DATABASE MANAGEMENT SYSTEMS LABchalapathiengg.ac.in/cse lab manual/dbms2021.pdfDBMS LAB 3 8. ETHICS:...

ACHARYA NAGARJUNA UNIVERSITY

DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

CHALAPATHI INSTITUTE OF ENGINEERING AND TECHNOLOGY

III/IV B. Tech I SEM, 2020-21

MASTER LAB MANUAL

For

DATABASE MANAGEMENT SYSTEMS LAB

Prepared by

G.RAMACHANDRA RAO Assistant Professor

CHALAPATHI INSTITUTE OF ENGINEERING AND TECHNOLOGY

(AUTONOMOUS)

CHALAPATHI NAGAR, LAM, GUNTUR-522034

INDEX

S.NO CONTENT Page

No

i Vision and Mission 1

ii POs and PEOs 2-3

iii Course Objectives and Outcomes 4

iv Mapping 4-5

iv Syllabus 6

1

Learn the Data Definition Language (DDL) commands in RDBMS, Data

Manipulation Language (DML) and Data Control Language (DCL). 7-18

2

Simple queries: selection, projection, sorting on a simple table, Small-large number

of attributes ,Distinct output values ,Renaming attributes , Computed attributes

Simple-complex conditions (AND, OR, NOT).

19-32

3

Partial Matching operators (LIKE, %, _, *, ?) ,ASC-DESC ordering combinations

Checking for Nulls. 33-36

4

Multi-table queries (JOIN OPERATIONS) ,Simple joins (no INNER JOIN)

Aliasing tables – Full/Partial name qualification ,Inner-joins (two and more

(different) tables) ,Inner-recursive-joins (joining to itself) ,Outer-joins (restrictions

as part of the WHERE and ON clauses) ,Using where & having clauses.

37-43

5

Nested queries :In, Not In Exists, Not Exists Dynamic relations (as part of

SELECT, FROM, and WHERE clauses). 44-50

6 Set Oriented Operations :Union, Difference, Intersection, Division.

51-55

7

PL/SQL Programming I :Programs using named and unnamed blocks, Programs

using Cursors, Cursor loops and records. 56-65

8 PL/SQL Programming II :Creating stored procedures, functions

66-72

9 Triggers and auditing triggers

73-77

ADDITIONAL EXPERIMENTS

1 To Check whether a number is Armstrong or not using functions.

78

2 To Perform Banking Operations Using Procedures

79-80

DBMS LAB

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CHALAPATHI INSTITUTE OF ENGINEERING & TECHNOLOGY


DEPARTMENT OF COMPUTER SCIENCE & ENGINEERING

COLLEGE VISION:

To emerge as an Institute of Excellence for Engineering and Technology and provide world-

class education and research opportunities to the students catering the needs of society.

COLLEGE MISSION:

Establishing a state-of-the-art Engineering Institute with continuously improving

infrastructure and produce students with innovative skills and global outlook.

DEPARTMENT VISION:

To produce professionally competent, research oriented and socially sensitive engineers and

technocrats in the emerging technologies.

DEPARTMENT MISSION:

DM 1: State of art laboratories to meet the needs of the continuous change.

DM 2: Provide a research environment to meet the societal issues.

DM 3: Facilitating collaborations/MOU’S towards emerging technologies.

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PEO’s:

PEO-1: To prepare students to excel in computer industry profession or higher studies

through quality education

PEO-2: To analyse real life problems, design computing systems appropriate to its solutions

by exhibiting professionalism and team work

PEO-3: To demonstrate their ability to a rapidly changing environment by engaging in life

long learning.

PO’s:

1. ENGINEERING KNOWLEDGE: Apply the knowledge of mathematics, science,

engineering fundamentals, and an engineering specialization to the solution of complex

engineering problems.

2. PROBLEM ANALYSIS: Identify, formulate, research literature, and analyze complex

engineering problems reaching substantiated conclusions using first principles of

mathematics, natural sciences, and engineering sciences.

3. DESIGN/DEVELOPMENT OF SOLUTIONS: Design solutions for complex

engineering problems and design system components or processes that meet the specified

needs with appropriate consideration for the public health and safety, and the cultural,

societal, and environmental considerations.

4. CONDUCT INVESTIGATIONS OF COMPLEX PROBLEMS: Use research-based

knowledge and research methods including design of experiments, analysis and interpretation

of data, and synthesis of the information to provide valid conclusions.

5. MODERN TOOL USAGE: Create, select, and apply appropriate techniques, resources,

and modern engineering and IT tools including prediction and modelling to complex

engineering activities with an understanding of the limitations.

6. THE ENGINEER AND SOCIETY: Apply reasoning informed by the contextual

knowledge to assess societal, health, safety, legal and cultural issues and the consequent

responsibilities relevant to the professional engineering practice.

7. ENVIRONMENT AND SUSTAINABILITY: Understand the impact of the professional

engineering solutions in societal and environmental contexts, and demonstrate the knowledge

of, and need for sustainable development.

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8. ETHICS: Apply ethical principles and commit to professional ethics and responsibilities

and norms of the engineering practice.

9. INDIVIDUAL AND TEAM WORK: Function effectively as an individual, and as a

member or leader in diverse teams, and in multidisciplinary settings.

10. COMMUNICATION: Communicate effectively on complex engineering activities with

the engineering community and with society at large, such as, being able to comprehend and

write effective reports and design documentation, make effective presentations, give and

receive clear instructions.

11. PROJECT MANAGEMENT AND FINANCE: Demonstrate knowledge and

understanding of the engineering and management principles and apply these to one’s own

work, as a member and leader in a team, to manage projects and in multidisciplinary

environments.

12. LIFE-LONG LEARNING: Recognize the need for, and have the preparation and ability

to engage in independent and life-long learning in the broadest context of technological

change.

PSO’s:

A graduate of the Computer Science and Engineering Program will demonstrate:

PSO1: Professional Skills: The ability to understand, analyze and develop computer

programs in the areas related to algorithms, system software, multimedia, web design, big

data analytics, and networking for efficient design of computer-based systems of varying

complexity.

PSO2: Problem-Solving Skills: The ability to apply standard practices and strategies in

software project development using open-ended programming environments to deliver a

quality product for business success.

PSO3: Successful Career and Entrepreneurship: The ability to employ modern computer

languages, environments, and platforms in creating innovative career paths to be an

entrepreneur, and a zest for higher studies.

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CHALAPATHI INSTITUTE OF ENGINEERING AND TECHNOLOGY CHALAPATHI NAGAR, LAM, GUNTUR

CLASS: III/IV B.Tech-I Sem BRANCH: CSE

DATABASE MANAGEMENT SYSTEMS LAB

Course Objectives

1. Describe the basics of SQL and construct queries using SQL.

2. To demonstrate use of constraints and relational algebra operations.

3. To emphasize the importance of cursors,packages,Triggers,procedures.

4. To facilitate students in Database design

Course Outcomes

After Completion of this Lab Students will be able

C352.1 Design and implement a database schema for a given problem domain.

