customer data in the alphabetic order of state.

SELECT state, first_name, last_name, pin

FROM Customer

ORDER BY state;

12) To sort in the descending order.

SELECT state, first_name, last_name, pin

FROM Customer

ORDER BY state DESC;

13) To sort the customer data, state wise and within state by the last name.

SELECT state, first_name, last_name, pin

FROM Customer

ORDER BY state, last_name;

14) To retrieve records of Karnataka customers who are valid.

SELECT *

FROM Customer

WHERE UPPER(state) = ‘KARNATAKA’

AND UPPER(status) = ‘V’;

15) To retrieve records of Karnataka/Kerala customers.

SELECT *

FROM Customer

WHERE UPPER(state) = ‘KARNATAKA’

OR UPPER(state) = ‘KERALA’;

16) To retrieve records of Karnataka/Kerala customers who are active.

SELECT *

FROM Customer

WHERE (UPPER(state) = ‘KARNATAKA’

OR UPPER(state) = ‘KERALA’)

AND UPPER(status) = ‘A’;

17) To retrieve records of Karnataka customers with pin code 576101.

SELECT *

FROM Customer

WHERE LOWER(state) = ‘karnataka’

AND pin = ‘576101’;

18) To retrieve rows where the state name begins with K and followed by any other character.

SELECT first_name, last_name, state

FROM Customer

WHERE state LIKE ‘K%’;

19) To retrieve rows where the first name contains the word RAJ embedded in it.

SELECT first_name, last_name, state

FROM Customer

WHERE first_name LIKE ‘%RAJ%’;

20) To retrieve rows where the address2 contains the word UDUPI or UDIPI in which the 3^rd character may be anything.

SELECT first_name, last_name, state

FROM Customer

WHERE address2 LIKE ‘UD_PI’;

21) To retrieve rows where the cust_no has data representing any value between 1003 and 1005, both numbers included.

SELECT *

FROM Customer

WHERE cust_no BETWEEN 1003 AND 1005;

22) To retrieve rows of persons born after 9-JAN-70 and before 1-AUG-96.

SELECT *

FROM Customer

WHERE birth_date BETWEEN ’10-JAN-70’ AND ’31-JUL-96’;

23) To retrieve rows where the city has data which is equal to UDP or MNG or BNG or PJM or MAR.

SELECT *

FROM Customer

WHERE city IN (‘UDP’, ‘MNG’, ‘BNG’, ‘PJM’, ‘MAR’);

TABLE DEFINITIONS

SQL> CREATE TABLE Emp (

emp_no NUMBER,

emp_name VARCHAR(20),

join_date DATE,

join_basic NUMBER(7, 2),

PRIMARY KEY (emp_no)

);

Table created.

Insert the following data:

EMP NO	EMP NAME	JOIN DATE	JOIN BASIC
1001	Subhas bose	01-JUN-96	3000
1002	Nadeem shah	01-JUN-96	2500
1003	Charles babbage	01-JUN-96	3000
1004	Shreyas kumar	01-JUL-96	2500
1005	George boole	01-JUL-96	2800

SQL> CREATE TABLE Salary (

emp_no NUMBER,

basic NUMBER(7, 2),

commission NUMBER(7, 2),

deduction NUMBER(7, 2),

salary_date DATE,

FOREIGN KEY (emp_no) REFERENCES Emp

);

Table created.

Insert the following data:

EMP NO	BASIC	COMMISSION	DEDUCTION	SALARY DATE
1001	3000	200	250	30-JUN-96
1002	2500	120	200	30-JUN-96
1003	3000	500	290	30-JUN-96
1004	2500	200	300	30-JUN-96
1005	2800	100	250	30-JUN-96
1001	3000	200	250	31-JUL-96
1002	2500	120	200	31-JUL-96
1003	3000	500	290	31-JUL-96
1004	2500	200	300	31-JUL-96
1005	2800	100	150	31-JUL-96

QUERIES

1) To sum the salary of each employee.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no;

2) To sum the salary of each employee and sort it on the sum of basic.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no

ORDER BY SUM(basic);

3) To sum the salary of each employee and sort it in descending order on the sum of basic.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no

ORDER BY SUM(basic) DESC;

4) To sum the salary of each employee and sort it in descending order on the sum of basic. Display name also

SELECT s.emp_no, e.emp_name, SUM(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_no

ORDER BY SUM(s.basic) DESC;

5) To group the data by average salary of each employee.

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_no

ORDER BY AVG(s.basic);

6) To group the basic by month.

SELECT TO_CHAR(salary_date, ‘MONTH’) “MONTH”, SUM(basic) “TOTAL BASIC”

FROM salary

GROUP BY TO_CHAR(salary_date, ‘MONTH’);

7) To group the data by average salary of each employee and display where average basic is more than 2000..

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_no

HAVING AVG(s.basic) >= 2000

ORDER BY AVG(s.basic);

SUBQUERIES

8) To list the employees who earn less than the average salary.

SELECT *

FROM salary

WHERE basic < (SELECT AVG(basic)

FROM salary);

9) To list the employees whose deduction is 150.

SELECT *

FROM salary

WHERE emp_no IN (SELECT emp_no

FROM salary

WHERE deduction = 150);

10) To list the names of employees and salary details, whose basic is less than the average salary.

SELECT s.*, e.emp_name

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

AND s.basic < (SELECT AVG(basic)

FROM salary);

WEEK 5

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4^th rank in the class.

3) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

4) Queries using Conversions, functions (to_char, to_num, and to_date), string function (Conactenation, lpad, rpad, ltrim, rtrim, lower, upper, initcap, length, substr, and instr), date functions (sysdate, next_day, add_months, last_day, months_between, least, greatest, trunk, round, to_char, to_date).

TABLES WITH QUARIES

2011-11-15T10:34:00.000-08:00

TABLE DEFINITIONS

SQL> CREATE TABLE Emp (

emp_no NUMBER,

emp_name VARCHAR(20),

join_date DATE,

join_basic NUMBER(7, 2),

PRIMARY KEY (emp_no)

);

Table created.

Insert the following data:

EMP NO	EMP NAME	JOIN DATE	JOIN BASIC
1001	Subhas bose	01-JUN-96	3000
1002	Nadeem shah	01-JUN-96	2500
1003	Charles babbage	01-JUN-96	3000
1004	Shreyas kumar	01-JUL-96	2500
1005	George boole	01-JUL-96	2800

SQL> CREATE TABLE Salary (

emp_no NUMBER,

basic NUMBER(7, 2),

commission NUMBER(7, 2),

deduction NUMBER(7, 2),

salary_date DATE,

FOREIGN KEY (emp_no) REFERENCES Emp

);

Table created.

Insert the following data:

EMP NO	BASIC	COMMISSION	DEDUCTION	SALARY DATE
1001	3000	200	250	30-JUN-96
1002	2500	120	200	30-JUN-96
1003	3000	500	290	30-JUN-96
1004	2500	200	300	30-JUN-96
1005	2800	100	250	30-JUN-96
1001	3000	200	250	31-JUL-96
1002	2500	120	200	31-JUL-96
1003	3000	500	290	31-JUL-96
1004	2500	200	300	31-JUL-96
1005	2800	100	150	31-JUL-96

QUERIES

1) To sum the salary of each employee.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no;

2) To sum the salary of each employee and sort it on the sum of basic.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no

ORDER BY SUM(basic);

3) To sum the salary of each employee and sort it in descending order on the sum of basic.

SELECT emp_no, SUM(basic)

FROM salary

GROUP BY emp_no

ORDER BY SUM(basic) DESC;

4) To sum the salary of each employee and sort it in descending order on the sum of basic. Display name also

SELECT s.emp_no, e.emp_name, SUM(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_name

ORDER BY SUM(s.basic) DESC;

5) To group the data by average salary of each employee.

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_no

ORDER BY AVG(s.basic);

6) To group the basic by month.

SELECT TO_CHAR(salary_date, ‘MONTH’) “MONTH”, SUM(basic) “TOTAL BASIC”

FROM salary

GROUP BY TO_CHAR(salary_date, ‘MONTH’);

7) To group the data by average salary of each employee and display where average basic is more than 2000..

SELECT s.emp_no, INITCAP(e.emp_name), AVG(s.basic)

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

GROUP BY s.emp_no, e.emp_no

HAVING AVG(s.basic) >= 2000

ORDER BY AVG(s.basic);

SUBQUERIES

8) To list the employees who earn less than the average salary.

SELECT *

FROM salary

WHERE basic < (SELECT AVG(basic)

FROM salary);

9) To list the employees whose deduction is 150.

SELECT *

FROM salary

WHERE emp_no IN (SELECT emp_no

FROM salary

WHERE deduction = 150);

10) To list the names of employees and salary details, whose basic is less than the average salary.

SELECT s.*, e.emp_name

FROM salary s, emp e

WHERE s.emp_no = e.emp_no

AND s.basic < (SELECT AVG(basic)

FROM salary);

WEEK 6

2) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4^th rank in the class.

3) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

TABLE DEFINITIONS

Branch Schema <branch-name, branch-city, assets>

Customer Schema <customer-name, customer-street, customer-city>

Loan Schema <loan-number, branch-name, amount>

Borrower Schema <customer-name, loan-number>

Account Scheme <account-number, branch-name, balance>

Depositor Scheme <customer-name, account-number>

BRANCH TABLE

Branch Name	Branch City	Assets
Brighton	Brooklyn	7100000
Downtown	Brooklyn	9000000
Mianus	Horseneck	400000
North Town	Rye	3700000
Perryridge	Horseneck	1700000
Pownal	Bennington	300000
Redwood	Palo Alto	2100000
Round Hill	Horseneck	800000

CUSTOMER TABLE

Customer Name	Customer Street	Customer City
Adams	Spring	Pittsfield
Brooks	Senator	Brooklyn
Curry	North	Rye
Glenn	Sand Hill	Woodside
Green	Walnut	Stamford
Hayes	Main	Harrison
Johnson	Alma	Palo Alto
Jones	Main	Harrison
Lindsay	Park	Pittsfield
Smith	North	Rye
Turner	Putnam	Stamford
Williams	Nassau	Princeton

LOAN TABLE

Loan Number	Branch Name	Amount
L-11	Round Hill	900
L-14	Downtown	1500
L-15	Perryridge	1500
L-16	Perryridge	1300
L-17	Downtown	1000
L-23	Redwood	2000
L-93	Mianus	500

BORROWER TABLE

Customer Name	Loan Number
Adams	l-16
Curry	L-93
Hayes	L-15
Jackson	L-14
Jones	L-17
Smith	L-11
Smith	L-23
Williams	L-17

ACCOUNT TABLE

Account Number	Branch Name	Balance
A-101	Downtown	500
A-102	Perryridge	400
A-201	Brighton	900
A-215	Mianus	700
A-217	Brighton	750
A-222	Redwood	700
A-305	Round Hill	350

DEPOSITOR TABLE

Customer Name	Account Number
Hayes	A102
Johnson	A-101
Johnson	A-201
Jones	A-217
Lindsay	A-222
Smith	A-215
Turner	A-305

QUERIES

1) To list all the fields from the table Customer.

SELECT branch_name

FROM Loan;

2) To list rows after eliminating duplicates.

SELECT distinct branch_name

FROM Loan;

3) To explicitly list rows, including duplicates.

SELECT all branch_name

FROM Loan;

4) To list fields after applying arithmetic operations.

SELECT loan_number, branch_name, amount *100

FROM Loan;

5) Find all loan numbers for loans made at the Perryridge branch with loan amounts

greater than Rs1200.

SELECT loan_number

FROM Loan

WHERE branch_name = ‘Perryridge’

AND amount > 1200;

6) Find all loan numbers for loans with loan amounts between Rs90,000 and Rs100,000.

SELECT loan_number

FROM Loan

WHERE amount BETWEEN 90000 AND 100000;

SELECT loan_number

FROM Loan

WHERE amount <= 100000

AND amount >= 90000;

7) Find all loan numbers for loans with loan amounts not between Rs90,000 and

Rs100,000.

SELECT loan_number

FROM Loan

WHERE amount NOT BETWEEN 90000 AND 100000;

8) For all customers who have a loan from the bank, find their names, loan numbers and loan amounts.

SELECT customer_name, Borrower.loan_number, amount

FROM Borrower, Loan

WHERE Borrower.loan_number = Loan.loan_number;

SELECT customer_name, Borrower.loan_number AS loan_id, amount

FROM Borrower, Loan

WHERE Borrower.loan_number = Loan.loan_number;

9) Find the customer names, loan numbers and loan amounts for all loans at the Perryridge branch.

SELECT customer_name, Borrower.loan_number, amount

FROM Borrower, Loan

WHERE Borrower.loan_number = Loan.loan_number

AND branch_name = ‘Perryridge’;

SELECT customer_name, T.loan_number, S.amount

FROM Borrower AS T, Loan AS S

WHERE T.loan_number = S.loan_number

AND branch_name = ‘Perryridge’;

10) Find the names of all branches that have assets greater than atleast one branch located in Brooklyn.

SELECT DISTINCT T.branch_name

FROM Branch as T, Branch as S

WHERE T.assets > S.assets

AND S.branch_city = ‘Brooklyn’;

11) Find the names of all customers whose street address includes the substring ‘Main’.

SELECT customer_name

FROM Customer

WHERE customer_street LIKE ‘%Main%’;

12) To list in alphabetic order all customers who have a loan at the Perryridge branch.

SELECT DISTINCT customer_name

FROM Borrower B, Loan L

WHERE B.loan_number = L.loan_number

AND branch_name = ‘Perryridge’

ORDER BY customer_name;

13) To list the entire loan info in descending order of amont.

SELECT *

FROM Loan

ORDER BY amount DESC, loan_number ASC;

14) To find all customers having a loan, an account or both at the bank, without duplicates.

(SELECT customer_name

FROM Depositor)

UNION

(SELECT customer_name

FROM Borrower);

15) To find all customers having a loan, an account or both at the bank, with duplicates.

(SELECT customer_name

FROM Depositor)

UNION ALL

(SELECT customer_name

FROM Borrower);

16) To find all customers having both a loan and an account at the bank, without duplicates.

(SELECT customer_name

FROM Depositor)

INTERSECT

(SELECT customer_name

FROM Borrower);

17) To find all customers having a loan, an account or both at the bank, with duplicates.

(SELECT customer_name

FROM Depositor)

INTERSECT ALL

(SELECT customer_name

FROM Borrower);

18) To find all customers who have an account but no loan at the bank, without duplicates.

(SELECT DISTINCT customer_name

FROM Depositor)

EXCEPT

(SELECT customer_name

FROM Borrower);

19) To find all customers who have an account but no loan at the bank, with duplicates.

(SELECT DISTINCT customer_name

FROM Depositor)

EXCEPT ALL

(SELECT customer_name

FROM Borrower);

20) Find the average account balance at the Perryridge branch

SELECT branch_name, AVG(balance)

FROM Account

WHERE branch_name = ‘Perryridge’;

21) Find the average account balance at the each branch

SELECT AVG(balance)

FROM Account

GROUP BY branch_name;

22) Find the number of depositors for each branch .

SELECT branch_name, COUNT(DISTINCT customer_name)

FROM Depositor D, Account A

WHERE D.account_number = A.account_number

GROUP BY branch_name;

23) Find the number of depositors for each branch where average account balance is more than Rs 1200.

SELECT branch_name, COUNT(DISTINCT customer_name)

FROM Depositor D, Account A

WHERE D.account_number = A.account_number

GROUP BY branch_name

HAVING AVG(balance) > 1200;

24) Find the average balance for all accounts.

SELECT AVG(balance)

FROM Account;

25) Find the number of tuples in the customer relation.

SELECT COUNT(*)

FROM Customer;

26) Find the average balance for each customer who lives in Harrision and has at least three accounts.

SELECT D.customer_name, AVG(balance)

FROM Depositor D, Account A, Customer C

WHERE D.account_number = A.account_number

AND D.customer_name = C.customer_name

AND C.customer_city = ‘Harrison’

GROUP BY D.customer_name

HAVING COUNT(DISTINCT D.account_number) >= 3;

27) Find all the loan number that appear in loan relation with null amount values.

SELECT loan_number

FROM Loan

WHERE amount IS NULL;

28) Find all customers who have both a loan and an account at the bank.

SELECT customer_name

FROM Borrower

WHERE customer_street IN (SELECT customer_name

FROM Depositor);

29) Find all customers who have both an account and a loan at the Perryridge branch

SELECT DISTINCT B.customer_name

FROM Borrower B, Loan L

WHERE B.loan_number L.loan_number

AND branch_name = ‘Perryridge’

AND (branch_name, customer_name) IN

(SELECT branch_name, customer_name

FROM Depositor D, Account A

WHERE D.account_number = A.account_number);

SELECT customer_name

FROM Borrower B

WHERE EXISTS (SELECT *

FROM Depositor D

WHERE D.customer_name = B.customer_name);

30) Find all customers who do not have a loan at the bank, but do not have an account the bank.

SELECT DISTINCT customer_name

FROM Borrower

WHERE customer_name NOT IN

(SELECT customer_name

FROM Depositor);

31) Find the names of customers who do have a loan at the bank, and whose names are neither Smith nor Jones.

SELECT DISTINCT customer_name

FROM Borrower

WHERE customer_name NOT IN (‘Smith’, ‘Jones’);

32) Find the names of all branches that have assets greater than those of at least one branch located in Brooklyn.

SELECT DISTINCT T.branch_name

FROM Branch AS T, Branch AS S

WHERE T.assets > S.assets

AND S.branch_city = ‘Brooklyn’;

33) Find the names of all branches that have assets greater than that of each branch located in Brooklyn.

SELECT branch_name

FROM Account

GROUP BY branch_name

HAVING AVG(balance) >= ALL (SELECT AVG(balance)

FROM Account

GROUP BY branch_name);

34) Find all customers who have an account at all the branches located in Brooklyn.

SELECT DISTINCT S.customer_name

FROM Depositor AS D

WHERE NOT EXISTS ((SELECT branch_name

FROM Branch

WHERE branch_city = ‘Brroklyn)

EXCEPT

(SELECT R.branch_name

FROM Depositor AS T, Account AS R

WHERE T.account_number =

R.account_number

AND D.customer_name = t.customer_name));

35) Find all customers who have at most one account at the Perryridge branch.

SELECT T.customer_name

FROM Depositor AS T

WHERE UNIQUE (SELECT R.customer_name

FROM Depositor AS R, Account AS A

WHERE T.customer_name = R.customer_name

AND R.account_number = A.account_number

AND A.branch_name = ‘Perryridge’);

36) Find all customers who have at least two accounts at the Perryridge branch.

SELECT DISTINCT T.customer_name

FROM Depositor AS T

WHERE NOT UNIQUE (SELECT R.customer_name

FROM Depositor AS R, Account AS A

WHERE T.customer_name = R.customer_name

AND R.account_number = A.account_number

AND A.branch_name = ‘Perryridge’);

37) Find the average account balance of those branches where the average account balance is greater than 1200.

SELECT branch_name, avg_balance

FROM (SELECT branch_name, AVG(balance)

FROM Account

GROUP BY branch_name)

AS Branch_avg(branch_name, avg_balance)

WHERE avg_balance > 1200;

38) Find the maximum across all branches of the total balance at each branch.

SELECT MAX(tot_balance)

FROM (SELECT branch_name, SUM(balance)

FROM Account

GROUP BY branch_name)

AS Branch_total(branch_name, tot_balance);

39) Find the all customers who have an account but no loan at the bank.

SELECT d-CN

FROM (Depositor LEFT OUTER JOIN Borrower

ON Depositor.customer_name = Borrower.customer_name)

AS db1(d-CN, account_number, b-CN, loan_number)

WHERE b-CN is null;

40) Find the all customers who have either an account or a loan (but not both) at the bank.

SELECT customer_name

FROM (Depositor NATURAL FULL OUTER JOIN Borrower)

WHERE account_number IS NULL

OR loan_number IS NULL;

WEEK 7

5) Queries (along with subqueries) using ANY, ALL, IN, EXISTS, NOT EXISTS, UNIQUE, INTERSECT, Constraints. Example: select the rollno and name of the student who secured 4^th rank in the class.

6) Queries using Aggregate functions (COUNT, SUM, AVG, MAX and MIN), GROUP BY, HAVING and Creation and Dropping of Views.

7) Queries using Conversions, functions (to_char, to_num, and to_date), string function (Conactenation, lpad, rpad, ltrim, rtrim, lower, upper, initcap, length, substr, and instr), date functions (sysdate, next_day, add_months, last_day, months_between, least, greatest, trunk, round, to_char, to_date).

DUAL (ORACLE WORK TABLE):

1) To display system date.

select sysdate from dual;

2) To display arithmetic calculations.

select 2*2 from dual;

3) To display the logged user.

select USER from dual;

4) To display system time.

select TO_CHAR(sysdate, ‘HH:MI:SS’) from dual;

5) To display current month.

select TO_CHAR(sysdate, ‘MONTH’) from dual;

6) To display system date in specified format.

select TO_CHAR(sysdate, ‘dd/mm/yy’) from dual;

7) To display system date in specified format.

select TO_CHAR(sysdate, ‘mm’) from dual;

8) To display date arithmetic.

select ADD_MONTHS(SYSDATE, 5) from dual;

9) To display date arithmetic.

select LAST_DAY(SYSDATE) from dual;

10) To display date arithmetic.

select MONTHS_BETWEEN(SYSDATE, ’01-APR-09’) from dual;

11) To display date arithmetic.

select NEXT_DAY(SYSDATE, ‘MON’) from dual;

GROUP FUNCTIONS:

12) To display average basic salary of the employees.

SELECT SUM(basic) FROM salary;

13) To display minimum basic salary of the employees.

SELECT MIN(basic) FROM salary;

14) To display maximum basic salary of the employees.

SELECT MAX(basic) FROM salary;

15) To display sum of basic salaries of all the employees.

SELECT SUM(basic) FROM salary;

16) To display the number of records in salary table.

SELECT COUNT(*) FROM salary;

STRING FUNCTIONS:

17) To display a field value after left padding.

select lpad('page-1', 10, '*') from dual;

18) To display a field value after left padding.

select Rpad('page-1', 10, '*') from dual;

19) To display a field value after converting to lower case.

select LOWER(‘A’) from dual;

20) To display a field value after converting to upper case.

select LOWER(‘a’) from dual;

21) To display a field value after converting to initial capital case.

select INITCAP(‘HOw are you?’) from dual;

22) To display a substring of a field value.

select substr(‘cse2a’, 4, 2) from dual;

23) To display the length of a field value.

select length(’how long am i?’) from dual;

24) To display a field value after trimming the right side.

select rtrim(‘cse2a’, ‘2a’) from dual;

25) To display a field value after trimming the left side.

SELEct ltrim(‘cse2a’, ‘cse’) from dual;

WEEK 8 (PL/SQL)

6) TheTk(i) Creation of simple PL/SQL program which includes declaration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

7) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIFTheTkjfdkjlkdfjskjlkfd and COALESCE functions.

8) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

9) Program development using creation of procedure, passing parameters IN and OUT procedures.

10) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

11) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

12) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CURSOR variables.

Syntax to write a sql program

Declare

Begin

[exception <exceptional stmts>]----- optional

End;

/---end of buffer

Example: 1

Create a file DBFOR.SQL, to execute the FOR loop and display the variable.

At SQL Prompt type, ed dbfor to open notepad and type the below program:

Program

declare

cnt number;

begin

dbms_output.put_line('This is a demo of FOR loop ');

for cnt in 1..5 loop

dbms_output.put_line('loop number ' || cnt);

end loop;

end;

set serveroutput off

Save the file and at SQL prompt run as:

Execution

SQL>set serveroutput on

SQL> start dbfor (press enter) OR

SQL> @dbfor

OUTPUT:-

This ia a demo of FOR loop

loop number 1

loop number 2

loop number 3

loop number 4

loop number 5

PS:

For syntax:

For <var> in <start_num> .. <endnum> loop

<statement(s);>

End loop;

Example: 2

Create a file DBREVFOR.SQL, to execute the REVERSE FOR loop and display the variable.

Program

begin

dbms_ouput.put_line(‘This is a demo of REVERSE FOR loop’);

for cnt in reverse 1..10 loop

if mod(cnt, 2) = 0 then

dbms_output.put_line(‘loop counter ‘ || cnt);

end if;

end loop;

end;

OUTPUT:-

This is a demo of REVERSE FOR loop

loop counter 10
loop counter 8
loop counter 6
loop counter 4
loop counter 2

PS:

Reverse For syntax:

For <var> in reverse <start_num> .. <endnum> loop

<statement(s);>

End loop;

Other forms of if syntax are:

If <condition> then

<action(s);>

End if;

If <condition> then

<action(s);>

Else

<action(s);>

End if;

If <condition> then

<action(s);>

Elsif <condition> then

<action(s);>

else

<action(s);>

End if;

Example: 3

Create a file DBLOOP.SQL, to execute the LOOP loop and display the variable.

Program

set serveroutput on

declare

cnt number(2) := 0;

begin

dbms_ouput.put_line(‘This is a demo of LOOP loop’);

loop

cnt := cnt + 1;

exit when cnt > 10;

dbms_output.put_line(‘loop counter ‘ || cnt);

end loop;

end;

set serveroutput off

OUTPUT:-

This is the demo of LOOP loop

loop counter 1

loop counter 2

loop counter 3
loop counter 4
loop counter 5
loop counter 6
loop counter 7
loop counter 8
loop counter 9
loop counter 10

PS:

Loop syntax:

loop

<statement(s);>

Exit when <condition>;

End loop;

Example: 4

Create a file DBWHILE.SQL, to execute the WHILE loop and display the variable.

Program

set serveroutput on

declare

cnt number(2) := 1;

begin

dbms_ouput.put_line(‘This is a demo of WHILE loop’);

while cnt <= 10 loop dbms_output.put_line(‘loop counter: ‘ || to_char(cnt, ‘999’));

cnt := cnt + 1;

end loop;

end;

set serveroutput off

PLSQL SQL SERVER PROGRAMMES

2011-11-14T10:27:00.000-08:00

PLSQL SQL SERVER PROGRAMMES

OUTPUT:-

This is a demo of WHILE loop

loop counter : 1

loop counter : 2

loop counter : 3
loop counter : 4
loop counter : 5
loop counter : 6
loop counter : 7
loop counter : 8
loop counter : 9
loop counter : 10

PS:

while syntax:

while <condition> loop

<statement(s);>

End loop;

Example: 4

Write a program EMPDATA.SQL, to retrieve the employee details of an employee whose number is input by the user .

Program

-- PROGRAM TO RETRIEVE EMP DETAILS

set serveroutput on

prompt Enter Employee Number:

accept n

declare

dname emp.emp_name%type;

dbasic emp.emp_basic%type;

ddesig emp.desig%type;

begin

select emp_name, basic, design

into dname, dbasic, ddesig

from emp

where emp_no = &n;

dbms_ouput.put_line(‘Employee Details:);

dbms_output.put_line(‘Name: ‘ || dname);

dbms_output.put_line(‘Basic: ‘ || dbasic);

dbms_output.put_line(‘Designation: ‘ || ddesig);

end;

OUTPUT:-

enter employee number:

old 9:where eno =&n;

new 9:where eno=13;

employee details

Name:allen

basic:9500

desig:mech

set serveroutput off

PS:

Similarly you can use other SQL statements in the PL/SQL block

Exercises:

1) Write a PL/SQL code, EX_INVNO.SQL, block for inverting a number using all forms of loops.

ANSWER:-

declare

n number(20):=123;

s number(13):=0;

d number(3):=1;

r number(3):=10;

begin

dbms_output.put_line('the number is :' || n);

while n>0 loop

d:=mod(n,10);

s:=(s*r)+d;

n:=n/r;

end loop;

dbms_output.put_line('inverted values' || s);

end;

OUTPUT:-

the number is:123

inverted value is:321

2) Write a PL/SQL code, EX_SUMNO.SQL that prints the sum of ‘n’ natural numbers.

ANSWER:-

prompt enter number:

accept number n

declare

isum number(2):=0;

i number;

n number:=&n;

begin

for i in 1..n loop

isum:=isum+i;

end loop;

dbms_output.put_line('sum is ' || isum);

end;

OUTPUT:-

enter the number:7

sum is 28

3) Write a PL/SQL code, EX_AREA.SQL, of block to calculate the area of the circle for the values of radius varying from 3 to 7. Store the radius and the corresponding values of calculated area in the table AREA_VALUES.

ANSWER:-

set serveroutput on

declare

area number(5);

rad number(3);

pi number(4):=3.14;

begin

for rad in 3..7 loop

area:=pi*rad*rad;

dbms_output.put_line('area is' || area);

insert into area_values values(area,rad);

end loop;

end;

OUTPUT:-

area is :27
area is :48
area is :75
area is :108
area is :147

SQL>select * from area_values;
area rad
____ ____
27      3
48     4
75      5
108    6
147    7

WEEK 9 (PL/SQL)

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

7) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIFTheTkjfdkjlkdfjskjlkfd and COALESCE functions.

8) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

9) Program development using creation of procedure, passing parameters IN and OUT procedures.

10) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

11) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

12) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CURSOR variables.

Example: 1

Create a file (NEWINS.SQL), to insert into a new table, NEWEMP, the record of any employee whose number is input by the user.

1. Create the table NEWEMP <emp_no, emp_name, join_date, basic).

2. Open an editor and type the following program.

Program

prompt Enter Employee Number:

accept userno number

declare

dno number(4);

dname varchar2(30);

ddate date;

dbasic number(10);

begin

select emp_no, emp_name, join_date, basic

into dno, dname, ddate, dbasic

from emp

where emp_no = &userno;

if sql%rowcount > 0

then

insert into newemp

values (dno, dname, ddate, dbaisc);

end if;

end;

3. Save the file as NEWINS

4. Execute the program as

SQL> start newins

Example: 2

Create a file (NEWINS2.SQL), to insert into a new table, NEWEMP, the record of any employee whose number is input by the user. Also display on the screen the employee details and to handle errors like user entering a number which does not exist in the table.

Program

prompt Enter Employee Number:

accept userno number

declare

dno number(4);

dname varchar2(30);

ddate date;

dbasic number(10);

begin

select emp_no, emp_name, join_date, basic

into dno, dname, ddate, dbasic

from emp

where emp_no = &userno;

if sql%rowcount > 0

then

insert into newemp

values (dno, dname, ddate, dbasic);

dbms_output.put_line(‘Record inserted into NEWEMP’);

dbms_output.put_line(DNO || ‘ ‘ || DNAME || ‘ ‘ || DDATE || ‘ ‘ || DBASIC);

end if;

exception

when no_data_found then

dbms_output.put_line (‘Record ‘ || &userno || ‘ does not exist’);

end;

Example: 3

Create a file (CALCTAX.SQL), to calculate tax for a specific employee and display name and tax.

Program

prompt Enter Employee Number:

accept userno number

declare

tot_basic number(10, 2);

tax number(10, 2);

name varchar2(30);

begin

select emp_name, basic

into name, tot_basic

from emp

where emp_no = &userno;

if tot_basic = 0 or tot_basic is null

then

dbms_output.put_line(‘NO BASIC’);

elsif tot_basic <= 2000

then

tax := tot_basic * .02;

dbms_output.put_line(NAME || ‘ TOTAL BASIC: ‘ || TOT_BASIC);

dbms_output.put_line(NAME || ‘ TOTAL TAX: ‘ || TAX);

else

tax := tot_basic * .04;

dbms_output.put_line(NAME || ‘ TOTAL BASIC: ‘ || TOT_BASIC);

dbms_output.put_line(NAME || ‘ TOTAL TAX: ‘ || TAX);

end if;

exception

when no_data_found then

dbms_output.put_line (‘Record ‘ || &userno || ‘ does not exist’);

end;

PS:

EXECPTIONS

When a program is executed certain errors are automatically recognized and certain error situations must be recognized by the program itself. Errors in general are referred to as Exceptions.

Exceptions can be either System defined or User defined.