C352.2 Use the basics of SQL and construct queries using SQL in database creation and

interaction.

C352.3 Implement PL/SQL Programs including stored procedures,functions & packages.

C352.4 Implement programs on Triggers.

Mapping of CO’s and PO’s CO’s PO’S

1 2 3 4 5 6 7 8 9 10 11 12

C352.1 2 3 2

C352.2

2 3 2

C352.3 3 2

C352.4 3 2

1-HIGH 2-MEDIUM 3-LOW

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CHALAPATHI NAGAR, LAM, GUNTUR-5220L4


EXP

NO

List of Experiments

Course

outcomes

attained

Program

outcomes

attained

1

Learn the Data Definition Language (DDL) commands in

RDBMS, Data Manipulation Language (DML) and Data

Control Language (DCL).

C352.1

C352.2

PO2,PO3,PO5

2

Simple queries: selection, projection, sorting on a simple table,

Small-large number of attributes ,Distinct output values

,Renaming attributes , Computed attributes Simple-complex

conditions (AND, OR, NOT).

C352.1

C352.2

PO2,PO3,

PO5

3

Partial Matching operators (LIKE, %, _, *, ?) ,ASC-DESC

ordering combinations Checking for Nulls.

C352.1

C352.2

PO2,PO3,

PO5

4

Multi-table queries (JOIN OPERATIONS) ,Simple joins (no

INNER JOIN) Aliasing tables – Full/Partial name qualification

,Inner-joins (two and more (different) tables) ,Inner-recursive-

joins (joining to itself) ,Outer-joins (restrictions as part of the

WHERE and ON clauses) ,Using where & having clauses.

C352.1

C352.2

PO2,PO3,

PO5

5

Nested queries :In, Not In Exists, Not Exists Dynamic relations

(as part of SELECT, FROM, and WHERE clauses).

C352.1

C352.2

PO2,PO3,

PO5

6

Set Oriented Operations :Union, Difference, Intersection,

Division.

C352.1

C352.2

PO2,PO3,

PO5

7

PL/SQL Programming I :Programs using named and unnamed

blocks, Programs using Cursors, Cursor loops and records.

C352.3 PO3, PO5

8

PL/SQL Programming II :Creating stored procedures,

functions

C352.3 PO3, PO5

9 Triggers and auditing triggers C352.4 PO3, PO5

ADDITIONAL PROGRAMS

1 To Check whether a number is Armstrong or not using

functions.

C352.3 PO3, PO5

2 To Perform Banking Operations Using Procedures

C352.3 PO3, PO5

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Syllabus as per University

EXP.NO List of Experiments

1 Learn the Data Definition Language (DDL) commands in RDBMS, Data

Manipulation Language (DML) and Data Control Language (DCL).

2

Simple queries: selection, projection, sorting on a simple table, Small-large

number of attributes ,Distinct output values ,Renaming attributes , Computed

attributes Simple-complex conditions (AND, OR, NOT).

3

Partial Matching operators (LIKE, %, _, *, ?) ,ASC-DESC ordering combinations

Checking for Nulls.

4

Multi-table queries (JOIN OPERATIONS) ,Simple joins (no INNER JOIN)

Aliasing tables – Full/Partial name qualification ,Inner-joins (two and more

(different) tables) ,Inner-recursive-joins (joining to itself) ,Outer-joins

(restrictions as part of the WHERE and ON clauses) ,Using where & having

clauses.

5

Nested queries :In, Not In Exists, Not Exists Dynamic relations (as part of

SELECT, FROM, and WHERE clauses).

6 Set Oriented Operations :Union, Difference, Intersection, Division.

7

PL/SQL Programming I :Programs using named and unnamed blocks, Programs

using Cursors, Cursor loops and records.

8 PL/SQL Programming II :Creating stored procedures, functions

9 Triggers and auditing triggers

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EXPERIMENT-1

Aim: Learn the Data Definition Language (DDL) commands in RDBMS, Data Manipulation

Language (DML) and Data Control Language (DCL).

Program:

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:

DBMS LAB

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drop table emp;

DML commands :

1. 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:

insert into <tablename> (<col1>,<col2>) values(<exp>,<exp>);

2. Delete operations.

a) Remove all rows

Syntax:

delete from <tablename>;

b) Removal of a specified row/s

Syntax:

delete from <tablename> where <condition>;

3. Updating the contents of a table.

a) Updating all rows

Syntax:

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

b) Updating selected records.

Syntax:

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

4. 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.

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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>)

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) Revoke permissions using the REVOKE statement:

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

Syntax:

REVOKE<object privilege> ON FROM<user name>;

Table Creation:

Syntax: create table <table_name> (c olumn1 datatype,column2 datatype...);

create table dept(deptnoint,dname char(10),location char(10),primary key(deptno));

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Output:

Table is created

Description of dept table

desc dept;

Table

Colu

mn

Data

Type

Len

gth

Preci

sion

Sc

ale

Primar

y Key

Null

able

Def

ault

Com

ment

DEPTS

DEPT

NO

Numbe

r - - 0 1 - - -

DNAM

E

Char 10 - - -

- -

LOCA

TION

Char 10 - - -

- -

1 - 3

Insert the values into the table:

insert into dept values(10,’Accounting’,’Newyork’);

Output:

1 row inserted

insert into dept values(20,’Research’,’Dallas’);

Output:

1 row inserted

insert into dept values(30,’sales’,’chicago’);

Output:

1 row inserted

insert into dept values(40,’operation’,’boston’);

Output:

1 row inserted

Display imformation of department table.

select * from dept;

DEPTNO DNAME LOCATION

10 accounting newyork

30 sales chicago

javascript:ret_Column('SYSTEM.DEPTS');

javascript:ret_Column('DEPTNO');

javascript:ret_Column('DEPTNO');

javascript:ret_Column('DNAME');

javascript:ret_Column('DNAME');

javascript:ret_Column('LOCATION');

javascript:ret_Column('LOCATION');

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20 research dallas

40 operation boston

Creation of Employee table

create table emp

(empno int,

empname char(10),

job char(10),

mgr int,

hiredate date,

salary int,

com int,

dno int,

primary key(empno),

foreign key(dno) references dept(deptno));

Output:

Table created

Description of dept table:

desc emp;

Output:

Tabl

e Column

Data

Type

Lengt

h

Preci

sion

Sca

le

Primar

y Key

Nullab

le Default

Com

ment

EMP EMPNO

Numbe

r - - 0 1 - - -

EMPNA

ME

Char 10 - - -

- -

JOB Char 10 - - -

- -

javascript:ret_Column('SYSTEM.EMP');

javascript:ret_Column('EMPNO');

javascript:ret_Column('EMPNAME');

javascript:ret_Column('EMPNAME');

javascript:ret_Column('JOB');

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Insertion of values into Employee table:

insert into emp values(7839,'king','president',NULL,'17-november-1981',5000,NULL,10);

Output:

1 row inserted

insert into emp values(7698,'blake','manager',7839,'1-may-1981',2850,NULL,30);

Output:

1 row inserted

insert into emp values(7782,'clark','manager',7839,'9-june-1981',2450,NULL,10);

Output:

1 row inserted

insert into emp values(7902,'ford','analyst',7566,'13-december-1981',3000,NULL,20);

Output:

1 row inserted

insert into emp values(7566,'jones','manager',7839,'2-april-1981',2975,NULL,20);

Output:

1 row inserted

insert into emp values(7788,'Scott','analyst',7566,'13-july-1987', 3000, NULL, 20);

Output:

1 row inserted

MGR

Numbe

r - - 0 -

- -

HIREDA

TE

Date 7 - - -

- -

SALARY

Numbe

r - - 0 -

- -

COM

Numbe

r - - 0 -

- -

DNO

Numbe

r - - 0 -

- -

1 - 8

javascript:ret_Column('MGR');

javascript:ret_Column('HIREDATE');

javascript:ret_Column('HIREDATE');

javascript:ret_Column('SALARY');

javascript:ret_Column('COM');

javascript:ret_Column('DNO');

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insert into emp values(7369, 'Smith','clerk',7902,'17-december-1980',800,NULL,20);

Output:

1 row inserted

insert into emp values(7499, 'Allen', 'salesman', 7698,'20-february-1981', 1600,300,30);

Output:

1 row inserted

insert into emp values(7521,'ward', 'salesman',7608, '22-february-1981', 1250,500,30);

Output:

1 row inserted

insert into emp values(7654, 'Martin', 'salesman',7698,'28-september-1981',1250,1400,30);

Output:

1 row inserted

insert into emp values(7844,'Turner','salesman',7698,'8-september-1981',1500,0,30);

Output:

1 row inserted

insert into emp values(7876,'Adams','clerk',7788,'13-july-1987', 1100, NULL, 20);

Output:

1 row inserted

insert into emp values(7900,'james','clerk',7698,'13-december-1981',950,NULL,30);

Output:

1 row inserted

insert into emp values(7934,'miller','clerk',7782,'21-march-1982',1300,NULL,10);

Output:

1 row inserted

Dispaly information of employee table

select* from emp;

Output:

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EMPNO EMPNAME JOB MGR HIREDATE SALARY

DNO

7839 king president - 17-NOV-81 5000

10

7698 blake manager 7839 01-MAY-81 2850

30

7782 clark manager 7839 09-JUN-81 2450

10

7902 ford analyst 7566 13-DEC-81 3000

20

7566 jones manager 7839 02-APR-81 2975

20

7788 Scott analyst 7566 13-JUL-87 3000

20

7369 Smith clerk 7902 17-DEC-80 800

20

7499 Allen salesman 7698 20-FEB-81 1600

30

7521 ward sales Man 7608 22-FEB-81 1250

30

7654 Martin sales man 7698 28-SEP-81 1250

30

7844 Turner salesman 7698 08-SEP-81 1500

30

7934 miller clerk 7782 21-MAR-82 1300

10

7876 Adams clerk 7788 13-JUL-87 1100

20

7900 james clerk 7698 13-DEC-81 950

30

Update row into a Department Table:

update dept

Set location =’Atlanta’

Where deptno=40;

Output:

ROWS UPDATED

Deletion of row into a Table

delete from emp

Where empno=7844;

Output:

STATEMENT PROCESSED

Alter Command

alter table dept add(NUM_OF_EMP int);

Output:

TABLE ALTERED

Drop Command

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drop table dept;

Output:

TABLE DROPPED

Creation Of Salgrade Table

create table salgrade(grade int,lowsal int,highsal int);

Output:

Table created

Insert Values into a salgrade Table

insert into salgrade values(1,700,1200);

Output:

1 row inserted


Output:

1 row inserted


Output:

1 row inserted

Description of salgrade table

desc salgrade;

Output:

Table

Colu

mn

Data

Type

Len

gth

Preci

sion

Sc

ale

Primar

y Key

Null

able

Def

ault

Com

ment

SALGRA

DE

GRA

DE

Numbe

r - - 0 -

- -

LOW

SAL

Numbe

r - - 0 -

- -

HIGH

SAL

Numbe

r - - 0 -

- -

1 - 3

javascript:ret_Column('SYSTEM.SALGRADE');

javascript:ret_Column('SYSTEM.SALGRADE');

javascript:ret_Column('GRADE');

javascript:ret_Column('GRADE');

javascript:ret_Column('LOWSAL');

javascript:ret_Column('LOWSAL');

javascript:ret_Column('HIGHSAL');

javascript:ret_Column('HIGHSAL');

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Display information of salgrade

Select * from salgrade;

Output:

GRADE LOWSAL HIGHSAL

1 700 1200

2 1200 1400

3 1400 2000

GRANT: Grant privileges to user using the GRANT statement.

session creation for user:

Permission given to user for create a table :

REVOKE: Revoke permissions using the REVOKE statement.

Remove given permission from user:

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Viva Questions:

1) What is SQL?

Structured Query Language SQL is a database tool that is used to create and access the

database to support software applications.

2) What are tables in SQL?

The table is a collection of record and its information at a single view.

3) What are the different types of statements supported by SQL?

4) Explain DML, DDL, DCL and TCL statements with examples?

DML: DML stands for Data Manipulation Language. DML is used to retrieve, store, modify,

delete, insert and update data in database.

Examples of DML statements: SELECT, UPDATE, INSERT, DELETE statements.

DDL: DDL stands for Data Definition Language. DDL is used to create and modify the

structure of database objects.

Examples: CREATE, ALTER, DROP statements.

DCL: DCL stands for Data Control Language. DCL is used to create roles, grant and revoke

permissions, establish referential integrity etc.

Examples: GRANT, REVOKE statements

TCL: TCL stands for Transactional Control Language. TCL is used to manage transactions

within a database.

Examples: COMMIT, ROLLBACK statements

5) What is the difference between Drop, Delete and Truncate statements in SQL

Server?

Drop, Delete and Truncate - All operations can be rolled back.

Delete is a logged operation, which means deleted rows are written to the transaction

log.Truncate is not a logged operation, which means deleted rows are not written to the

transaction log.

Hence, truncate is a little faster than Delete. You can have a where clause in Delete

statement where as Truncate statement cannot have a where clause. Truncate will delete all

https://www.softwaretestinghelp.com/wp-content/qa/uploads/2016/07/statements-supported-by-SQL.jpg

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the rows in a Table, but the structure of the table remains. Drop would delete all the rows

including the structure of the Table.

6) What is Cascading referential integrity constraint?

Cascading referential integrity constraints allow you to define the actions Microsoft SQL

Server should take when a user attempts to delete or update a key to which an existing

foreign keys point.

You can instruct SQL Server to do the following:

a. No Action: This is the default behaviour. No Action specifies that if an attempt is made to

delete or update a row with a key referenced by foreign keys in existing rows in other tables,

an error is raised and the DELETE or UPDATE is rolled back.

b. Cascade: Specifies that if an attempt is made to delete or update a row with a key

referenced by foreign keys in existing rows in other tables, all rows containing those foreign

keys are also deleted or updated.

c. Set NULL: Specifies that if an attempt is made to delete or update a row with a key


keys are set to NULL.

d. Set Default: Specifies that if an attempt is made to delete or update a row with a key


keys are set to default values.