Certain system exceptions raise the following flags:

CURSOR_ALREADY_OPEN – Displayed when the user tries to open a cursor that is already open

DUP_VAL_ON_INDEX – when user tries to insert a duplicate value into a unique column

INVALID_CURSOR – when user references an invalid cursor or attempts an illegal cursor operation

INVALID_NUMBER – when user tries to use something other than a number where one is called for

LOGIN_DENIED – when connect request for user has been denied

NO_DATA_FOUND – this flag becomes TRUE when SQL select statement failed to retrieve any rows

NOT_LOGGED_ON – user is not connected to ORACLE

PROGRAM_ERROR – user hits a PL/SQL internal error

STORAGE_ERROR – user hits a PL/SQL memory error

TIMEOUT_ON_RESOURCE – user has reached timeout while waiting for an Oracle resource

TRANSACTION_BACKED_OUT – a remote server has rolled back the transaction

TOO_MANY_ROWS – the flag becomes TRUE when SQL select statement retrieves more than one row and it was supposed to retrieve only 1 row

VALUE_ERROR – user encounters an arithmetic, conversion, truncation or constraint error

ZERO_DIVIDE – flag becomes TRUE if SQL select statement tries to divide a number by 0

OTHERS – this flag is used to catch any error situations not coded by the programmer

In the exception section and must appear last in the exception section

User defined exceptions must be declared in the declare section with the reserved word, EXCEPTION.

Syntax for user defined exception:

<exception-name> EXCEPTION;

This exception can be brought into action by the command,

RAISE <exception-name>

When the exception is raised, processing control is passed to the EXCEPTION section of the PL/SQL block.

The code for the exception must be defined in the EXCEPTION section of the PL/SQL block.

WHEN <exception-name> THEN

<action>;

Exercises:

1) Write a PL/SQL code block that will accept an account number from the user and debit an amount of RS2000 from the account. If the account has a minimum balance of 500 after amount is debited the process should set a freeze on the account by setting the status to F.

(use table schema Accounts (acno, balance, status)

2) Write a PL/SQL block of code to achieve the following:

If the price of the product is >4000 then change the price to 4000. The price change is to be recorded in the old price table along with product number and date on which the price was last changed.

(use table schemas Product(pno, price) and Old_Price(pno, date_of_change, oldprice)

WEEK 10 (PL/SQL)

13) TheTk(i) Creation of simple PL/SQL program which includes declaration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

14) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIFTheTkjfdkjlkdfjskjlkfd and COALESCE functions.

15) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

16) Program development using creation of procedure, passing parameters IN and OUT procedures.

17) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

18) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

19) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CURSOR variables.

Example: 1

Create a PL/SQL program using cursors, to retrieve first tuple from the department relation.

(use table dept(dno, dname, loc))

Program

declare

vdno dept.deptno%type;

vdname dept.dname%type;

vloc dept.loc%type;

cursor c1 is select * from dept;

or // cursor c1 is select * from dept where rowno = 1;

begin

open c1;

fetch c1

into vdno,vdname,vloc;

dbms_output.put_line('vdno = ' ||vdno|| ' vdname = '||vdname||' vloc = '||vloc);

close c1;

end;

PS:

Cursors are used when the SQL select statement is expected to return more than 1 row.

A cursor must be declared and its definition contains a query and is defined in the DECLARE section of the program.

A cursor must be opened before processing and closed after processing. (Similar to how files are opened and closed in a C program).

Syntax to define a cursor:

CURSOR <CURSOR-NAME> IS <SELECT STATEMENT>

Syntax to open the cursor:

OPEN <CURSOR-NAME>

Syntax to store data in the cursor:

FETCH <CURSOR-NAME> INTO <VAR1>, <VAR2>, <VAR3>….

FETCH <CURSOR-NAME> INTO <RECORD-NAME>

Syntax to close the cursor:

CLOSE <CURSOR-NAME>

Example: 2

Create a PL/SQL program using cursors, to retrieve each tuple from the department relation.

(use table dept(dno, dname, loc))

Program

declare

vdept dept%rowtype;

cursor c1 is select * from dept;

begin

for vdept in c1 loop

dbms_output.put_line('vdno = ' ||vdept.deptno|| ' vdname = '||vdept.dname||' vloc = '||vdept.loc);

end loop;

end;

PS:

The cursor for loop can be used to process multiple records. The advantage of cursor for loop is that the loop itself will open the cursor, read the records into the cursor from the table until end of file and close the cursor.

Syntax for cursor FOR LOOP:

FOR <VARIABLE> IN <CURSOR-NAME> LOOP

END LOOP;

Example: 3

Create a PL/SQL program using cursors, to display the number, name, salary of the three highest paid employees.

(use table emp(empno, ename,sal))

Program

declare

no emp.empno%type;

name emp.ename%type;

salary emp.sal%type;

cursor c1 is select empno, ename, sal from emp order by sal desc;

begin

open c1;

loop

fetch c1 into no,name,salary;

exit when c1 %notfound;

exit when c1 %rowcount >3;

dbms_output.put_line(no||name||salary);

end loop;

close c1;

end;

PS:

Cursors Attributes:

There are 4 cursor attributes used to provide information on the status of a cursor.

%NOTFOUND – To determine if a row was retrieved

Used after FETCH

NOTFOUND is TRUE if row is not retrieved

NOTFOUND is FALSE if row is retrieved

%FOUND – To determine if a row was retrieved.

Used after FETCH

FOUND is TRUE if row is retrieved

FOUND is FALSE if row is not retrieved

%ROWCOUNT – To determine the number of rows retrieved

ROWCOUNT is 0 when cursor is opened

ROWCOUNT returns the number of rows retrieved

%ISOPEN – To determine the cursor is open

ISOPEN is TRUE if a cursor is open

ISOPEN is FALSE if a cursor is not open

Example: 4

Create a PL/SQL program using cursors, to delete the employees whose salary is more than 3000.

Program

declare

vrec emp%rowtype;

cursor c1 is select * from emp where sal>3000 for update;

begin

open c1;

loop

fetch c1 into vrec;

exit when c1 %notfound;

delete from emp where current of c1;

dbms_output.put_line('Record deleted');

end loop;

close c1;

end;

PS:

In order to DELETE or UPDATE rows, the cursor must be defined with the FOR UPDATE clause.

Example: 5

Create a PL/SQL program using cursors, to update the salary of each employee by the avg salary if their salary is less than avg salary.

Program

declare

vrec emp%rowtype;

avgsal number(10,2);

cursor c1 is select * from emp for update;

begin

select avg(sal) into avgsal from emp;

for vrec in c1 loop

if vrec.sal < avgsal then

vrec.sal := avgsal;

update emp set sal = vrec.sal where current of c1;

dbms_output.put_line('Record updated');

end if;

end loop;

end;

PS:

Variable Attributes:

%TYPE - is used in PL/SQL to declare a variable to be of the same type as a previously declared variable or to be of the same type as a column in a table.

TOTBASIC SALARY.BASIC%TYPE;

will declare TOTBASIC of the same type as BASIC column from the table SALARY.

%ROWTYPE – declares a variable which is actually a record which has the same structure as a row from a table.

SALREC SALARY%ROWTYPE;

will declare SALREC as a record variable equivalent to the row from the table SALARY.

Example: 6

Create a PL/SQL program using cursors, to insert into a table, NEWEMP, the record of ALL MANAGERS. Also DISPLAY on the screen the NO, NAME, JOIN_DATE. Handle any user defined exceptions.

(use table emp(emp_no, emp_name, join_date, desig))

Program

set serveroutput on

declare

ctr number(2) := 2;

dno number(4);

dname varchar2(30);

ddate date;

cursor cur_mgr is

select emp_no, emp_name, join_date

from emp

where upper(desig) = ‘MGR’;

no_manager_found exception;

begin

open cur_mgr;

loop

fetch cur_mgr

into dno, dname, ddate;

exit when cur_mgr%notfound;

ctr := ctr + 1;

dbms_output.put_line(ctr || ‘Record inserted into NEWEMP’);

dbms_output.put_line(dno || ‘ ‘ || dname || ‘ ‘ ddate);

insert into new emp

values (dno, dname, ddate);

end loop;

if cur_mgr%rowcount = 0

then

close cur_mgr;

raise no_manager_found;

end if;

dbms_output.put_line(‘TOTAL number of records’ || ctr);

close cur_mgr;

exception

when no_manager_found then

dbms_output.put_line(‘NO RECORS FOUND’);

end;

Exercises:

1) Create a PL/SQL program using cursors, to insert into a table, NEWEMP, for any designation input by the user from the keyboard. Handle any user defined exceptions.

2) Code a program to calculate Tax for any employee whose number is input from the keyboard. Display appropriate error message if data does not exist in the table.

WEEK 11 (PL/SQL)

20) TheTk(i) Creation of simple PL/SQL program which includes declaration section, executable section and exception handling section ( ex: Student marks can be selected from the table and printed for those who secured first class and an exception can be raised if no records were found).

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

21) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIFTheTkjfdkjlkdfjskjlkfd and COALESCE functions.

22) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

23) Program development using creation of procedure, passing parameters IN and OUT procedures.

24) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

25) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

26) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CURSOR variables.

Example: 1

Code a procedure to calculate the sales made to a particular customer.

{ create table trn (itmid number(10),

cstid number(10),

trnqty number(10));

create table itmmast (itmid number(10),

itmprice number(10,2));

create table cstmast ( cstid number(10),

name varchar2(30));}

Step 1: Open the editor

Step 2: Type the code below in a file named, TOTSALES.

Program

CREATE OR REPLACE PROCEDURE TOTSALES

(CID IN CSTMAST.CSTID%TYPE, SAL OUT NUMBER)

id TRN.ITMID%TYPE;

qty TRN.TRNQTY%TYPE;

price ITMMAST.ITMPRICE%TYPE;

sales NUMBER(10, 2) := 0;

cursor cur_tr is

select trn.itmid, trnqty, itmprice

from trn, itmmast

where trn.cstid = cid

and trn.itmid = itmmast.itmid;

begin

open cur_tr;

loop

fetch cur_tr into id, qty, price;

if cur_tr%rowcount = 0

then

raise_application_error(-20020, ‘ERREOR!!!THERE IS NO DATA’);

end if;

exit when cur_tr%notfound;

sales := sales + qty * price;

end loop;

close cur_tr;

sal := sales;

end;

Step 3: Save the TOTSALES.SQL file.

Step 4: Return to SQL Prompt and compile as

SQL> start TOTSALES; (press enter)

Step 5: On the screen you will get the message Procedure created. If you have errors type

SQL> show errors

Step 6: To execute the procedure at SQL prompt type

SQL> variable sl number

SQL> execute totsales(2001, :sl)

SQL> print sl

PS:

Procedural Objects

Groups of SQL and PL/SQL statements can be stored in the database. The code stored once in the database can be used by multiple applications. Since the code is in the database, which is in the server, processing is faster.

Procedures and functions are also referred to as sub-programs as they can take parameters and be invoked.

Various types of procedural objects are: Procedures, Functions, Packages.

Procedures:

Procedures are sub-programs, which will perform an action and functions are subprograms that are generally coded to compute some value.

The clients execute the procedure or function and the processing is done in the server.

Procedures can receive and return values from and to the caller.

Communication is passed to a procedure through a parameter and communication is passed out of a procedure through a parameter.

When calling a procedure, the parameters passed can be declared to be IN, OUT or IN OUT.

The IN parameter is used to pass values to the procedure being called. It behaves like a constant inside the procedure, i.e., cannot be assigned values inside the procedure.

The OUT parameter is used to pass values out of a procedure to the caller of the procedure. It behaves like a uninitialized variable inside the procedure.

The IN OUT parameter is used to pass values to the procedure being called and it is used to pass values to the caller of the procedure. The IN OUT variable behaves like a regular variable inside the procedure.

Functions:

Functions are also a collection of SQL and PL/SQL code which can return a value to the caller.

Unlike procedures, functions can return a value to the caller. This value is returned through the use of the RETURN keyword within the function. A function can return a single value to the caller. Functions do not allow the OUT and IN OUT arguments.

Packages:

Packages are groups of procedures, functions, variables and SQL statements in a single unit.

It consists of the package definition/specification and package body.

A package specification consists of the list of functions, procedures, variables, constants, cursors and exceptions that will be available to users of the package.

A package body consists of the PL/SQL blocks and specifications for all of the public objects listed in the package specification. It may also include code that is run every time the package is invoked, regardless of the part of the package that is executed.

The name of the package body should be the same as the name of the package specification.

To delete procedural objects:

SQL> drop procedure <procedure-name>

SQL> drop function <function-name>

SQL> drop package<package-name>

SQL PLSQL SERVER PROGRAMMES

2011-11-13T09:06:00.000-08:00

SQL PLSQL SERVER PROGRAMMES

Example: 2

Code a function to return the square of a given number.

Step 1: Open the editor

Step 2: Type the code below in a file named, SQR.

Program

CREATE OR REPLACE FUNCTION SQR

(NO NUMBER)

RETURN NUMBER

BEGIN

return no*no;

END;

Step 3: To test the function:

a. At SQL prompt type:

SQL> select sqr(10) from dual;

b. At SQL prompt, type the following

SQL> variable sq number

SQL> execute :sq := sqr(10)

c. In the editor, type the following

set serveroutput on

begin

dbms_output.put_line(‘Square of 10 is ‘ || sqr(10));

end;

set serveroutput off

Example: 3

Code a function to return the net salary of a given employee.

Program

CREATE OR REPLACE FUNCTION NETSAL

(id in salary.emp_no%type)

RETURN NUMBER

netsal salary.basic%type;

BEGIN

select sum(basic) + sum(commission) – sum(deduction)

into netsal

from salary

where emp_no = id;

return (netsal);

end;

To test the function:

At SQL prompt, type

SQL> variable sal number

SQL> execute :sal := netsal(1001)

SQL> print sal

Example: 4

Code a package

Step 1: Open the editor and create the package

Step 2: Type the code below in a file named, MAHEPACK.

Program

CREATE OR REPLACE PACKAGE MAHEPACK

function netsal

(id in salary.emp_no%type)

return number;

procedure tax

(id in salary.emp_no%type, tax out number);

procedure totsales

(cid in cstmast.cstid%type, sal out number);

END;

Step 3: Save the above file and open the editor to create the package body

Program

CREATE OR REPLACE PACKAGE BODY MAHEPACK

function netsal

(id in salary.emp_no%type)

return number

netsal salary.basic%type;

begin

select sum(basic) + sum(commission) – sum(deduction)

into netsal

from salary

where emp_no = id;

return (netsal);

end;

procedure tax

(id in salary.emp_no%type, tax out number)

netsalary number(10, 2);

begin

netsalary := netsal(id);

if netsalary < 2000

then

tax := netsalary * 0.02;

elsif netsalary < 4000

then

tax := netsal * 0.04;

else

tax := netsalary * 0.01;

end if;

end;

procedure totsales

(cid in cstmast.cstid%type, sal out number)

id TRN.ITMID%TYPE;

qty TRN.TRNQTY%TYPE;

price ITMMAST.ITMPRICE%TYPE;

sales NUMBER(10, 2) := 0;

cursor cur_tr is

select trn,itmid, trnqty, itmprice

from trn, itmmast

where trn.cstid = cid

and trn.itmid = itmmast.itmid;

begin

sales := 0;

open cur_tr;

loop

fetch cur_tr into id, qty, price;

if cur_tr%rowcount = 0

then

raise_application_error(-20020, ‘ERREOR!!!THERE IS NO DATA’);

end if;

exit when cur_tr%notfound;

sales := sales + qty * price;

end loop;

close cur_tr;

sal := sales;

end;

END;

Step 4: Save the above file and to create the package, at SQL prompt type

SQL> start mpack

SQL> start mpackb

Step 5: To execute, at SQL prompt type

SQL> variable tx number

SQL> execute mahepack.tax(1001, :tx)

SQL> print tx

SQL> variable nsal number

SQL> execute :nsal := mahepack.netsal(1001)

SQL> print nsal

PS:

Show Errors:

SHOW ERRORS is used to display the line number and error of the most recent compilation errors.

SQL> SHOW ERRORS

Raise Application Error:

RAISE_APPLICATION_ERROR procedure is one of Oracles utilities which help the user to manage the error conditions in the applications by specifying user-defined error numbers and messages.

It takes 2 input parameters – the error number (which must be between -20000 and -20999) and the error message to display.

It terminates the procedure execution, rolls back any effects of the procedure, returns any user-specified error number and error message.

WEEK 12 (PL/SQL)

(ii) Insert data into student table and use COMMIT, ROLLBACK and SAVEPOINT in SQL block.

7) Develop a program that includes the features NESTED IF, CASE and CASE expression. The program can be extended using the NULLIFTheTkjfdkjlkdfjskjlkfd and COALESCE functions.