6) Why do we use SQL constraints? Which constraints we can use while creating a

database in SQL?

Constraints are used to set the rules for all records in the table. If any constraints get violated

then it can abort the action that caused it.

Constraints are defined while creating the database itself with the CREATE TABLE

statement or even after the table is created once with the ALTER TABLE statement.

There are 5 major constraints are used in SQL, such as NOT NULL: That indicates that the column must have some value and cannot be left

NULL.

UNIQUE: This constraint is used to ensure that each row and column has a unique

value and no value is being repeated in any other row or column.

PRIMARY KEY: This constraint is used in association with NOT NULL and

UNIQUE constraints such as on one or the combination of more than one column to

identify the particular record with a unique identity.

FOREIGN KEY: It is used to ensure the referential integrity of data in the table. It

matches the value in one table with another using the PRIMARY KEY.

CHECK: It ensures whether the value in columns fulfills the specified condition.

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EXPERIMENT-2

Aim: Simple queries: selection, projection, sorting on a simple table, Small-large number of

attributes ,Distinct output values ,Renaming attributes , Computed attributes Simple-complex

conditions (AND, OR, NOT).

Program:

1. Display the unique department with jobs

select unique dno,jobs from emp;

Output:

DNO JOB

10 Manager

30 SalesMan

20 Clerk

10 Clerk

30 Manager

20 Manager

10 President

20 Analyst

30 Clerk

2. Display the details of the clerk.

select * from emp

Where empname=’Clerk’;

Output:

EMP

NO

EMPNA

ME JOB

MG

R

HIREDA

TE

SALA

RY

CO

M

DN

O

ANNUALINC

OME

7782 Clark Manag

er 7839

09-JUN-

81 2450 - 10 29400

3. List all the clerks.

select empname from emp

where job='Clerk';

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Output:

EMPNAME

Adams

Smith

James

Miller

4. List all the employees join on may 1st 1981.


where hiredate='01-May-1981';

Output:

EMPNAME

Blake

5. List the employees whose salaries are less than 3500


where salary < 3500;

Output:

EMPNAME

Ward

Adams

Turner

Blake

Clark

Ford

Jones

Scott

Smith

Allen

Martin

James

Miller

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6. List the empno,empname,salary of all employees joined before 1st april 81.

select empno,empname,salary from emp

where hiredate <'01-April=1981';

Output:

EMPNO EMPNAME SALARY

7521 Ward 1250

7. List the employees whose experience is more than 10 years.

select empno,hiredate from emp

where (months_between(sysdate,hiredate))/12>10;

Output:

EMPNO HIREDATE

7521 22-FEB-81

7876 13-AUG-81

7844 08-SEP-81

7839 17-NOV-81

7698 01-MAY-81

7782 09-JUN-81

7902 13-DEC-81

7566 02-APR-81

7788 13-JUL-81

7369 17-DEC-81

7499 20-DEC-81

7654 28-SEP-81

7900 13-DEC-81

7934 21-MAR-82

8. List the employees working as manager.

select empno,empname,job from emp

Where job = ‘Manager’;

Output:

EMPNO EMPNAME JOB

7698 Blake Manager

7782 Clark Manager

DBMS LAB

22

7566 Jones Manager

9. List the employes whose annual salary ranging from 23000 to 40000.

select empno,empname,salary from emp

where salary>23000 and salary<40000;

Output:

NO DATA FOUND

10. List the employees who joined in the year 1981 and who joined in month of January.

select empno,empname from emp

where hiredate>('01-Jan-81')and hiredate<('01-Jan-82');

Output:

EMPNO EMPNAME

7521 Ward

7876 Adams

7844 Turner

7839 King

7698 Blake

7782 Clark

7902 Ford

7566 Jones

7788 Scott

7369 Smith

7499 Allen

7654 Martin

7900 James

11. List the employees whose salary is 4 digit number.

select empname,salary from emp

where salary between 999 and 9999;

Output:

EMPNAME SALARY

Ward 1250

Adams 1100

Turner 1500

King 5000

DBMS LAB

23

Blake 2850

Clark 2450

Ford 3000

Jones 2975

Scott 3000

Allen 1600

Martin 1250

Miller 1300

12. List the empno,empname,location,deptname of all the depts. of 10 or 20.

select empno,empname,location,dname from emp,depts

where depts.deptno=emp.dno and (deptno=10 or deptno=20);

Output:

EMPNO EMPNAME LOCATION DNAME

7876 Adams Dallas Research

7839 King NewYork Accounting

7782 Clark NewYork Accounting

7902 Ford Dallas Research

7566 Jones Dallas Research

7788 Scott Dallas Research

7369 Smith Dallas Research

7934 Miller NewYork Accounting

13. List the mgr of department 10 or 20.

select dno,mgr from emp

where dno=10 or dno=20;

Output:

DNO MGR

20 7788

10 -

10 7839

20 7566

20 7839

20 7566

20 7902

DBMS LAB

24

10 7782

14. List the empno,empname,deptname of employees.

select empno,empname,dname

from emp,depts

where emp.dno=depts.deptno;

Output:

EMPNO EMPNAME DNAME

7521 Ward Sales

7876 Adams Research

7844 Turner Sales

7839 King Accounting

7698 Blake Sales

7782 Clark Accounting

7902 Ford Research

7566 Jones Research

7788 Scott Research

7369 Smith Research

7499 Allen Sales

7654 Martin Sales

7900 James Sales

7934 Miller Accounting

15.List the dname,location,empname from employee and department tables.

Select dname,location,empname

from emp,depts

Where depts..deptno=emp.dno;

Output:

DNAME LOCATION EMPNAME

Sales Chicago Ward

Research Dallas Adams

Sales Chicago Turner

Accounting NewYork King

Sales Chicago Blake

DBMS LAB

25

Accounting NewYork Clark

Research Dallas Ford

Research Dallas Jones

Research Dallas Scott

Research Dallas Smith

Sales Chicago Allen

Sales Chicago Martin

Sales Chicago James

Accounting NewYork Miller

16. Display all the information of the employee table.

select * from emp;

Output:

EMP

NO

EMPNA

ME JOB

MG

R

HIREDA

TE

SALA

RY

CO

M

DN

O

ANNUALINC

OME

7521 Ward SalesM

an 7698

22-FEB-

81 1250 500 30 21000

7876 Adams Clerk 7788 13-AUG-

81 1100 - 20 13200

7844 Turner SalesM

an 7698 08-SEP-81 1500 0 30 18000

7839 King Preside

nt -

17-NOV-

81 5000 - 10 60000

7698 Blake Manag

er 7839

01-MAY-

81 2850 - 30 34200

7782 Clark Manag

er 7839

09-JUN-

81 2450 - 10 29400

7902 Ford Analys

t 7566

13-DEC-

81 3000 - 20 36000

7566 Jones Manag

er 7839

02-APR-

81 2975 - 20 35700

7788 Scott Analys

t 7566 13-JUL-81 3000 - 20 36000

7369 Smith Clerk 7902 17-DEC-

81 800 - 20 9600

7499 Allen SalesM

an 7698

20-DEC-

81 1600 300 30 22800

7654 Martin SalesM

an 7698 28-SEP-81 1250 1400 30 31800

7900 James Clerk 7698 13-DEC-

81 950 - 30 11400

DBMS LAB

26

7934 Miller Clerk 7782 21-MAR-

82 1300 - 10 15600

17. Display unique jobs from employee table.

select unique job from emp;