8) Program development using WHILE LOOPS, numeric FOR LOOPS, nested loops using ERROR handling, BUILT IN exceptions, USER defined exceptions, RAISE APPLICATION ERROR.

9) Program development using creation of procedure, passing parameters IN and OUT procedures.

10) Program development using creation of stored function, invoke functions in SQL statements and write complex functions.

11) Program development using creation of package specification, package bodies, private objects, package variables and cursors and calling stored packages.

12) Develop programs using features of parameters in a CURSOR, FOR UPDATE CURSOR, WHERE CURRENT of clause and CURSOR variables.

Example: 1

Write a row trigger to insert the existing values of the salary table into a new table when the salary table is updated.

(Salary < emp_no, basic, commission, deduction, salary_date, department>

Salaryaud < emp_no, basic, commission, deduction, salary_date, department>)

Step 1: Open the editor

Step 2: Type the code below in a file named, TRSAL.

Program

CREATE TRIGGER UPDSAL

BEFORE UPDATE ON SALARY

FOR EACH ROW

BEGIN

insert intosalaryaud

values (:old.emp_no, :old.basic, :old.commission, :old.deduction, :old.salary_date, :old.department);

END

Step 3: Save the above file

Step 4: To create the trigger, at SQL prompt type

SQL> start trsal

Step 5: To test the trigger, update values in salary table and see if data is inserted in salaryaud table.

Example: 2

Write a trigger to restrict the user from using the emp table on Tuesday.

Program

create or replace trigger tr2

before insert or update or delete

on emp

begin

if (rtrim(to_char(sysdate, 'day')) = 'tuesday')

then

raise_application_error(-20121, 'Cannot delete on Tuesday');

end if;

end;

Example: 3

Write a PL/SQL block of code that first inserts a record in an Emp table. Update salaries of emp 1001 and emp 1002 by Rs 2000 and Rs 1500. Then check to see that the total salary does not exceed Rs 20000. If total salary is greater than Rs 20000 then undo the updates made to emp 1001 and emp 1002.

Program

DECLARE

total_sal number(9);

BEGIN

insert into emp

values (‘1009’, ‘Ram’, 1000);

SAVEPOINT no_update;

update emp

set sal = sal + 2000

where emp_id = 1001;

update emp

set sal = sal + 1500

where emp_id = 1002;

select sum(sal)

into total_sal

from emp;

if total_sal > 20000

then

ROLLBACK TO SQVEPOINT no_update;

end if;

COMMIT;

END;

PS:

The above program first inserts a record into emp table. It then marks and saves the current position in the transaction by using the SAVEPOINT. It updates the salaries, if the salaries are exceeding 20000 it rollsback to the save point, ie ignores the 2 updates and only commits the insert. If the salaries do not exceed 20000 then the insert and 2 updates are committed.

Example: 4

Write a PL/SQL code of block, to calculate the area of the circle for the values of radius varying from 1 to 10. Store the odd radius values and the corresponding areas in a table.

Program

declare

pi constant number(4,2) := 3.14;

radius number(5);

areaa number(14,2);

begin

radius := 1;

while radius <= 10 loop

areaa := pi*power(radius,2);

case

when radius = 1

then

insert into area values (radius, areaa);

when radius = 3

then

insert into area values (radius, areaa);

when radius = 53

then

insert into area values(radius,areaa);

when radius = 7

then

insert into area values(radius,areaa);

when radius = 9

then

insert into area values(radius,areaa);

else

dbms_output.put_line(‘EVEN RADIUS, NOT INSERTING!!!’);

end case;

radius:=radius+1;

end loop;

end;

Overview of SQL 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 ar

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

Basic SQL DDL Commands.

To practice basic SQL DDL Commands such as CREATE, DROP, etc.

1. SQL - CREATE TABLE

Syntax: CREATE TABLE tablename (column_name data_ type constraints, …)

Example:

INPUT:

SQL> CREATE TABLE Emp ( EmpNo short CONSTRAINT PKey PRIMARY KEY,

EName VarChar(15), Job Char(10) CONSTRAINT Unik1 UNIQUE,

Mgr short CONSTRAINT FKey1 REFERENCES EMP (EmpNo),

Hiredate Date, DeptNo short CONSTRAINT FKey2 REFERENCES DEPT(DeptNo));

RESULT: Table created.

SQL>Create table prog20 (pname varchar2(20) not null), doj date not null,dob date not null, sex varchar(1) not null, prof1 varchar(20),prof2 varchar(20),salary number(7,2) not null);

RESULT:

Table created.

SQL>desc prog20;

Name Null? Type

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

PNAME NOT NULL VARCHAR2(20)

DOJ NOT NULL DATE

DOB NOT NULL DATE

SEX NOT NULL VARCHAR2(1)

PROF1 VARCHAR2(20)

PROF2 VARCHAR2(20)

SALARY NOT NULL NUMBER(7,2)

2. SQL - ALTER TABLE

INPUT:

SQL>ALTER TABLE EMP ADD CONSTRAINT Pkey1 PRIMARY KEY (EmpNo);

RESULT: Table Altered.

Similarly, ALTER TABLE EMP DROP CONSTRAINT Pkey1;

3. SQL - DROP TABLE

– Deletes table structure – Cannot be recovered – Use with caution

INPUT:

SQL> DROP TABLE EMP; Here EMP is table name

RESULT: Table Dropped.

4. TRUNCATE TRUNCATE TABLE <TABLE NAME>;

Basic SQL DML Commands.

To practice basic SQL DML Commands such as INSERT, DELETE, etc.

1. SQL - INSERT INTO

Syntax: INSERT INTO tablename VALUES (value list)

l Single-row insert

INSERT INTO S VALUES(‘S3’,’SUP3’,’BLORE’,10)

l Inserting one row, many columns at a time

INSERT INTO S (SNO, SNAME) VALUES (‘S1’, ‘Smith’);S1’ Smith’

l Inserting many rows, all/some columns at a time.

INSERT INTO NEW_SUPPLIER (SNO, SNAME)

SELECT SNO, SNAME FROM S

WHERE CITY IN (‘BLORE’,’MADRAS’)

Other Examples:

INPUT:

SQL>Insert into prog values (‘kkk’,’05-may-56’);

RESULT: 1 row created.

INPUT:

SQL>Insert into prog20 values(‘Hema’,’25-sept-01’28-jan-85’,’f’,’c’,’c++’,’25000’);

RESULT: 1 row created.

INPUT:

SQL>Insert into prog values(‘&pname’,’&doj’);

SQL> Insert into prog values('&pname','&doj');

Enter value for pname: ravi

Enter value for doj: 15-june-81

RESULT:

old 1: Insert into prog values('&pname','&doj')

new 1: Insert into prog values('ravi','15-june-81')

1 row created.

2. SQL - UPDATE

Syntax: UPDATE tablename SET column_name =value [ WHERE condition]

Examples:

UPDATE S SET CITY = ‘KANPUR’ WHERE SNO=‘S1’

UPDATE EMP SET SAL = 1.10 * SAL

SQL> update emp set sal=20000 where empno=7369;

1 row updated.

3. SQL - DELETE FROM

Syntax: DELETE FROM tablename WHERE condition

Examples:

DELETE FROM SP WHERE PNO= ‘P1’

DELETE FROM SP

INPUT:

SQL>Delete from emp where empno=7369;

RESULT: 1 row deleted.

Basic SQL DCL Commands.

To practice basic SQL DCL Commands such as COMMIT, ROLLBACK etc.

1. COMMIT

Save changes (transactional).

Syntax:

COMMIT [WORK] [COMMENT 'comment_text']

COMMIT [WORK] [FORCE 'force_text' [,int] ]

FORCE - will manually commit an in-doubt distributed transaction
force_text - transaction identifier (see the DBA_2PC_PENDING view)
int - sets a specific SCN.
If a network or machine failure prevents a distributed transaction from committing properly, Oracle will store any commit comment in the data dictionary along with the transaction ID.

INPUT:

SQL>commit;

RESULT: Commit complete.

2. ROLLBACK

Undo work done (transactional).
Syntax:

ROLLBACK [WORK] [TO [SAVEPOINT]'savepoint_text_identifier'];

ROLLBACK [WORK] [FORCE 'force_text'];

FORCE - will manually rollback an in-doubt distributed transaction

INPUT:

SQL>rollback;

RESULT:Rollback complete.

3. SAVEPOINT

Save changes to a point (transactional).

Syntax:

SAVEPOINT text_identifier

Example:

UPDATE employees

SET salary = 95000

WHERE last_name = 'Smith';

SAVEPOINT justsmith;

UPDATE employees

SET salary = 1000000;

SAVEPOINT everyone;

SELECT SUM(salary) FROM employees;

ROLLBACK TO SAVEPOINT justsmith;

COMMIT;

Writing and Practice of Simple Queries.

To write simple queries and practice them.

1. Get the description of EMP table.

SQL>desc emp;

RESULT:

Name Null? Type

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

EMPNO NOT NULL NUMBER(4)

ENAME VARCHAR2(10)

JOB VARCHAR2(9)

MGR NUMBER(4)

HIREDATE DATE

SAL NUMBER(7,2)

COMM NUMBER(7,2)

DEPTNO NUMBER(3)

AGE NUMBER(3)

ESAL NUMBER(10)

2. Get the description DEPT table.

SQL>desc dept;

RESULT:

Name Null? Type

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

DEPTNO NOT NULL NUMBER(2)

DNAME VARCHAR2(14)

LOC VARCHAR2(13)

3.List all employee details.

SQL>select * from emp;

RESULT:

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO AGE ESAL

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

7369 SMITH CLERK 7902 17-DEC-80 800 0 20 25 0

7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30 25 0

7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30 25 0

7566 JONES MANAGER 7839 02-APR-81 2975 500 20 25 0

7698 BLAKE MANAGER 7839 01-MAY-81 2850 1400 30 25 0

SQL PLSQL SERVER PROGRAMMES

2011-11-12T10:09:00.000-08:00

SQL PLSQL SERVER PROGRAMMES

4.List all employee names and their salaries, whose salary lies between 1500/- and 3500/- both inclusive.

INPUT

SQL>select ename from emp where sal between 1500 and 3500;

RESULT

ENAME

----------

ALLEN

JONES

BLAKE

CLARK

SCOTT

TURNER

FORD

russel

greg

9 rows selected.

5. List all employee names and their and their manager whose manager is 7902 or 7566 0r 7789.

INPUT SQL>select ename from emp where mgr in(7602,7566,7789);

RESULT

ENAME

-------

SCOTT

FORD

6. List all employees which starts with either J or T.

INPUT SQL>select ename from emp where ename like ‘J%’ or ename like ‘T%’;

RESULT:

ENAME

---------

JONES

TURNER

JAMES

7. List all employee names and jobs, whose job title includes M or P.

INPUT SQL>select ename,job from emp where job like ‘M%’ or job like ‘P%’;

RESULT:

ENAME JOB

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

JONES MANAGER

BLAKE MANAGER

CLARK MANAGER

KING PRESIDENT

8. List all jobs available in employee table.

INPUT SQL>select distinct job from emp;

RESULT:

JOB

---------

ANALYST

CLERK

MANAGER

PRESIDENT

SALESMAN

assistant

clerk

7 rows selected.

9. List all employees who belongs to the department 10 or 20.

INPUT SQL>select ename from emp where deptno in (10,20);

RESULT:

ENAME

----------

SMITH

JONES

CLARK

SCOTT

KING

ADAMS

FORD

MILLER

8 rows selected.

10. List all employee names , salary and 15% rise in salary.

INPUT SQL>select ename , sal , sal+0.15* sal from emp;

RESULT:

ENAME SAL SAL+0.15*SAL

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

SMITH 800 920

ALLEN 1600 1840

WARD 1250 1437.5

JONES 2975 3421.25

MARTIN 1250 1437.5

BLAKE 2850 3277.5

CLARK 2450 2817.5

7 rows selected.

11. List minimum , maximum , average salaries of employee.

INPUT SQL>select min(sal),max(sal),avg(sal) from emp;

RESULT:

MIN(SAL) MAX(SAL) AVG(SAL)

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

3 5000 1936.94118

12. Find how many job titles are available in employee table.

INPUT SQL>select count (distinct job) from emp;

RESULT:

COUNT(DISTINCTJOB)

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

13. What is the difference between maximum and minimum salaries of employees in the organization?

INPUT SQL>select max(sal)-min(sal) from emp;

RESULT:

MAX(SAL)-MIN(SAL)

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

4997

14. Display all employee names and salary whose salary is greater than minimum salary of the company and job title starts with ‘M’.

INPUT SQL>select ename,sal from emp where job like ‘M%’ and sal > (select min (sal) from emp);

RESULT

ENAME SAL

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

JONES 2975

BLAKE 2850

CLARK 2450

15. Find how much amount the company is spending towards salaries.

INPUT SQL>select sum (sal) from emp;

RESULT

SUM(SAL)

---------

32928

16. Display name of the dept. with deptno 20.

INPUT SQL>select ename from emp where deptno = 20;

RESULT

ENAME

----------

SMITH

JONES

SCOTT

ADAMS

17. List ename whose commission is NULL.

INPUT SQL>select ename from emp where comm is null;

ENAME

RESULT ----------

CLARK

SCOTT

KING

ADAMS

JAMES

FORD

6 rows selected.

18. Find no.of dept in employee table.

INPUT SQL>select count (distinct ename) from emp;

RESULT

COUNT(DISTINCTENAME

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

19. List ename whose manager is not NULL.

INPUT SQL>select ename from emp where mgr is not null;

RESULT

ENAME

----------

SMITH

ALLEN

WARD

JONES

MARTIN

5 rows selected.

Writing Queries using GROUP BY and other clauses.

To write queries using clauses such as GROUP BY, ORDER BY, etc. and retrieving information by joining tables.

Source tables: emp, dept, programmer, software, study.

Order by : The order by clause is used to display the results in sorted order.

Group by : The attribute or attributes given in the clauses are used to form groups. Tuples with the same value on all attributes in the group by clause are placed in one group.

Having: SQL applies predicates (conditions) in the having clause after groups have been formed, so aggregate function be used.

1. Display total salary spent for each job category.

INPUT SQL>select job,sum (sal) from emp group by job;

RESULT

JOB SUM(SAL)

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

ANALYST 6000

CLERK 23050

MANAGER 8275

PRESIDENT 5000

SALESMAN 5600

assistant 2200

clerk 2003

7 rows selected.

2. Display lowest paid employee details under each manager.

INPUT SQL>select ename, sal from emp where sal in (select min(sal) from emp group by mgr);

RESULT

ENAME SAL

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

chai 3

JAMES 950

MILLER 1000

ADAMS 1100

russel 2200

5 rows selected.

3. Display number of employees working in each department and their department name.

INPUT SQL> select dname, count (ename) from emp, dept where emp.deptno=dept.deptno group by dname;

RESULT

DNAME COUNT(ENAME)

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

ACCOUNTING 3

RESEARCH 5

SALES 9

4. Display the sales cost of package developed by each programmer.

INPUT SQL>select pname, sum(scost) from software group by pname;

RESULT

PNAME SUM(SCOST)

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

john 12000

kamala 12000

raju 12333

3 rows selected.

5. Display the number of packages sold by each programmer.

INPUT SQL>select pname, count(title) from software group by pname;

RESULT

PNAME COUNT(TITLE)

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

john 1

kamala 1

raju 1

ramana 1

rani 1

5 rows selected.

6. Display the number of packages in each language for which the development cost is less than thousand.

INPUT SQL>select devin, count(title) from software where dcost < 1000 group by devin;

RESULT

DEVIN COUNT(TITLE)

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

cobol 1

7. Display each institute name with number of students.

INPUT SQL>select splace, count(pname) from study group by splace;

RESULT

SPLACE COUNT(PNAME)

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

BDPS 2

BITS 1

BNRILLIANI 1

COIT 1

HYD 1

5 rows selected.

8. How many copies of package have the least difference between development and selling cost, were sold?

INPUT SQL>select sold from software where scost – dcost=(select min(scost – dcost) from software);

RESULT

SOLD

---------

9. Which is the costliest package developed in Pascal.

INPUT SQL>select title from software where devin = ‘PASCAL’ and dcost = (select max(dcost)from software where devin = ‘PASCAL’);

RESULT

no rows selected

10. Which language was used to develop most no .of packages.

INPUT SQL>select devin, count (*) from software group by devin having count(*) = (select max(count(*) ) from software group by devin);

RESULT

DEVIN COUNT(*)

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

jsp 2

11.Who are the male programmers earning below the average salary of female programmers?

INPUT SQL>select pname from programmer where sal < (select avg(sal) from programmer where sex = ‘F’) and sex = ‘M’;

RESULT

PNAME

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

vijay

12. Display the details of software developed by the male programmers earning more than 3000/-.

INPUT SQL>select programmer.pname, title, devin from programmer, software where sal > 3000 and sex = ‘M’ and programmer.pname = software.pname;

RESULT

no rows selected

13. Display the details of software developed in c language by female programmers of pragathi.

INPUT SQL>select software.pname, title, devin, scost, dcost, sold from programmer, software, study where devin = ‘c’ and sex =’F’ and splace = ‘pragathi’ and programmer.pname = software.pname and software.pname = study.pname;

14. Which language has been stated by the most of the programmers as proficiency one?

INPUT SQL>select prof1, count(*) from programmer group by prof1 having count (*) = (select max (count (*) ) from programmer group by prof1);

Writing Nested Queries.

To write queries using Set operations and to write nested queries.

Set Operations:

UNION - OR

INTERSECT - AND

EXCEPT - - NOT

NESTED QUERY:- A nested query makes use of another sub-query to compute or retrieve the information.

1. Find the name of the institute in which the person studied and developed the costliest package.

INPUT SQL>select splace, pname from study where pname = (select pname from software where scost = (select max (scost) from software);

RESULT

SPLACE PNAME

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

SAHBHARI MARY

2. Find the salary and institute of a person who developed the highest selling package.

INPUT SQL> select study.pname, sal, splace from study, programmer where study.pname = programmer.pname and study.pname = (select pname from software where scost = (select max (scost) from software));

RESULT

PNAME SAL SPLACE

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

MARY 4500 SABHARI

3. How many packages were developed by the person who developed the cheapest package.

INPUT SQL>select pname, count (title) from software where dcost = (select min(dcost) from software) group by pname;

RESULT

PNAME COUNT(TITLE)

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

VIJAY 1

4. Calculate the amount to be recovered for those packages whose development cost has not yet recovered.

INPUT SQL>select title , (dcost-scost) from software where dcost > scost;

5. Display the title, scost, dcost, difference of scost and dcost in the descending order of difference.

INPUT SQL> select title, scost, dcost, (scost - dcost) from software descending order by (scost-dcost);

6. Display the details of those who draw the same salary.

INPUT SQL> select p.pname, p.sal from programmer p, programmer t where p.pname <> t.pname and p.sal = t.sal;(or)

INPUT SQL>select pname,sal from programmer t where pname<>t.pname and sal= t.sal;

Writing Queries using functions.

AIM: To write queries using single row functions and group functions.

1. Display the names and dob of all programmers who were born in january.

INPUT SQL>select pname , dob from programmer where to_char (dob,’MON’)=’JAN’;

2. Calculate the experience in years of each programmer and display along with programmer name in descending order.

INPUT SQL> select pname, round (months_between(sysdate, doj)/12, 2) "EXPERIENCE" from programmer order by months_between (sysdate, doj) desc;

3. List out the programmer names who will celebrate their birthdays during current month.

INPUT SQL>select pname from programmer where to_char(dob,’MON’) like to_char (sysdate, ‘MON’);

4. Display the least experienced programmer’s details.

INPUT SQL>select * from programmer where doj = (select max (doj) from programmer);

5. Who is the most experienced programmer knowing pascal.

INPUT SQL>select pname from programmer where doj = (select min (doj) from programmer);

6. Who is the youngest programmer born in 1965.

INPUT SQL> select pname , dob from programmer where dob = (select max (dob) from programmer where to_char (dob,'yy') = 65);

7. In which year, most of the programmers are born.

INPUT SQL>select to_char (dob , ‘YY’) from programmer group by to_char (dob, ‘YY’) having count(*) = (select max (count(*)) from programmer group by to_char(dob,’YY’);

8. In which month most number of programmers are joined.

INPUT SQL>select to_char (doj,’YY’) from programmer group by to_char (doj,’YY’) having count (*) = (select max (count(*)) from programmer group by to_char (doj,’YY’);

9. What is the length of the shortest name in programmer table ?

INPUT SQL>select length (pname) from programmer where length (pname) = select min ( length (pname) from programmer);

10. Display the names of the programmers whose name contains up to 5 characters.

INPUT SQL>select pname from programmer where length (pname) <=5;

11. Display all packages names in small letters and corresponding programmer names in uppercase letters.

INPUT SQL>select lower (title), upper (pname) from software;

Writing Queries on views.

AIM: To write queries on views.

1. Create a view from single table containing all columns from the base table.

SQL>create view view1 as (select * from programmer);

2. Create a view from single table with selected columns.

SQL>create a view view2 as (select pname,dob,doj,sex,sal from programmer);

3. Create a view from two tables with all columns.

SQL>create view xyz as select * from programmer full natural join software;

4. Create a view from two tables with selected columns.

SQL> create view lmn as (select programmer, pname, title, devin from programmer, software where sal < 3000 and programmer.pname = software.pname);

5. Check all DML commands with above 4 views.

INPUT SQL> insert into view1 values (‘ramu’,’12-sep-03’,’28-jan-85’,’f’,’dbase’,’oracle’,74000);

RESULT

1 row created;

INPUT SQL>update view1 set salary =50000 where pname like ‘raju’;

RESULT 1 row updated.

Note: update command does not works for all queries on views.

INPUT SQL>delete from view1 where pname like ‘raju’;

RESULT 1 row deleted.

6. Drop views which you generated.

INPUT SQL>drop view view1;

RESULT View dropped;

INPUT SQL>drop view view2;

RESULT View dropped;

INPUT SQL>drop view xyz;

Writing PL/SQL block for insertion into a table.

To write a PL/SQL block for inserting rows into EMPDET table with the following Calculations:

HRA=50% OF BASIC

DA=20% OF BASIC

PF=7% OF BASIC

NETPAY=BASIC+DA+HRA-PF

INPUT

DECLARE

ENO1 empdet.eno%type;

ENAME1 empdet.name%type;

DEPTNO1 empdet.deptno%type;

BASIC1 empdet.basic%type;

HRA1 empdet.HRA%type;

DA1 empdet.DA%type;

PF1 empdet.pf%type;

NETPAY1 empdet.netpay%type;

BEGIN

ENO1:=&ENO1;

ENAME1:='&ENAME1';

DEPTNO1:=&DEPTNO1;

BASIC1:=&BASIC1;

HRA1:=(BASIC1*50)/100;

DA1:=(BASIC1*20)/100;

PF1:=(BASIC1*7)/100;

NETPAY1:=BASIC1+HRA1+DA1-PF1;

INSERT INTO EMPDET VALUES (ENO1, ENAME1, DEPTNO1, BASIC1, HRA1, DA1, PF1, NETPAY1);

END;

RESULT:

SQL> @BASIC

Enter value for eno1: 104

old 11: ENO1:=&ENO1;

new 11: ENO1:=104;

Enter value for ename1: SRINIVAS REDDY

old 12: ENAME1:='&ENAME1';

new 12: ENAME1:='SRINIVAS REDDY';

Enter value for deptno1: 10

old 13: DEPTNO1:=&DEPTNO1;

new 13: DEPTNO1:=10;

Enter value for basic1: 6000

old 14: BASIC1:=&BASIC1;

new 14: BASIC1:=6000;

PL/SQL procedure successfully completed.

SQL>/

Enter value for eno1: 105

old 11: ENO1:=&ENO1;

new 11: ENO1:=105;

Enter value for ename1: CIRAJ

old 12: ENAME1:='&ENAME1';

new 12: ENAME1:='CIRAJ';

Enter value for deptno1: 10

old 13: DEPTNO1:=&DEPTNO1;

new 13: DEPTNO1:=10;

Enter value for basic1: 6000

old 14: BASIC1:=&BASIC1;

new 14: BASIC1:=6000;

PL/SQL procedure successfully completed.

SQL> SELECT * FROM EMPDET;

RESULT

ENO NAME DEPTNO BASIC HRA DA PF NETPAY

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

101 SANTOSH 10 5000 2500 1000 350 8150

102 SHANKAR 20 5000 2500 1000 350 8150

103 SURESH 20 5500 2750 1100 385 8965

104 SRINIVASA REDDY 10 6000 3000 1200 420 9780

105 CIRAJ 10 6000 3000 1200 420 9780

SQL PLSQL SERVER PROGRAMMES

2011-11-11T07:12:00.000-08:00

SQL PLSQL SERVER PROGRAMMES

Writing PL/SQL block for checking armstrong number

To write a PL/SQL block to check whether given number is Armstrong or not.

INPUT

DECLARE

num number(5);

rem number(5);

s number(5):=0;

num1 number(5);

BEGIN

num:=&num;

num1:=num;

while(num>0)

loop

rem:=mod(num,10);

s:=s+power(rem,3);

num:=trunc(num/10);

End loop;

if (s=num1)then

dbms_RESULT.put_line(num1||' IS ARMSTRONG NUMBER ');

else

dbms_RESULT.put_line(num1||' IS NOT ARMSTRONG NUMBER ');

End if;

END;

RESULT:

SQL>@arm

Enter value for num: 153

old 7: num:=&num;

new 7: num:=153;

153 IS ARMSTRONG NUMBER

PL/SQL procedure successfully completed.

SQL> /

Enter value for num: 123

old 7: num:=&num;

new 7: num:=123;

123 IS NOT ARMSTRONG NUMBER

PL/SQL procedure successfully completed.

Writing a PL/SQL block for checking a number even or odd.

AIM: To write a PL/SQL block to check whether a given number is Even or Odd.

INPUT

DECLARE

num number(5);

rem number;

BEGIN

num:=&num;

rem:=mod(num,2);

if rem=0

then

dbms_RESULT.put_line(' Number '||num||' is Even');

else

dbms_RESULT.put_line(' Number '||num||' is Odd');

end if;

END;

RESULT:

SQL>start even

Enter value for num: 6

old 5: num:=&num;

new 5: num:=6;

Number 6 is Even

PL/SQL procedure successfully completed.

SQL> /

Enter value for num: 3

old 5: num:=&num;

new 5: num:=3;

Number 3 is Odd

PL/SQL procedure successfully completed.

Writing PL/SQL block to find sum of digits of a given number.

To write a PL/SQL block to find Sum of Digits of a given Number.

INPUT

DECLARE

num number(5);

rem number(5);

sm number(5):=0;

num1 number(5);

BEGIN

num:=&num;

num1:=num;

while(num>0) loop

rem:=mod(num,10);

sm:=sm+rem;

num:=trunc(num/10);

end loop;

dbms_RESULT.put_line('SUM OF DIGITS OF '||num1||' IS: '||sm);

end;

RESULT:

SQL> @sum

INPUT truncated to 2 characters

Enter value for num: 123

old 7: num:=&num;

new 7: num:=123;

SUM OF DIGITS OF 123 IS: 6

PL/SQL procedure successfully completed.

SQL> @sum

INPUT truncated to 2 characters

Enter value for num: 456

old 7: num:=&num;

new 7: num:=456;

SUM OF DIGITS OF 456 IS: 15

PL/SQL procedure successfully completed.

Writing PL/SQL block for generating Fibonacci series.

To write a PL/SQL block to Generate Fibonacci Series

INPUT

DECLARE

num number(5);

f1 number(5):=0;

f2 number(5):=1;

f3 number(5);

i number(5):=3;

BEGIN

num:=&num;

dbms_RESULT.put_line('THE FIBONACCI SERIES IS:');

dbms_RESULT.put_line(f1);

dbms_RESULT.put_line(f2);

while(i<=num) loop

f3:=f1+f2;

dbms_RESULT.put_line(f3);

f1:=f2;

f2:=f3;

i:=i+1;

end loop;

END;

RESULT:

SQL> start fib

Enter value for num: 10

old 8: num:=&num;

new 8: num:=10;

THE FIBONACCI SERIES IS:

PL/SQL procedure successfully completed.

Writing PL/SQL block for checking palendrome.

To write a PL/SQL block to Check the Given String is Palindrome or Not.

INPUT

DECLARE

name1 varchar2(20);

name2 varchar2(20);

l number(5);

BEGIN

name1:='&name1';

l:=length(name1);

while l>0 loop

name2:=name2||substr(name1,l,1);

l:=l-1;

end loop;

dbms_RESULT.put_line('REVERSE OF STRING IS:'||NAME2);

if(name1=name2) then

dbms_RESULT.put_line(name1||' IS PALINDROME ');

else

dbms_RESULT.put_line(name1||' IS NOT PALINDROME ');

end if;

END;

RESULT

Enter value for name1: LIRIL

old 6: name1:='&name1';

new 6: name1:='LIRIL';

REVERSE OF STRING IS:LIRIL

LIRIL IS PALINDROME

PL/SQL procedure successfully completed.

SQL> /

Enter value for name1: MADAM

old 6: name1:='&name1';

new 6: name1:='MADAM';

REVERSE OF STRING IS:MADAM

MADAM IS PALINDROME

PL/SQL procedure successfully completed.

Writing PL/SQL block to demonstrate Cursors.

To write a Cursor to display the list of Employees and Total Salary Department wise.

INPUT

DECLARE

cursor c1 is select * from dept;

cursor c2 is select * from emp;

s emp.sal%type;

BEGIN

for i in c1 loop

s:=0;

dbms_RESULT.put_line('----------------------------------------------'); dbms_RESULT.put_line('Department is :' || i.deptno ||' Department name is:' || i.dname);

dbms_RESULT.put_line('-------------------------------------------');

for j in c2 loop

if ( i.deptno=j.deptno) then

s:=s+j.sal;

dbms_RESULT.put_line(j.empno|| ' '|| j.ename || ' '|| j.sal );

end if;

end loop;

dbms_RESULT.put_line('----------------------------------------------');

dbms_RESULT.put_line('Total salary is: '|| s);

dbms_RESULT.put_line('----------------------------------------------');

end loop;

END;

RESULT:

SQL> @abc

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

Department is :10 Department name is : ACCOUNTING

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

7782 CLARK 2450

7839 KING 5000

7934 MILLER 1300

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

Total salary is: 8750

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

Department is :20 Department name is:RESEARCH

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

7369 SMITH 800

7566 JONES 2975

7788 SCOTT 3000

7876 ADAMS 1100

7902 FORD 3000

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

Total salary is: 10875

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

Department is :30 Department name is:SALES

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

7499 ALLEN 1600

7521 WARD 1250

7654 MARTIN 1250

7698 BLAKE 2850

7844 TURNER 1500

7900 JAMES 950

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

Total salary is: 9400

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

Department is :40 Department name is:OPERATIONS

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

Total salary is: 0

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

PL/SQL procedure successfully completed.

Writing PL/SQL CURSOR

To write a Cursor to display the list of employees who are Working as a Managers or Analyst.

INPUT

DECLARE

cursor c(jb varchar2) is select ename from emp where job=jb;

em emp.job%type;

BEGIN

open c('MANAGER');

dbms_RESULT.put_line(' EMPLOYEES WORKING AS MANAGERS ARE:');

loop

fetch c into em;

exit when c%notfound;

dbms_RESULT.put_line(em);

end loop;

close c;

open c('ANALYST');

dbms_RESULT.put_line(' EMPLOYEES WORKING AS ANALYST ARE:');

loop

fetch c into em;

exit when c%notfound;

dbms_RESULT.put_line(em);

end loop;

close c;

END;

RESULT:

EMPLOYEES WORKING AS MANAGERS ARE:

JONES

BLAKE

CLARK

EMPLOYEES WORKING AS ANALYST ARE:

SCOTT

FORD

PL/SQL procedure successfully completed.

Writing PL/SQL CURSOR

To write a Cursor to display List of Employees from Emp Table in PL/SQL block

INPUT

DECLARE

cursor c is select empno, ename, deptno, sal from emp ;

i emp.empno%type;

j emp.ename%type;

k emp.deptno%type;

l emp.sal%type;

BEGIN

open c;

dbms_RESULT.put_line('Empno, name, deptno, salary of employees are:= ');

loop

fetch c into i, j, k, l;

exit when c%notfound;

dbms_RESULT.put_line(i||' '||j||' '||k||' '||l);

end loop;

close c;

END;

RESULT:

SQL> @EMP

Empno,name,deptno,salary of employees are:=

7369 SMITH 20 800

7499 ALLEN 30 1600

7521 WARD 30 1250

7566 JONES 20 2975

7654 MARTIN 30 1250

7698 BLAKE 30 2850

7782 CLARK 10 2450

7788 SCOTT 20 3000

7839 KING 10 5000

7844 TURNER 30 1500

7876 ADAMS 20 1100

7900 JAMES 30 950

7902 FORD 20 3000

7934 MILLER 10 1300

PL/SQL procedure successfully completed.