Output:

JOB

Manager

Clerk

President

SalesMan

Analyst

18. List the employees who joined before 1981.

select empno,empname,hiredate from emp

where hiredate<('01-Jan-81');

Output:

EMPNO EMPNAME HIREDATE

7369 Smith 17-DEC-1980

19. List the empno,name,salary,experience of all employees working for mgr 7369.

select empno,empname,salary,(months_between(sysdate,hiredate))/12 from emp

where mgr =7369;

Output:

NO DATA FOUND

20. Display all the details of the employees whose comm is more than their salary.

select * from emp

where salary<com;

Output:

DBMS LAB

27

EMP

NO

EMPNA

ME JOB

MG

R

HIREDA

TE

SALA

RY

CO

M

DN

O

ANNUALINC

OME

7654 Martin SalesM

an 7698 28-SEP-81 1250 1400 30 31800

21.List theemployees along with their exp & daily salary is more than 100.

select empname,(months_between(sysdate,hiredate)/12) exp,(salary/30) DS

from emp where (salary/30)>100;

Output:

EMPNA

ME EXP DS

King 39.29704431750298685782556750298

68578256

166.6666666666666666666666666666

66666667

22. List the employees who are working for department 10 or 20.

select empname,dno from emp

where dno=10 or dno=20;

Output:

EMPNAME DNO

Adams 20

King 10

Clark 10

Ford 20

Jones 20

Scott 20

Smith 20

Miller 10

23. List employees who joined in month of august 1980.


where hiredate>=('01-Aug-1980') and hiredate<=('31-Aug-1980');

Output:

NO DATA FOUND

24. List employees whose annual salary ranging from 22000 to 45000.

DBMS LAB

28

select empname,annualincome from emp

where annualincome<=22000 or annualincome>=45000;

Output:

EMPNAME ANNUALINCOME

Ward 21000

Adams 13200

Turner 18000

King 60000

Smith 9600

James 11400

Miller 15600

25.List the employees who joined in january.

Select * from emp whereto_char(hiredate,'Mon')='Jan';

Output:

EMPN

O

EMPNAM

E

JO

B

MG

R

HIREDAT

E

SALAR

Y

COM

M

DN

O

7934 miller clerk 7782 23-JAN-82 1300 - 10

26. List the employees who does not belongs to deptno 20.

select empno,empname from emp where dno!=20;

Output:

EMPNO EMPNAME

7893 Tim

7499 allen

7900 james

7934 miller

7698 blake

7782 clerk

7654 martin

7521 ward

7844 turter

DBMS LAB

29

27. List employees who are working under manager.

select empno,empname from emp where job='manager';

Output:

EMPNO EMPNAME

7698 blake

7782 clerk

7566 jones

28. List all the clerks of deptno is 20

select empname from emp where empname = 'clerk' and dno=20;

Output:

no data found

29.List the employees whose deptno 30 or 10 joined in the year 1981.

selectempno,empname from emp

where to_char(hiredate,'YYYY')=1981 and dno = 10 or dno = 30;

Output:

EMPNO EMPNAME

7893 Tim

7499 allen

7900 james

7698 blake

7782 clerk

7654 martin

7521 ward

7844 turter

30. Display details of smith.

select * from emp where empname = 'smith';

Output:

EMPN

O

EMPNAM

E

JO

B

MG

R

HIREDAT

E

SALAR

Y

COM

M

DN

O

7369 smith clerk 7902 17-DEC-80 800 - 20

DBMS LAB

30

31.Display location of smith.

select empno,empname,job,mgr,hiredate,salary,comm,dno,location from emp,dept

where dept.deptno=emp.dno and empname='smith';

Output:

EMP

NO

EMPNA

ME

JO

B

MG

R

HIREDA

TE

SALA

RY

COM

M

DN

O

LOCATI

ON

7369 smith cler

k 7902

17-DEC-

80 800 - 20 dallas

32. Displayempname,job who are without manager.

selectempno,empname,job from emp where job!='manager';

Output:

EMPNO EMPNAME JOB

7893 Tim President

7369 smith clerk

7499 allen salesman

7876 adams clerk

7900 james clerk

7934 miller clerk

7788 scott analyst

7654 martin salesman

7521 ward salesman

7844 turter salesman

33. List the emp whose salaries are odd value.

selectempname,salary from emp where mod(salary,2)=1;

Output:

EMPNAME SALARY

jones 2795

34. List the emp whose salary contain 3 digitno.

selectempno,empname from emp where salary<999;

DBMS LAB

31

Output:

EMPNO EMPNAME

7369 smith

7900 james

35. List the amp who joined in the month of dec.

selectempno,empname from emp where to_char(hiredate,'Mon')='Dec';

Output:

EMPNO EMPNAME

7369 smith

7900 james

7788 scott

36. List the emp who joined before 15 th of month.

selectempname from emp where to_char(hiredate,'dd')='15';

Output:

no data found

37. Following output from emp ‘EMP-AND-JOB’ for emp and job

selectempname||'and'||job as "EMP_AND_JOB" from emp;

Output:

EMP_AND_JOB

Tim andPresident

smith andclerk

allenandsalesman

adamsandclerk

jamesandclerk

miller andclerk

blakeandmanager

clerk andmanager

scottandanalyst

DBMS LAB

32

martin andsalesman

jones andmanager

ward andsalesman

turterandsalesman

38. List the employees in date in format of 9-JUN-81

selectempname from emp where hiredate = '09-JUN-81';

Output:

EMPNAME

clerk

39. Procedure for the following output for empclark and adams

selectempname ||'c'||(job)||')' as "emp(job)" from emp where empname = 'clark' and empname

= 'adams';

Output:

no data found

Viva questions:

1. What is CLAUSE?

SQL clause is defined to limit the result set by providing condition to the query. This usually

filters some rows from the whole set of records.

Example – Query that has WHERE condition

Query that has HAVING condition.

2. What is an ALIAS command?

ALIAS name can be given to a table or column. This alias name can be referred in WHERE

clause to identify the table or column.

DBMS LAB

33

EXPERIMENT-3

Aim: Partial Matching operators (LIKE, %, _, *, ?) ,ASC-DESC ordering combinations

Checking for Nulls.