PL/SQL CURSOR

2011-11-10T08:47:00.000-08:00

Writing PL/SQL CURSOR

To write a Cursor to find employee with given job and deptno.

INPUT

DECLARE

cursor c1(j varchar2, dn number) is select empno, ename from emp where job=j and deptno=dn;

row1 emp%rowtype;

jb emp.job%type;

d emp.deptno%type;

BEGIN

jb:='&jb';

d:=&d;

open c1(jb,d);

fetch c1 into row1.empno,row1.ename;

if c1%notfound then

dbms_RESULT.put_line('Employee does not exist');

else

dbms_RESULT.put_line('empno is:'||row1.empno||' ' ||'employee name is:'||row1.ename);

end if;

END;

RESULT:

SQL> @CUR

Enter value for jb: MANAGER

old 7: jb:='&jb';

new 7: jb:='MANAGER';

Enter value for d: 20

old 8: d:=&d;

new 8: d:=20;

empno is:7566 employee name is:JONES

PL/SQL procedure successfully completed.

SQL> /

Enter value for jb: CLERK

old 7: jb:='&jb';

new 7: jb:='CLERK';

Enter value for d: 40

old 8: d:=&d;

new 8: d:=40;

Employee does not exist

PL/SQL procedure successfully completed.

Writing PL/SQL BLOCK using string functions.

To write a PL/SQL block to apply String Functions on a given input String.

INPUT

DECLARE

a varchar2(20);

l number(5);

BEGIN

a:='&a';

l:=length(a);

dbms_RESULT.put_line('Using Lower Function:' || lower(a));

dbms_RESULT.put_line('Using UPPER Function:' || upper(a));

dbms_RESULT.put_line('Using Initcap Function:' || initcap(a));

dbms_RESULT.put_line('Using Substring Function:' || substr(a,l,1));

dbms_RESULT.put_line('Using Substring Function:' || substr(a,1,3));

dbms_RESULT.put_line('Using Ltrim function for xxxabcxxxx:' || ltrim('xxxabcxxxx','x'));

dbms_RESULT.put_line('Using Rtrim function for xxxabcxxxx:'|| rtrim('xxxabcxxxx','x'));

dbms_RESULT.put_line('Using Lpad function :'|| lpad(a,l+4,'*'));

dbms_RESULT.put_line('Using Rpad function :'|| rpad(a,l+4,'*'));

END;

RESULT:

SQL>@STR

Enter value for a: santosh reddy

old 5: a:='&a';

new 5: a:='santosh reddy';

Using Lower Function:santosh reddy

Using UPPER Function:SANTOSH REDDY

Using Initcap Function:Santosh Reddy

Using Substring Function:y

Using Substring Function:san

Using Ltrim function for xxxabcxxxx:abcxxxx

Using Rtrim function for xxxabcxxxx:xxxabc

Using Lpad function :****santosh reddy

Using Rpad function :santosh reddy****

PL/SQL procedure successfully completed.

SQL> /

Enter value for a: UMA SHANKAR

old 5: a:='&a';

new 5: a:='UMA SHANKAR';

Using Lower Function:uma shankar

Using UPPER Function:UMA SHANKAR

Using Initcap Function:Uma Shankar

Using Substring Function:R

Using Substring Function:UMA

Using Ltrim function for xxxabcxxxx:abcxxxx

Using Rtrim function for xxxabcxxxx:xxxabc

Using Lpad function :****UMA SHANKAR

Using Rpad function :UMA SHANKAR****

PL/SQL procedure successfully completed

Writing PL/SQL triggers

To write a TRIGGER to ensure that DEPT TABLE does not contain duplicate of null values in DEPTNO column.

INPUT

CREATE OR RELPLACE TRIGGER trig1 before insert on dept for each row

DECLARE

a number;

BEGIN

if(:new.deptno is Null) then

raise_application_error(-20001,'error::deptno cannot be null');

else

select count(*) into a from dept where deptno=:new.deptno;

if(a=1) then

raise_application_error(-20002,'error:: cannot have duplicate deptno');

end if;

END;

RESULT:

SQL> @trigger

Trigger created.

SQL> select * from dept;

DEPTNO DNAME LOC

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

10 ACCOUNTING NEW YORK

20 RESEARCH DALLAS

30 SALES CHICAGO

40 OPERATIONS BOSTON

SQL> insert into dept values(&deptnp,'&dname','&loc');

Enter value for deptnp: null

Enter value for dname: marketing

Enter value for loc: hyd

old 1: insert into dept values(&deptnp,'&dname','&loc')

new 1: insert into dept values(null,'marketing','hyd')

insert into dept values(null,'marketing','hyd')

ERROR at line 1:

ORA-20001: error::deptno cannot be null

ORA-06512: at "SCOTT.TRIG1", line 5

ORA-04088: error during execution of trigger 'SCOTT.TRIG1'

SQL> /

Enter value for deptnp: 10

Enter value for dname: manager

Enter value for loc: hyd

old 1: insert into dept values(&deptnp,'&dname','&loc')

new 1: insert into dept values(10,'manager','hyd')

insert into dept values(10,'manager','hyd')

ERROR at line 1:

ORA-20002: error:: cannot have duplicate deptno

ORA-06512: at "SCOTT.TRIG1", line 9

ORA-04088: error during execution of trigger 'SCOTT.TRIG1'

SQL> /

Enter value for deptnp: 50

Enter value for dname: MARKETING

Enter value for loc: HYDERABAD

old 1: insert into dept values(&deptnp,'&dname','&loc')

new 1: insert into dept values(50,'MARKETING','HYDERABAD')

1 row created.

SQL> select * from dept;

DEPTNO DNAME LOC

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

10 ACCOUNTING NEW YORK

20 RESEARCH DALLAS

30 SALES CHICAGO

40 OPERATIONS BOSTON

50 MARKETING HYDE

Locking Table.

AIM: To learn commands related to Table Locking

LOCK TABLE Statement Manually lock one or more tables.

Syntax:

LOCK TABLE [schema.] table [options] IN lockmode MODE [NOWAIT]

LOCK TABLE [schema.] view [options] IN lockmode MODE [NOWAIT]

Options:

PARTITION (partition)

SUBPARTITION (subpartition)

@dblink

lockmodes:

EXCLUSIVE

ROW EXCLUSIVE

SHARE ROW EXCLUSIVE

ROW SHARE* | SHARE UPDATE*

If NOWAIT is omitted Oracle will wait until the table is available.

Several tables can be locked with a single command - separate with commas

e.g. LOCK TABLE table1,table2,table3 IN ROW EXCLUSIVE MODE;

Default Locking Behaviour :

A pure SELECT will not lock any rows.

INSERT, UPDATE or DELETE's - will place a ROW EXCLUSIVE lock.

SELECT...FROM...FOR UPDATE NOWAIT - will place a ROW EXCLUSIVE lock.

Multiple Locks on the same rows with LOCK TABLE

Even when a row is locked you can always perform a SELECT (because SELECT does not lock any rows) in addition to this, each type of lock will allow additional locks to be granted as follows.

ROW SHARE = Allow ROW EXCLUSIVE or ROW SHARE or SHARE locks to be granted to the locked rows.

ROW EXCLUSIVE = Allow ROW EXCLUSIVE or ROW SHARE locks to be granted to the locked rows.

SHARE ROW EXCLUSIVE = Allow ROW SHARE locks to be granted to the locked rows.

SHARE = Allow ROW SHARE or SHARE locks to be granted to the locked rows.

EXCLUSIVE = Allow SELECT queries only

Although it is valid to place more than one lock on a row, UPDATES and DELETE's may still cause a wait if a conflicting row lock is held by another transaction.

Generation of Forms using ORACLE FORM BUILDER

To design a form using Oracle Developer 2000

Introduction

Use Form Builder to simplify for the creation of data-entry screens, also known as Forms. Forms are the applications that connect to a database, retrieve information requested by the user, present it in a layout specified by Form designer, and allow the user to modify or add information. Form Builder allows you to build forms quickly and easily.

In this Hands-On, you learn how to: Create a Data block for the “Customer” table, Create a layout, Use “content” canvas, Use “execute query”, Navigate a table, Use next, previous record, Enter query, Manipulate table’s record, Insert, Update, Delete and Save record.

Form Builder Tool

Open the "Form Builder" tool.

Welcome window

You will get the ‘Welcome to the Form Builder’ window. If you don’t want to get this window anymore uncheck the ‘Display at startup’ box. You can start your work with any of the following options:

· Use the data Block Wizard

· Build a new form manually

· Open an existing form

· Build a form based on a template

The default is ‘Use the data Block Wizard.’ If you want to build a new form manually, click on "Cancel” or check ‘Build a new form manually’ and click ‘OK.’

Connect to database

In the ‘Object Navigator’ window, highlight "Database Objects." Go to the Main menu and choose "File," then "Connect."

In the ‘Connect’ window, login in as “scott” password “tiger,” then click “CONNECT.”

Notice that the box next to ‘Database Objects’ is not empty anymore and it has a ‘+’ sign in it. That will indicate that this item is expandable and you are able to see its entire objects.

Click on the ‘+’ sign next to the ‘Database Objects’ to expand all database schemas.

Create a Module

In the ‘Object Navigator’ window, highlight module1. This is a default name. Go to the Main menu and choose “File,” select “Save as” to store the new object in the “iself” folder and save it as customer data entry. "c:_de." In this example the ‘DE’ abbreviation stands for Data Entry.

Create a Data Block

In the ‘Object Navigator’ window, highlight "Data Blocks,” and click on the "create” icon. The ‘Create’ icon is in the vertical tool bar in the ‘Object Navigator’ window. It is a green ‘+’ sign. If you drag your cursor on the icon a tooltip will show ‘Create.’

New Data Block

In the ‘New Data Block’ window, choose the default option “Data Block Wizard” and click "OK."

Welcome Data Block

In the ‘Welcome Data Block Wizard’ window click on the “NEXT” icon.

Type of Data Block

Select the type of data block you would like to create by clicking on a radio button. Select the default option ‘Table or View’ and then click “NEXT” again.

Selecting Tables

Click on “browse.” In the ‘Tables’ window, highlight the "cust11” table; then click "OK."

SELECTING COLUMNS FOR THE DATA BLOCK WIZARD

2011-11-09T08:19:00.000-08:00

Selecting columns for the Data Block Wizard

To choose all columns, click on the two arrow signs in the ‘Data Block Wizard’ window. To choose selected columns, click on the one arrow sign. And then select all columns, and click “next.”

Layout Wizard

End of the Data Block Wizard and beginning of the Layout Wizard

In the ‘Congratulations’ screen, use the default checkmark radio button (Create the data block, then call the Layout Wizard), and click "Finish." You can also use the Data Block Wizard to modify your existing data block. Simply select the data block in the Object Navigator and click the Data Block Wizard toolbar button, or choose ‘Data Block wizard’ from the ‘Tools’ menu.

Welcome screen

In the ‘Welcome to the Layout Wizard’ window, click ”Next.”

Selecting canvas

In the ‘Layout Wizard’ window, select the "new canvas" option. Canvas is a place that you will have your objects such as columns, titles, pictures, etc. If you have already had your canvas, select the canvas and then click on the next. The following are different types of canvases: Content, Stacked, Vertical Toolbar, Horizontal Toolbar, and Tab.

Think of the ‘Content’ canvas as one flat place to have all your objects. In the stacked canvas, you can have multiple layers of objects and it is the same as the tab canvas. You use the vertical or horizontal toolbar canvases for your push buttons. Check the different types of canvases by clicking on the ‘down arrow’ box next to the ‘Type’ field. Select "content," then click “Next.”

Selecting Columns for the Layout Wizard

In the ‘Layout Wizard’ window, select all the columns. These are the columns that you want to be displayed on the canvas. Then click “Next.”

Change your objects appearances

Change size or prompt if needed. In this window, you can enter a prompt, width, and height for each item on the canvas. You can change the measurement units. As a default the default units for item width and height are points. You can change it to inch or centimeter. When you change size, click “Next.”

Selecting a layout style

Select a layout style for your frame by clicking a radio button. Select "Form," if you want one record at a time to be displayed. Select “Tabular,” if you want more than one record at a time to be displayed. Select "Forms," and then click “next.”

Record layout

Type the "Frame Title" and click "next." Checkmark the ‘Display Scrollbar’ box when you use multiple records or the ‘Tabular’ option.

Congratulation Screen

In the ‘Congratulations’ window, click "Finish."

You will see the output layout screen.

Make some window adjustments and then run the form. To run the form, click on the ‘Run’ icon. The ‘Run’ icon is on the horizontal toolbar in the ‘CUSTOMER_DE’ canvas.

The object module should be compiled successfully before executing the Form.

Execute Query

Click on the "Execute Query" icon below the main menu. If you drag the cursor on the toolbar in the ‘Forms Runtime’ window, a tooltip will be displayed and you see ‘Execute Query.’

So to know all your option, drag your cursor to view all the icon descriptions.

Next Record

Click on the "Next Record" icon to navigate to the next record.

Previous Record

Click on the "Previous Record" icon to navigate to the previous record.

This is an easy way to navigate through the “Customer” table.

Enter Query

Click on the "Enter Query" icon to query selected records.

Insert Record

Click "Insert Record" to add new customer. All items on the forms will be blanked. You can either type all the customer information or duplicate it from pervious record.

Duplicate Record

To duplicate the previous record, go to the main menu and select the ‘Record’ sub-menu. A drop down menu will be displayed. Select the ‘Duplicate’ option in the sub-menu.

Apply the changes. Remember in this stage, your record was inserted but not committed yet.

Next and Previous Record

Click "next record" and "previous record" to navigate through the records and the one was added.

Save transactions

Click "Save" to commit the insert statement.

Delete Record

Click "Remove Record" to delete the record.

Lock a Record

You can also lock the record.

Exit from Form Runtime

Exit the FORM Runtime. If you have not committed any transaction, you will be prompted to save changes. Click “YES” to save changes.

Click “OK” for acknowledgement.

Don’t forget to save the Form.

REPORTING TOOLS

2011-11-08T09:05:00.000-08:00

Generating REPORTS using Oracle Developer 2000

AIM: To design reports using Oracle Developer 2000

Introduction

Tabular report shows data in a table format. It is similar in concept to the idea of an Oracle table. Oracle, by default, returns output from your select statement in tabular format.

Hands-on

In this Hands-On, your client is a stock broker that keeps track of its customer stock transactions. You have been assigned to write the reports based on their reports layout requirements.

Your client wants you to create a simple listing report to show list of the stock trades by using stocks table for their brokerage company

Your tasks are:

1- Write a tabular report.

2- Apply user layout Format mask.

3- Run the report.

4- Test the repot.

You will learn how to: use report wizard, object navigator, report builder, “date model”, property palette, work on query and group box, see report style, use tabular style, navigating through report’s record, change the format mask for dollar, numeric and date items.

Open Report Builder tool

Open the "Report Builder" tool.

Connect to database

In the Object Navigator, highlight "Database Objects,” choose "File," then select the "Connect" option.

In the ‘Connect’ window, login as “iself” password schooling, then click “CONNECT.”

Save a report

In the Object Navigator, highlight the "untitled" report, choose “File,” and select the “Save as” option.

In the ‘Save as’ window, make sure to save the report in the ISELF folder and name it "rpt01_stock_history,” report number 1 stock history.

Data Model

In the Object Navigator, double click on the "Data Model" icon.

Create SQL box

In the Data Model window, click on the "SQL Query" icon. Then drag the plus sign cursor and click it anywhere in the “Data Model” screen where you wish your object to be.

In the ‘SQL Query Statement’ window, write a query to read all the stocks record sorted by their symbol.

(SQL Query Statement)

SELECT * FROM stocks

ORDER BY symbol

Click “OK.”

Change SQL box’s name

In the Data Model window, in the “SQL” box, right click on the ‘Q_1’ and open its property palette.

In its property palette, change the name to Q_STOCKS. Then close the window.

Change GROUP box’s name

In the Data Model, right click on the group box (G_SYMBOL) and open its property palette.

In the Group property palette, change the name to ‘G_STOCKS,’ and close the window.

Open Report Wizard

In the Data Model, click on the ‘Report Wizard’ icon on the horizontal tool bar.

In the Style tab, on the Report Wizard window, type ‘Stock History’ in the Title box and choose the report style as ‘Tabular.’

Notice that when you change the report style a layout of that report will be displayed on the screen.

Choose a different style to display its layout of its report style.

Data, Fields, Totals, Labels and Template tabs

Click “NEXT” to go to the Data tab. In the ‘SQL Query Statement’ verify your query.

Click “NEXT” to navigate to the Fields tab, select the fields that you would like to be display in your report. Select all the columns to be display.

Click “NEXT” to navigate to Totals tab, select the fields for which you would like to calculate totals. We have none in this hands-on exercise.

Click “NEXT” to open the Labels tab, modify the labels and widths for your fields and totals as desired.

Click “NEXT” again to go to the Template tab, and choose a template for your report. Your report will inherit the template’s colors, fonts, line widths, and structure.

Use the default template and click “finish.”

Running a report

Now, you should have your output report on the screen.

Resize an object

Maximize the output report and format the report layout. To resize an object , select it and drag its handler to the preferred size.

Move an object

To move an object, select and drag it while the cursor is on the object.

This is a simple report.

Navigate through the output

To navigate through the output report in the Report Editor - Live Pre-viewer, click on the "next page" or "previous page" icon on the horizontal toolbar.

Do the same with the "first page" or "last page" icon.

Use the “zoom in” and “zoom out” icon to preview the report.

Know report’s functions

To know each icon functionalities, drag your cursor on it and a tooltip will display its function.

Change Format Mask

To change the "format mask" of a column, the column should be selected. Then go to the toolbar and click on the “$” icon, "add decimal place," and the “right justify” format to the all currency columns (Todays Low, Todays High, and current price)

Select the “traded today” column, and click on the ‘,0’ icon (apply commas), and make it right justify.

Also, you can change any attributes of field by opening its property palette. To open an object’s property palette, right click on it and select the Property Palette option.

Right click on the "trade date" column and open its "property palette."

Change the date "Format Mask" property and make it “year 2000 complaint (MM-DD-RR).”

PROVIDING SECURITY USING GRANT AND REVOKE

2011-11-07T07:55:00.000-08:00

Providing Security using GRANT and REVOKE

AIM: To learn GRANT and REVOKE commands to restrict privileges.

(1) GRANT Statement

Grant privileges to a user (or to a user role)

Syntax:

Grant System-wide Privs:

GRANT system_priv(s) TO grantee

[IDENTIFIED BY password] [WITH ADMIN OPTION]

GRANT role TO grantee

[IDENTIFIED BY password] [WITH ADMIN OPTION]

GRANT ALL PRIVILEGES TO grantee

[IDENTIFIED BY password] [WITH ADMIN OPTION]

Grant privs on specific objects:

GRANT object_priv [(column, column,...)]

ON [schema.]object

TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT ALL PRIVILEGES [(column, column,...)]

ON [schema.]object

TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT object_priv [(column, column,...)]

ON DIRECTORY directory_name

TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

GRANT object_priv [(column, column,...)]

ON JAVA [RE]SOURCE [schema.]object

TO grantee [WITH GRANT OPTION] [WITH HIERARCHY OPTION]

grantee:

user

role

PUBLIC

system_privs:

CREATE SESSION - Allows user to connect to the database

UNLIMITED TABLESPACE - Use an unlimited amount of any tablespace.

SELECT ANY TABLE - Query tables, views, or mviews in any schema

UPDATE ANY TABLE - Update rows in tables and views in any schema

INSERT ANY TABLE - Insert rows into tables and views in any schema

Also System Admin rights to CREATE, ALTER or DROP:

cluster, context, database, link, dimension, directory, index,

materialized view, operator, outline, procedure, profile, role,

rollback segment, sequence, session, synonym, table, tablespace,

trigger, type, user, view. (full list of system privs)

object_privs:

SELECT, UPDATE, INSERT, DELETE, ALTER, DEBUG, EXECUTE, INDEX, REFERENCES

roles:

SYSDBA, SYSOPER, OSDBA, OSOPER, EXP_FULL_DATABASE, IMP_FULL_DATABASE

SELECT_CATALOG_ROLE, EXECUTE_CATALOG_ROLE, DELETE_CATALOG_ROLE

AQ_USER_ROLE, AQ_ADMINISTRATOR_ROLE - advanced queuing

SNMPAGENT - Enterprise Manager/Intelligent Agent.

RECOVERY_CATALOG_OWNER - rman

HS_ADMIN_ROLE - heterogeneous services

plus any user defined roles you have available

Notes:

Several Object_Privs can be assigned in a single GRANT statement
e.g.
GRANT SELECT (empno), UPDATE (sal) ON scott.emp TO emma

WITH HIERARCHY OPTION will grant the object privilege on all subobjects, including any created after the GRANT statement is issued.

WITH GRANT OPTION will enable the grantee to grant those object privileges to other users and roles.

"GRANT ALL PRIVILEGES..." may also be written as "GRANT ALL..."

(ii) REVOKE Statement

Revoke privileges from users or roles.

Syntax:

Roles:

REVOKE role FROM {user, | role, |PUBLIC}

System Privs:

REVOKE system_priv(s) FROM {user, | role, |PUBLIC}

REVOKE ALL FROM {user, | role, |PUBLIC}

Object Privs:

REVOKE object_priv [(column1, column2..)] ON [schema.]object

FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON [schema.]object

FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON DIRECTORY directory_name

FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

REVOKE object_priv [(column1, column2..)] ON JAVA [RE]SOURCE [schema.]object

FROM {user, | role, |PUBLIC} [CASCADE CONSTRAINTS] [FORCE]

key:

object_privs

ALTER, DELETE, EXECUTE, INDEX, INSERT,

REFERENCES, SELECT, UPDATE, ALL PRIVILEGES

system_privs

ALTER ANY INDEX, BECOME USER, CREATE TABLE, DROP ANY VIEW

RESTRICTED SESSION, UNLIMITED TABLESPACE, UPDATE ANY TABLE

plus too many others to list here

roles

Standard Oracle roles -

SYSDBA, SYSOPER, OSDBA, OSOPER, EXP_FULL_DATABASE, IMP_FULL_DATABASE

plus any user defined roles you have available

FORCE, will revoke all privileges from a user-defined-type and mark it's dependent objects INVALID.

The roles CONNECT, RESOURCE and DBA are now deprecated (supported only for backwards compatibility) unless you are still running Oracle 6.0

Error ORA-01927 "cannot REVOKE privileges you did not grant" - This usually means you tried revoking permission from the table owner, e.g.
Oracle will not allow REVOKE select on USER1.Table1 from USER1 Owners of objects ALWAYS have full permissions on those objects. This is one reason it makes sense to place tables in one schema and the packaged prodecures used to access those tables in a separate schema.

VIVA VOICE QUESTIONS AND ANSWERS

2011-11-06T08:51:00.000-08:00

VIVA VOICE QUESTIONS AND ANSWERS

1. What is database?
A database is a logically coherent collection of data with some inherent meaning, representing some aspect of real world and which is designed, built and populated with data for a specific purpose.

2. What is DBMS?
It is a collection of programs that enables user to create and maintain a database. In other words it is general-purpose software that provides the users with the processes of defining, constructing and manipulating the database for various applications.

3. What is a Database system?
The database and DBMS software together is called as Database system.

4. Advantages of DBMS?
Ø Redundancy is controlled.
Ø Unauthorised access is restricted.
Ø Providing multiple user interfaces.
Ø Enforcing integrity constraints.
Ø Providing backup and recovery.

5. Disadvantage in File Processing System?
Ø Data redundancy & inconsistency.
Ø Difficult in accessing data.
Ø Data isolation.
Ø Data integrity.
Ø Concurrent access is not possible.
Ø Security Problems.

6. Describe the three levels of data abstraction?
The are three levels of abstraction:
Ø Physical level: The lowest level of abstraction describes how data are stored.
Ø Logical level: The next higher level of abstraction, describes what data are stored in database and what relationship among those data.
Ø View level: The highest level of abstraction describes only part of entire database.
7. Define the "integrity rules"
There are two Integrity rules.
Ø Entity Integrity: States that “Primary key cannot have NULL value”
Ø Referential Integrity: States that “Foreign Key can be either a NULL value or should be Primary Key value of other relation.

8. What is extension and intension?
Extension -
It is the number of tuples present in a table at any instance. This is time dependent.
Intension -
It is a constant value that gives the name, structure of table and the constraints laid on it.

9. What is System R? What are its two major subsystems?
System R was designed and developed over a period of 1974-79 at IBM San Jose Research Center. It is a prototype and its purpose was to demonstrate that it is possible to build a Relational System that can be used in a real life environment to solve real life problems, with performance at least comparable to that of existing system.
Its two subsystems are
Ø Research Storage
Ø System Relational Data System.

10. How is the data structure of System R different from the relational structure?
Unlike Relational systems in System R
Ø Domains are not supported
Ø Enforcement of candidate key uniqueness is optional
Ø Enforcement of entity integrity is optional
Ø Referential integrity is not enforced

11. What is Data Independence?
Data independence means that “the application is independent of the storage structure and access strategy of data”. In other words, The ability to modify the schema definition in one level should not affect the schema definition in the next higher level.
Two types of Data Independence:
Ø Physical Data Independence: Modification in physical level should not affect the logical level.
Ø Logical Data Independence: Modification in logical level should affect the view level.
NOTE: Logical Data Independence is more difficult to achieve
12. What is a view? How it is related to data independence?
A view may be thought of as a virtual table, that is, a table that does not really exist in its own right but is instead derived from one or more underlying base table. In other words, there is no stored file that direct represents the view instead a definition of view is stored in data dictionary.
Growth and restructuring of base tables is not reflected in views. Thus the view can insulate users from the effects of restructuring and growth in the database. Hence accounts for logical data independence.

13. What is Data Model?
A collection of conceptual tools for describing data, data relationships data semantics and constraints.

14. What is E-R model?
This data model is based on real world that consists of basic objects called entities and of relationship among these objects. Entities are described in a database by a set of attributes.

15. What is Object Oriented model?
This model is based on collection of objects. An object contains values stored in instance variables with in the object. An object also contains bodies of code that operate on the object. These bodies of code are called methods. Objects that contain same types of values and the same methods are grouped together into classes.

16. What is an Entity?
It is a 'thing' in the real world with an independent existence.

17. What is an Entity type?
It is a collection (set) of entities that have same attributes.

18. What is an Entity set?
It is a collection of all entities of particular entity type in the database.

19. What is an Extension of entity type?
The collections of entities of a particular entity type are grouped together into an entity set.
20. What is Weak Entity set?
An entity set may not have sufficient attributes to form a primary key, and its primary key compromises of its partial key and primary key of its parent entity, then it is said to be Weak Entity set.

21. What is an attribute?
It is a particular property, which describes the entity.

22. What is a Relation Schema and a Relation?
A relation Schema denoted by R(A1, A2, …, An) is made up of the relation name R and the list of attributes Ai that it contains. A relation is defined as a set of tuples. Let r be the relation which contains set tuples (t1, t2, t3, ..., tn). Each tuple is an ordered list of n-values t=(v1,v2, ..., vn).

23. What is degree of a Relation?
It is the number of attribute of its relation schema.

24. What is Relationship?
It is an association among two or more entities.

25. What is Relationship set?
The collection (or set) of similar relationships.

VIVA VOICE QUESTIONS AND ANSWERS

2011-11-05T08:20:00.000-07:00

26. What is Relationship type?
Relationship type defines a set of associations or a relationship set among a given set of entity types.

27. What is degree of Relationship type?
It is the number of entity type participating.

28. What is DDL (Data Definition Language)?
A data base schema is specifies by a set of definitions expressed by a special language called DDL.

29. What is VDL (View Definition Language)?
It specifies user views and their mappings to the conceptual schema.
30. What is SDL (Storage Definition Language)?
This language is to specify the internal schema. This language may specify the mapping between two schemas.

31. What is Data Storage - Definition Language?
The storage structures and access methods used by database system are specified by a set of definition in a special type of DDL called data storage-definition language.

32. What is DML (Data Manipulation Language)?
This language that enable user to access or manipulate data as organised by appropriate data model.
Ø Procedural DML or Low level: DML requires a user to specify what data are needed and how to get those data.
Ø Non-Procedural DML or High level: DML requires a user to specify what data are needed without specifying how to get those data.

33. What is DML Compiler?
It translates DML statements in a query language into low-level instruction that the query evaluation engine can understand.

34. What is Query evaluation engine?
It executes low-level instruction generated by compiler.

35. What is DDL Interpreter?
It interprets DDL statements and record them in tables containing metadata.

36. What is Record-at-a-time?
The Low level or Procedural DML can specify and retrieve each record from a set of records. This retrieve of a record is said to be Record-at-a-time.

37. What is Set-at-a-time or Set-oriented?
The High level or Non-procedural DML can specify and retrieve many records in a single DML statement. This retrieve of a record is said to be Set-at-a-time or Set-oriented.

38. What is Relational Algebra?
It is procedural query language. It consists of a set of operations that take one or two relations as input and produce a new relation.

39. What is Relational Calculus?
It is an applied predicate calculus specifically tailored for relational databases proposed by E.F. Codd. E.g. of languages based on it are DSL ALPHA, QUEL.

40. How does Tuple-oriented relational calculus differ from domain-oriented relational calculus
The tuple-oriented calculus uses a tuple variables i.e., variable whose only permitted values are tuples of that relation. E.g. QUEL
The domain-oriented calculus has domain variables i.e., variables that range over the underlying domains instead of over relation. E.g. ILL, DEDUCE.

41. What is normalization?
It is a process of analysing the given relation schemas based on their Functional Dependencies (FDs) and primary key to achieve the properties
Ø Minimizing redundancy
Ø Minimizing insertion, deletion and update anomalies.

42. What is Functional Dependency?
A Functional dependency is denoted by X Y between two sets of attributes X and Y that are subsets of R specifies a constraint on the possible tuple that can form a relation state r of R. The constraint is for any two tuples t1 and t2 in r if t1[X] = t2[X] then they
have t1[Y] = t2[Y]. This means the value of X component of a tuple uniquely determines the value of component Y.

43. When is a functional dependency F said to be minimal?
Ø Every dependency in F has a single attribute for its right hand side.
Ø We cannot replace any dependency X A in F with a dependency Y A where Y is a proper subset of X and still have a set of dependency that is equivalent to F.
Ø We cannot remove any dependency from F and still have set of dependency that is equivalent to F.

44. What is Multivalued dependency?
Multivalued dependency denoted by X Y specified on relation schema R, where X and Y are both subsets of R, specifies the following constraint on any relation r of R: if two tuples t1 and t2 exist in r such that t1[X] = t2[X] then t3 and t4 should also exist in r with the following properties
Ø t3[x] = t4[X] = t1[X] = t2[X]
Ø t3[Y] = t1[Y] and t4[Y] = t2[Y]
Ø t3[Z] = t2[Z] and t4[Z] = t1[Z]
where [Z = (R-(X U Y)) ]
45. What is Lossless join property?
It guarantees that the spurious tuple generation does not occur with respect to relation schemas after decomposition.

46. What is 1 NF (Normal Form)?
The domain of attribute must include only atomic (simple, indivisible) values.

47. What is Fully Functional dependency?
It is based on concept of full functional dependency. A functional dependency X Y is full functional dependency if removal of any attribute A from X means that the dependency does not hold any more.

48. What is 2NF?
A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully functionally dependent on primary key.

49. What is 3NF?
A relation schema R is in 3NF if it is in 2NF and for every FD X A either of the following is true
Ø X is a Super-key of R.
Ø A is a prime attribute of R.
In other words, if every non prime attribute is non-transitively dependent on primary key.

50. What is BCNF (Boyce-Codd Normal Form)?
A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that for every FD X A, X must be a candidate key.

VIVA VOICE QUESTIONS AND ANSWERS

2011-11-04T09:11:00.000-07:00

51. What is 4NF?
A relation schema R is said to be in 4NF if for every Multivalued dependency X Y that holds over R, one of following is true
Ø X is subset or equal to (or) XY = R.
Ø X is a super key.