Programs:

1.Write a query to retrieve all employees who are joined during the 1987?

select empno,empname,hiredate

from emp

where hiredate like ‘%87’;

Output:

EMPNO EMPNAME HIREDATE

7788 SCOTT 13-JUL-1987

7876 ADAMS 13-JUL-1987

2.Find employee names start with A.

select empno,empname

from emp

where empname like 'A%';

Output:

EMPNO EMPNAME

7499 ALLEN

7876 ADAMS

3. Find employee names whose second letter start with A and after that any number of

letters.


from emp

where empname like '_A%';

DBMS LAB

34

Output:

EMPNO EMPNAME

7521 WARD

7654 MARTIN

7900 JAMES

4. Find the employee names and annual salary whose employee belongs to dept 10.

select empno,empname,deptno,salary*12 as annual sal

from emp

where deptno=10;

Output:

EMPNO EMPNAME DEPTNO

ANNUAL

SALARY

7839 KING 10 60000

7782 CLARK 10 29400

7934 MILLER 10 15600

5. Find employee names in ascending order whose deptno is 10.

select empno,empname,job,salary

from emp

where deptno=10

order by empname ASC;

Output:

EMPNO EMPNAME JOB SALARY

7782 CLARK MANAGER 2450

7839 KING PRESIDENT 5000

7934 MILLER CLERK 1300

6. Find employee salaries in descening order whose deptno=10.

DBMS LAB

35

select empno,empname,job,salary

from emp

where deptno=10

order by salary desc;

Output:

EMPNO EMPNAME SALARY DEPTNO

7839 king 5000 president

7782 clark 2450 manager

7934 miller 1300 clerk

7. Find employee details whose comm is null.

select empname,comm

from emp

where comm is null;

Output:

EMPNO COMM

KING -

BLAKE -

CLARK -

JONES -

SCOTT -

FROD -

SMITH -

ADAMS -

JAMES -

MILLER -

8. Find the details of employee name must starts with ‘a’ and 4th character must be ‘e’

and ends with ‘n’.


from emp

where empname like ‘A__e%n’;

Output:

DBMS LAB

36

EMPNO EMPNAME

7499 ALLEN

9. Find details of employee whose name ends with ‘th’.


from emp

where empname like ‘%th’;

Output:

EMPNO EMPNAME

7369 SMITH

10.Find the details of employee whose name can contain 3rd letter ‘i’.


from emp

where empname like ‘__i%’;

Output:

EMPNO EMPNAME

7369 SMITH

Viva Questions:

1. Which operator is used in query for pattern matching?

LIKE operator is used for pattern matching, and it can be used as -.

1. % - Matches zero or more characters.

2. _(Underscore) – Matching exactly one character.

2. What is Query evaluation engine?

It executes low-level instruction generated by compiler.

3. What is DDL Interpreter?

It interprets DDL statements and record them in tables containing metadata.

DBMS LAB

37

EXPERIMENT-4

Aim: Multi-table queries (JOIN OPERATIONS) ,Simple joins (no INNER JOIN) Aliasing

tables – Full/Partial name qualification ,Inner-joins (two and more (different) tables) ,Inner-

recursive-joins (joining to itself) ,Outer-joins (restrictions as part of the WHERE and ON

clauses) ,Using where & having clauses.

Program:

1.Find dept names of the employees

select e.empname,d,dname

from emp e,dept d

where e.dno=d.deptno;

Output:

EMPNAME DNAME

KING ACCOUNTING

BLAKE SALES

CLARK ACCOUNTING

JONES RESEARCH

SCOTT RESEARCH

FROD RESEARCH

SMITH RESEARCH

ALLEN SALES

WARD SALES

MARTIN SALES

TURNER SALES

ADAMS RESEARCH

JAMES SALES

MILLER ACCOUNTING

2. Find deptname of ALLEN.

select e.empname,d.dname

from emp e,dept d

where e.dno=d.deptno and empname=’allen’;

Output:

EMPNAME DNAME

ALLEN SALES

DBMS LAB

38

3.Find the names of employees and deptno and their location where employee working

under dname sales.

select e.empno,e.empname,e.deptno,d.dname,d.location

from emp e,dept d

where e.dno=d.deptno and dname=’sales’;

Output:

EMPNO EMPNAME DEPTNO LOCATION DNAME

7698 BLAKE 30 CHICAGO SALES

7499 ALLEN 30 CHICAGO SALES

7521 WARD 30 CHICAGO SALES

7654 MARTIN 30 CHICAGO SALES

7844 TURNER 30 CHICAGO SALES

7900 JAMES 30 CHICAGO SALES

4.Find the empno,empname,dname whose employees belongs to the same department.

select empno,empname,dname,emp.deptno

from emp inner join dept

on emp.deptno=dept.deptno;

Output:

EMPNO EMPNAME DNAME DEPTNO

7566 jones research 20

7369 smith research 20

7499 allen sales 30

7521 ward sales 30

7654 martin sales 30

7934 miller accounting 10

7698 blake sales 30

7782 clark accounting 10

7788 scott research 20

7844 turner sales 30

7900 james sales 30

7839 king accounting 10

DBMS LAB

39

7902 frod research 20

7876 adams research 20

5.Find the details of employees and dept whose dept numbers are not equals to each

other.

select empno,empname,job,emp.deptno,dname

from emp,dept

where emp.deptno!=dept.deptno;

Output:

EMPNO EMPNAME JOB DEPTNO DNAME

7499 allen salesman 30 research

7521 ward salesman 30 research

7654 martin salesman 30 research

7934 miller clerk 10 research

7698 blake manager 30 research

7782 clark manager 10 research

7844 turner salesman 30 research

7900 james clerk 30 research

7839 king president 10 research

7566 jones manager 20 operations

7369 smith clerk 20 operations

7499 allen salesman 30 operations

7521 ward salesman 30 operations

7654 martin salesman 30 operations

7934 miller clerk 10 operations

7698 blake manager 30 operations

7782 clark manager 10 operations

7788 scott analyst 20 operations

7844 turner salesman 30 operations

7900 james clerk 30 operations

7839 king president 10 operations

7902 frod analyst 20 operations

7876 adams clerk 20 operations

7566 jones manager 20 accounting

7369 smith clerk 20 accounting

7499 allen salesman 30 accounting

7521 ward salesman 30 accounting

7654 martin salesman 30 accounting

7698 blake manager 30 accounting

7788 scott analyst 20 accounting

DBMS LAB

40

7844 turner salesman 30 accounting

7900 james clerk 30 accounting

7902 frod analyst 20 accounting

7876 adams clerk 20 accounting

7566 jones manager 20 sales

7369 smith clerk 20 sales

7934 miller clerk 10 sales

7782 clark manager 10 sales

7788 scott analyst 20 sales

7839 king president 10 sales

7902 frod analyst 20 sales

7876 adams clerk 20 sales

6.Find empnames,salaries and grades of employees whose deptno is 10.

Select e.ename,e.salary,A.grade

From emp e,salgrade A

Where e.sal between lowsalary and highsalary and deptno=10;

Output:

EMPNAME SALARY GRADE

MILLER 1300 2

7.Find empnames of each employee manager.

select e.empname||'works for'||m.empname

from emp e,emp m

where e.mgr=m.empno;

Output:

E.EMPNAME||'WORKSFOR'||M.EMPNAME

frod works forjones

scott works forjones

james works forblake

turner works forblake

DBMS LAB

41

martin works forblake

ward works forblake

allen works forblake

miller works forclark

adams works forscott

clark works forking

blake works forking

jones works forking

smith works forfrod

8. Find the empno and deptname for the employees whose belons to deptno is 10 using

‘on’ clause.

select empno,dname,emp.deptno

from emp inner join dept

on emp.deptno=dept.deptno and emp.deptno=10;

Output:

EMPNO DNAME DEPTNO

7934 accounting 10

7782 accounting 10

7839 accounting 10

9. Display all employees and dname by using left outer join.


from emp left outer join dept

on emp.deptno=dept.deptno;

Output:

EMPNO EMPNAME DNAME

7876 adams Research

7902 frod Research

7788 scott Research

7369 smith Research

7566 jones Research

7839 king Accounting

7782 clark Accounting

7934 miller Accounting

DBMS LAB

42

7900 james Sales

7844 turner Sales

7698 blake Sales

7654 martin Sales

7521 ward Sales

7499 allen Sales

10. Display all employees and dname by using right outer join.


from emp right outer join dept

on dept.deptno=emp.deptno;

Output:

EMPNO EMPNAME DNAME

7566 jones Research

7369 smith Research

7499 allen Sales

7521 ward Sales

7654 martin Sales

7934 miller Accounting

7698 blake Sales

7782 clark Accounting

7788 scott Research

7844 turner Sales

7900 james Sales

7839 king Accounting

7902 frod Research

7876 adams Research

- - Operations

Viva Questions:

1. What is a join?

This is a keyword used to query data from more tables based on the relationship between the

fields of the tables. Keys play a major role when JOINs are used.

2. What are the types of join and explain each?

DBMS LAB

43

There are various types of join which can be used to retrieve data and it depends on the

relationship between tables.

Inner Join.

Inner join return rows when there is at least one match of rows between the tables.

Right Join.

Right join return rows which are common between the tables and all rows of Right hand side

table. Simply, it returns all the rows from the right hand side table even though there are no

matches in the left hand side table.

Left Join.

Left join return rows which are common between the tables and all rows of Left hand side

table. Simply, it returns all the rows from Left hand side table even though there are no

matches in the Right hand side table.

Full Join.

Full join return rows when there are matching rows in any one of the tables. This means, it

returns all the rows from the left hand side table and all the rows from the right hand side

table.

3. What is Self-Join?

Self-join is set to be query used to compare to itself. This is used to compare values in a

column with other values in the same column in the same table. ALIAS ES can be used for

the same table comparison.

4. What is Cross-Join?

Cross join defines as Cartesian product where number of rows in the first table multiplied by

number of rows in the second table. If suppose, WHERE clause is used in cross join then the

query will work like an INNER JOIN.

DBMS LAB

44

EXPERIMENT-5

Aim: Nested queries :In, Not In Exists, Not Exists Dynamic relations (as part of SELECT,

FROM, and WHERE clauses).

Program:

1.Find the Nth (second) highest salary drawing employee details.

select empname,salary

from emp e1

where 1=(select count(distinct salary)

from emp e2

where e1.salary<e2.salary);

Output:

EMPNAME SALARY

scott 3000

frod 3000

2.List all employees who work in Chicago.

select empname,deptno,job

from emp

where deptno=(select deptno

from dept

where location='chicago');

Output:

EMPNAME DEPTNO JOB

allen 30 Salesman

ward 30 Salesman

martin 30 Salesman

blake 30 Manager

turner 30 Salesman

james 30 Clerk

DBMS LAB

45

3. List out employees whose earning is above their manager’s salary.

select empno,empname,salary

from emp e1

where salary>(select salary

from emp e2

where e1.deptno=e2.deptno and e2.job='manager');

Output:

EMPNO EMPNAME SALARY

7788 scott 3000

7839 king 5000

7902 frod 3000

4. List employees with either same job as ‘Jones’ or the salary greater than that of

‘Ford’.

select empname,job

from emp

where job=(select job

from emp

where empname='jones' or salary>

(select salary

from emp

where empname='ford'));

Output:

EMPNAME JOB

Jones manager

Blake manager

Clark Manager

5. Display all the departments that have a minimum sal greater than that of department

20.

select deptno,min(salary)

DBMS LAB

46

from emp

group by deptno

having min(salary)>(select min(salary)

from emp

where deptno=20);

Output:

DEPTNO MIN(SALARY)

30 950

10 1300

6. Find the job with lowest average salary

select job,avg(salary)

from emp

group by job

having avg(salary)=(select min(avg(salary))

from emp

group by job);

Output:

JOB AVG(SALARY)

clerk 1037.5

7. Find the employees who earn the same salary as the minimum salary for each

department.

select empname,salary ,deptno

from emp

where salary in(select min(salary)

from emp

group by deptno);

Output:

DBMS LAB

47

EMPNAME SALARY DEPTNO

James 950 30

Smith 800 20

Miller 1300 10

8. Find the list of all employees who work for ‘Blakes’.

select empno,empname,job

from emp

where mgr=(select empno

from emp

where empname='blake');

Output:

EMPNO EMPNAME JOB

7499 allen salesman

7521 ward salesman

7654 martin Salesman

7844 turner Salesman

7900 james Clerk

9. Display all the employees who do not have any subordinates

select empname

from emp

where empno not in (select mgr

from emp

where mgr is not null);

Output:

EMPNAME

Turner

Ward

Martin

Allen

Miller

Smith

DBMS LAB

48

adams

james

10. Find departments which are not in emp table

select *

from dept

where deptno not in (select deptno

from emp);

Output:

DEPTNO DNAME LOCATION

40 operations boston

11. Display the details of employees who are managed by the same manager and work

in the same department as the employees with empno in 7499or 7900.

select empno,mgr,deptno

from emp

where(mgr,deptno) in (select mgr,deptno

from emp

where empno in(7499,7900));

Output:

EMPNO MGR DEPTNO

7900 7698 30

7844 7698 30

7654 7698 30

7521 7698 30

7499 7698 30

12. Find the employees details from emp table such that the salary of employee

is greater than the average salary of his own department

select empno,empname,job,deptno

from emp e1

DBMS LAB

49

where exists(select *

from emp e2

where e1.mgr=e2.empno);

Output:

EMPNO EMPNAME JOB DEPTNO

7566 jones manager 20

7369 smith clerk 20

7499 allen salesman 30

7521 ward salesman 30

7654 martin salesman 30

7934 miller clerk 10

7698 blake manager 30

7782 clark manager 10

7788 scott analyst 20

7844 turner salesman 30

7900 james clerk 30

7902 frod analyst 20

7876 adams clerk 20

13.Find all departments that do not have any employees.

select deptno,dname

from dept d

where not exists(select *

from emp e

where e.deptno=d.deptno);

Output:

DEPTNO DNAME

40 operations

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Viva Questions:

1. What is subquery?

A subquery is a query within another query. The outer query is called as main query, and

inner query is called subquery. SubQuery is always executed first, and the result of subquery

is passed on to the main query.