52. What is 5NF?
A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn} that holds R, one the following is true
Ø Ri = R for some i.
Ø The join dependency is implied by the set of FD, over R in which the left side is key of R.

53. What is Domain-Key Normal Form?
A relation is said to be in DKNF if all constraints and dependencies that should hold on the the constraint can be enforced by simply enforcing the domain constraint and key constraint on the relation.

54. What are partial, alternate,, artificial, compound and natural key?
Partial Key:
It is a set of attributes that can uniquely identify weak entities and that are related to same owner entity. It is sometime called as Discriminator.
Alternate Key:
All Candidate Keys excluding the Primary Key are known as Alternate Keys.
Artificial Key:
If no obvious key, either stand alone or compound is available, then the last resort is to simply create a key, by assigning a unique number to each record or occurrence. Then this is known as developing an artificial key.
Compound Key:
If no single data element uniquely identifies occurrences within a construct, then combining multiple elements to create a unique identifier for the construct is known as creating a compound key.
Natural Key:
When one of the data elements stored within a construct is utilized as the primary key, then it is called the natural key.

55. What is indexing and what are the different kinds of indexing?
Indexing is a technique for determining how quickly specific data can be found.
Types:
Ø Binary search style indexing
Ø B-Tree indexing
Ø Inverted list indexing
Ø Memory resident table
Ø Table indexing

56. What is system catalog or catalog relation? How is better known as?
A RDBMS maintains a description of all the data that it contains, information about every relation and index that it contains. This information is stored in a collection of relations maintained by the system called metadata. It is also called data dictionary.

57. What is meant by query optimization?
The phase that identifies an efficient execution plan for evaluating a query that has the least estimated cost is referred to as query optimization.

58. What is join dependency and inclusion dependency?
Join Dependency:
A Join dependency is generalization of Multivalued dependency.A JD {R1, R2, ..., Rn} is said to hold over a relation R if R1, R2, R3, ..., Rn is a lossless-join decomposition of R . There is no set of sound and complete inference rules for JD.
Inclusion Dependency:
An Inclusion Dependency is a statement of the form that some columns of a relation are contained in other columns. A foreign key constraint is an example of inclusion dependency.

59. What is durability in DBMS?
Once the DBMS informs the user that a transaction has successfully completed, its effects should persist even if the system crashes before all its changes are reflected on disk. This property is called durability.

60. What do you mean by atomicity and aggregation?
Atomicity:
Either all actions are carried out or none are. Users should not have to worry about the effect of incomplete transactions. DBMS ensures this by undoing the actions of incomplete transactions.
Aggregation:
A concept which is used to model a relationship between a collection of entities and relationships. It is used when we need to express a relationship among relationships.

61. What is a Phantom Deadlock?
In distributed deadlock detection, the delay in propagating local information might cause the deadlock detection algorithms to identify deadlocks that do not really exist. Such situations are called phantom deadlocks and they lead to unnecessary aborts.

62. What is a checkpoint and When does it occur?
A Checkpoint is like a snapshot of the DBMS state. By taking checkpoints, the DBMS can reduce the amount of work to be done during restart in the event of subsequent crashes.

63. What are the different phases of transaction?
Different phases are
Ø Analysis phase
Ø Redo Phase
Ø Undo phase

64. What do you mean by flat file database?
It is a database in which there are no programs or user access languages. It has no cross-file capabilities but is user-friendly and provides user-interface management.

65. What is "transparent DBMS"?
It is one, which keeps its Physical Structure hidden from user.

66. Brief theory of Network, Hierarchical schemas and their properties
Network schema uses a graph data structure to organize records example for such a database management system is CTCG while a hierarchical schema uses a tree data structure example for such a system is IMS.

67. What is a query?
A query with respect to DBMS relates to user commands that are used to interact with a data base. The query language can be classified into data definition language and data manipulation language.

68. What do you mean by Correlated subquery?
Subqueries, or nested queries, are used to bring back a set of rows to be used by the parent query. Depending on how the subquery is written, it can be executed once for the parent query or it can be executed once for each row returned by the parent query. If the subquery is executed for each row of the parent, this is called a correlated subquery.
A correlated subquery can be easily identified if it contains any references to the parent subquery columns in its WHERE clause. Columns from the subquery cannot be referenced anywhere else in the parent query. The following example demonstrates a non-correlated subquery.
E.g. Select * From CUST Where '10/03/1990' IN (Select ODATE From ORDER Where CUST.CNUM = ORDER.CNUM)

69. What are the primitive operations common to all record management systems?
Addition, deletion and modification.

70. Name the buffer in which all the commands that are typed in are stored
‘Edit’ Buffer

71. What are the unary operations in Relational Algebra?
PROJECTION and SELECTION.

72. Are the resulting relations of PRODUCT and JOIN operation the same?
No.
PRODUCT: Concatenation of every row in one relation with every row in another.
JOIN: Concatenation of rows from one relation and related rows from another.

73. What is RDBMS KERNEL?
Two important pieces of RDBMS architecture are the kernel, which is the software, and the data dictionary, which consists of the system-level data structures used by the kernel to manage the database
You might think of an RDBMS as an operating system (or set of subsystems), designed specifically for controlling data access; its primary functions are storing, retrieving, and securing data. An RDBMS maintains its own list of authorized users and their associated privileges; manages memory caches and paging; controls locking for concurrent resource usage; dispatches and schedules user requests; and manages space usage within its table-space structures.

74. Name the sub-systems of a RDBMS
I/O, Security, Language Processing, Process Control, Storage Management, Logging and Recovery, Distribution Control, Transaction Control, Memory Management, Lock Management

75. Which part of the RDBMS takes care of the data dictionary? How
Data dictionary is a set of tables and database objects that is stored in a special area of the database and maintained exclusively by the kernel.

VIVA VOICE QUESTIONS AND ANSWERS

2011-11-03T08:48:00.000-07:00

76. What is the job of the information stored in data-dictionary?
The information in the data dictionary validates the existence of the objects, provides access to them, and maps the actual physical storage location.

77. Not only RDBMS takes care of locating data it also
determines an optimal access path to store or retrieve the data
76. How do you communicate with an RDBMS?
You communicate with an RDBMS using Structured Query Language (SQL)

78. Define SQL and state the differences between SQL and other conventional programming Languages
SQL is a nonprocedural language that is designed specifically for data access operations on normalized relational database structures. The primary difference between SQL and other conventional programming languages is that SQL statements specify what data operations should be performed rather than how to perform them.

79. Name the three major set of files on disk that compose a database in Oracle
There are three major sets of files on disk that compose a database. All the files are binary. These are
Ø Database files
Ø Control files
Ø Redo logs
The most important of these are the database files where the actual data resides. The control files and the redo logs support the functioning of the architecture itself.
All three sets of files must be present, open, and available to Oracle for any data on the database to be useable. Without these files, you cannot access the database, and the database administrator might have to recover some or all of the database using a backup, if there is one.

80. What is an Oracle Instance?
The Oracle system processes, also known as Oracle background processes, provide functions for the user processes—functions that would otherwise be done by the user processes themselves
Oracle database-wide system memory is known as the SGA, the system global area or shared global area. The data and control structures in the SGA are shareable, and all the Oracle background processes and user processes can use them.
The combination of the SGA and the Oracle background processes is known as an Oracle instance

81. What are the four Oracle system processes that must always be up and running for the database to be useable
The four Oracle system processes that must always be up and running for the database to be useable include DBWR (Database Writer), LGWR (Log Writer), SMON (System Monitor), and PMON (Process Monitor).

82. What are database files, control files and log files. How many of these files should a database have at least? Why?
Database Files
The database files hold the actual data and are typically the largest in size. Depending on their sizes, the tables (and other objects) for all the user accounts can go in one database file—but that's not an ideal situation because it does not make the database structure very flexible for controlling access to storage for different users, putting the database on different disk drives, or backing up and restoring just part of the database.
You must have at least one database file but usually, more than one files are used. In terms of accessing and using the data in the tables and other objects, the number (or location) of the files is immaterial.
The database files are fixed in size and never grow bigger than the size at which they were created Control Files
The control files and redo logs support the rest of the architecture. Any database must have at least one control file, although you typically have more than one to guard against loss. The control file records the name of the database, the date and time it was created, the location of the database and redo logs, and the synchronization information to ensure that all three sets of files are always in step. Every time you add a new database or redo log file to the database, the information is recorded in the control files.
Redo Logs
Any database must have at least two redo logs. These are the journals for the database; the redo logs record all changes to the user objects or system objects. If any type of failure occurs, the changes recorded in the redo logs can be used to bring the database to a consistent state without losing any committed transactions. In the case of non-data loss failure, Oracle can apply the information in the redo logs automatically without intervention from the DBA.
The redo log files are fixed in size and never grow dynamically from the size at which they were created.

83. What is ROWID?
The ROWID is a unique database-wide physical address for every row on every table. Once assigned (when the row is first inserted into the database), it never changes until the row is deleted or the table is dropped.
The ROWID consists of the following three components, the combination of which uniquely identifies the physical storage location of the row.
Ø Oracle database file number, which contains the block with the rows
Ø Oracle block address, which contains the row
Ø The row within the block (because each block can hold many rows)
The ROWID is used internally in indexes as a quick means of retrieving rows with a particular key value. Application developers also use it in SQL statements as a quick way to access a row once they know the ROWID

84. What is Oracle Block? Can two Oracle Blocks have the same address?
Oracle "formats" the database files into a number of Oracle blocks when they are first created—making it easier for the RDBMS software to manage the files and easier to read data into the memory areas.
The block size should be a multiple of the operating system block size. Regardless of the block size, the entire block is not available for holding data; Oracle takes up some space to manage the contents of the block. This block header has a minimum size, but it can grow.
These Oracle blocks are the smallest unit of storage. Increasing the Oracle block size can improve performance, but it should be done only when the database is first created.
Each Oracle block is numbered sequentially for each database file starting at 1. Two blocks can have the same block address if they are in different database files.

85. What is database Trigger?
A database trigger is a PL/SQL block that can defined to automatically execute for insert, update, and delete statements against a table. The trigger can e defined to execute once for the entire statement or once for every row that is inserted, updated, or deleted. For any one table, there are twelve events for which you can define database triggers. A database trigger can call database procedures that are also written in PL/SQL.

86. Name two utilities that Oracle provides, which are use for backup and recovery.
Along with the RDBMS software, Oracle provides two utilities that you can use to back up and restore the database. These utilities are Export and Import.
The Export utility dumps the definitions and data for the specified part of the database to an operating system binary file. The Import utility reads the file produced by an export, recreates the definitions of objects, and inserts the data
If Export and Import are used as a means of backing up and recovering the database, all the changes made to the database cannot be recovered since the export was performed. The best you can do is recover the database to the time when the export was last performed.

87. What are stored-procedures? And what are the advantages of using them.
Stored procedures are database objects that perform a user defined operation. A stored procedure can have a set of compound SQL statements. A stored procedure executes the SQL commands and returns the result to the client. Stored procedures are used to reduce network traffic.

88. How are exceptions handled in PL/SQL? Give some of the internal exceptions' name
PL/SQL exception handling is a mechanism for dealing with run-time errors encountered during procedure execution. Use of this mechanism enables execution to continue if the error is not severe enough to cause procedure termination.
The exception handler must be defined within a subprogram specification. Errors cause the program to raise an exception with a transfer of control to the exception-handler block. After the exception handler executes, control returns to the block in which the handler was defined. If there are no more executable statements in the block, control returns to the caller.
User-Defined Exceptions
PL/SQL enables the user to define exception handlers in the declarations area of subprogram specifications. User accomplishes this by naming an exception as in the following example:
ot_failure EXCEPTION;
In this case, the exception name is ot_failure. Code associated with this handler is written in the EXCEPTION specification area as follows:
EXCEPTION
when OT_FAILURE then
out_status_code := g_out_status_code;
out_msg := g_out_msg;
The following is an example of a subprogram exception:
EXCEPTION
when NO_DATA_FOUND then
g_out_status_code := 'FAIL';
RAISE ot_failure;
Within this exception is the RAISE statement that transfers control back to the ot_failure exception handler. This technique of raising the exception is used to invoke all user-defined exceptions.
System-Defined Exceptions
Exceptions internal to PL/SQL are raised automatically upon error. NO_DATA_FOUND is a system-defined exception. Table below gives a complete list of internal exceptions.

PL/SQL internal exceptions.
PL/SQL internal exceptions.

Exception Name Oracle Error
CURSOR_ALREADY_OPEN ORA-06511
DUP_VAL_ON_INDEX ORA-00001
INVALID_CURSOR ORA-01001
INVALID_NUMBER ORA-01722
LOGIN_DENIED ORA-01017
NO_DATA_FOUND ORA-01403
NOT_LOGGED_ON ORA-01012
PROGRAM_ERROR ORA-06501
STORAGE_ERROR ORA-06500
TIMEOUT_ON_RESOURCE ORA-00051
TOO_MANY_ROWS ORA-01422
TRANSACTION_BACKED_OUT ORA-00061
VALUE_ERROR ORA-06502
ZERO_DIVIDE ORA-01476

In addition to this list of exceptions, there is a catch-all exception named OTHERS that traps all errors for which specific error handling has not been established.

89. Does PL/SQL support "overloading"? Explain
The concept of overloading in PL/SQL relates to the idea that you can define procedures and functions with the same name. PL/SQL does not look only at the referenced name, however, to resolve a procedure or function call. The count and data types of formal parameters are also considered.
PL/SQL also attempts to resolve any procedure or function calls in locally defined packages before looking at globally defined packages or internal functions. To further ensure calling the proper procedure, you can use the dot notation. Prefacing a procedure or function name with the package name fully qualifies any procedure or function reference.

90. Tables derived from the ERD
a) Are totally unnormalised
b) Are always in 1NF
c) Can be further denormalised
d) May have multi-valued attributes

(b) Are always in 1NF

91. Spurious tuples may occur due to
i. Bad normalization
ii. Theta joins
iii. Updating tables from join
a) i & ii b) ii & iii
c) i & iii d) ii & iii

(a) i & iii because theta joins are joins made on keys that are not primary keys.

92. A B C is a set of attributes. The functional dependency is as follows
AB -> B
AC -> C
C -> B
a) is in 1NF
b) is in 2NF
c) is in 3NF
d) is in BCNF

(a) is in 1NF since (AC)+ = { A, B, C} hence AC is the primary key. Since C B is a FD given, where neither C is a Key nor B is a prime attribute, this it is not in 3NF. Further B is not functionally dependent on key AC thus it is not in 2NF. Thus the given FDs is in 1NF.

93. In mapping of ERD to DFD
a) entities in ERD should correspond to an existing entity/store in DFD
b) entity in DFD is converted to attributes of an entity in ERD
c) relations in ERD has 1 to 1 correspondence to processes in DFD
d) relationships in ERD has 1 to 1 correspondence to flows in DFD

(a) entities in ERD should correspond to an existing entity/store in DFD

94. A dominant entity is the entity
a) on the N side in a 1 : N relationship
b) on the 1 side in a 1 : N relationship
c) on either side in a 1 : 1 relationship
d) nothing to do with 1 : 1 or 1 : N relationship

(b) on the 1 side in a 1 : N relationship

95. Select 'NORTH', CUSTOMER From CUST_DTLS Where REGION = 'N' Order By
CUSTOMER Union Select 'EAST', CUSTOMER From CUST_DTLS Where REGION = 'E' Order By CUSTOMER
The above is
a) Not an error
b) Error - the string in single quotes 'NORTH' and 'SOUTH'
c) Error - the string should be in double quotes
d) Error - ORDER BY clause

(d) Error - the ORDER BY clause. Since ORDER BY clause cannot be used in UNIONS

96. What is Storage Manager?
It is a program module that provides the interface between the low-level data stored in database, application programs and queries submitted to the system.

97. What is Buffer Manager?
It is a program module, which is responsible for fetching data from disk storage into main memory and deciding what data to be cache in memory.

98. What is Transaction Manager?
It is a program module, which ensures that database, remains in a consistent state despite system failures and concurrent transaction execution proceeds without conflicting.

99. What is File Manager?
It is a program module, which manages the allocation of space on disk storage and data structure used to represent information stored on a disk.

100. What is Authorization and Integrity manager?
It is the program module, which tests for the satisfaction of integrity constraint and checks the authority of user to access data.