2. What are the types of subquery?

There are two types of subquery – Correlated and Non-Correlated.

A correlated subquery cannot be considered as independent query, but it can refer the column

in a table listed in the FROM the list of the main query.

A Non-Correlated sub query can be considered as independent query and the output of

subquery are substituted in the main query.

3. What is a nested table and how is it different from a normal table?

A nested table is a database collection object, which can be stored as a column in a table.

While creating a normal table, an entire nested table can be referenced in a single column.

Nested tables have only one column with no restriction of rows.

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EXPERIMENT-6

Aim: Set Oriented Operations :Union, Difference, Intersection, Division.

Program:

1. Using union display department numbers in emp and dept table.

select deptno

from emp

union

select dno

from dept;

Output

2.Find the common departments in employee and department table.

select deptno

from emp

intersect

select dno

from department;

Output :

3. Find the names of employees whose manager is 7698 and commission is null.

select ename

from emp

where mgr=7698

union

select ename

from emp

DNO

10

20

30

DNO

10

20

30

40

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where comm is null;

Output :

EMPNAME

Tim

adams

allen

blake

clerk

james

jones

martin

miller

scott

smith

turter

ward

4.Find all department in employee and department tables.

select deptno

from emp

unionall

select dno

from dept;

Output :

DNO

10

20

30

20

30

10

30

10

20

30

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20

30

30

30

40

20

10

5. Find the employees whose deptno is 20 but their salary is not greater than 2000

select ename,deptno

from emp

where deptno=20

minus

select ename,deptno

from emp

where sal<2000;

Output :

EMPNAME DNO

jones 20

scott 20

6. Using union matching select statement display deptno,location and hiredate of all

employees.

select dno,dloc,to_date(null) hiredate

from dept

union

select hiredate,deptno,to_char(null) dloc

from emp;

Output :

DEPTNO LOCATION HIREDATE

10 newyork -

10 - 09-JUN-81

10 - 17-NOV-81

10 - 23-JAN-82

20 dallas -

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20 - 17-DEC-80

20 - 12-APR-81

20 - 03-DEC-81

20 - 13-JUL-87

30 chicago -

30 - 20-FEB-81

30 - 22-FEB-81

30 - 01-MAY-81

30 - 08-SEP-81

30 - 28-SEP-81

30 - 13-DEC-81

40 boston -

7. Matching the select statement using union for the employee table.

select deptno,sal

from emp

union

select dno,0

from emp;

Output :

DNO SALARY

10 0

10 1300

10 2450

10 5000

20 0

20 800

20 1100

20 2795

20 3000

30 0

30 950

30 1250

30 1500

30 1600

30 2850

DBMS LAB

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Viva Questions:

1 ) Define UNION, MINUS, UNION ALL, INTERSECT?

MINUS – returns all distinct rows selected by the first query but not by the second.

UNION – returns all distinct rows selected by either query

UNION ALL – returns all rows selected by either query, including all duplicates.

INTERSECT – returns all distinct rows selected by both queries.

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EXPERIMENT-7

Aim: PL/SQL Programming I :Programs using named and unnamed blocks, Programs using

Cursors, Cursor loops and records.

Syntax of PL/SQL program:

Declare

Variable declaration;

Begin

Executable statements;

end;

PROGRAMS:

Unnamed blocks

1. Write a program to fetch salary from Emp table where empno is 7782

declare

num emp.empno%type;

salary emp.sal%type;

begin

select sal into salary from emp where empno=7782;

dbms_output.put_line('salary='||salary);

end ;

Output :

salary=2450

Statement processed.

2. Write a program using bind variable

variable g_sal NUMBER

begin

select sal into :g_sal from emp

where empno=7782;

end;

Output :

g_sal

----------

2450

3. Sum of salaries of all employees in specified department

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declare

sum_sal number(10,2);

dept_no number not null:=30;

begin

select sum(sal) into sum_sal

from emp where deptno=dept_no;

dbms_output.put_line('the sum salaryis'|| to_char(sum_sal));

end;

Output :

the sum salary is9400


4. Write a program to calculate maximum of 2 salaries,fetch the salaries of any 2

employees and find the maximum salary.

declare

sal1 emp.sal%type;

sal2 emp.sal%type;

begin

select sal into sal1 from emp where empno=7698;

select sal into sal2 from emp where empno=7521;

if sal1>sal2 then

dbms_output.put_line('max salary='||sal1);

else

dbms_output.put_line('max salary='||sal2);

end if;

end;

Output :

max salary=2850


5. Write a program to delete a row in a given table

declare

dept_no dept.deptno%type:=70;

begin

delete from dept where deptno=dept_no;

end;

Output :

1 row(s) deleted.

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6. Write a program using Case statement(like switch)

declare

grade char(1) := UPPER('&p_grade');

appraisal varchar2(20);

begin

appraisal:=case

when grade='A' then 'Excellent'

when grade='B' then 'very good'

when grade='C' then 'good'

else 'No such grade'

end;

dbms_output.put_line('Grade:'||grade||'Appraisal'||appraisal);

end;

Output :

Grade:A Appraisal Excellent


ii.Programs using Cursors,Cursor loops Records,Record type

7. Write a program to fetch data from one table and keep it in a record and assign the

fetched data to another record.

declare

type deptrec is record

(no dept.deptno%type,name dept.dname%type,local dept.loc%type);

cursor c1 is select * from dept;

d deptrec;

begin

for drec in c1 loop

d.no:=drec.deptno;

d.name:=drec.dname;

d.local:=drec.loc;

dbms_output.put_line('dno'||''||'dname'||''||'loc');

dbms_output.put_line(d.no||''||d.name||''||d.local);

end loop;

end;

Output :

dno dname location

30 sales chicago

dno dname location

40 operating boston

dno dname location

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20 research dallas

dno dname location



8. Write a program to calculate the sum of sal and commission and if commission is

having NULL values prevent it in calculation using Record Concept. Fetch the data

based on empno.

declare

type emp2 is record

(num emp.empno%type,salary emp.sal%type,co emp.comm%type);

em emp2;

net number;

begin

select empno,sal,comm into em from emp where empno=7499;

net:=em.salary+nvl(em.co,0);

dbms_output.put_line('empno'||' '||'esal'||' '||'net');

dbms_output.put_line(em.num||' '||em.salary||' '||net);

exception

when no_data_found then

dbms_output.put_line('invalid empno');

end;

Output :

empno esalary net

7499 1600 1900


9. Write a program to fetch the data from the department and store it in a PL/SQL

table and display it on the screen.

declare


type depttab is table of dept %rowtype

index by binary_integer;

dept_tab depttab;

i binary_integer:=0;

n number:=0;

begin

for deptrec in c1

loop

dept_tab(n):=deptrec;

n:=n+1;

end loop;

for i in 0.. n-1 loop

dbms_output.put_line(dept_tab(i).deptno||''||dept_tab(i).dname||''||dept_tab(i).loc);

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end loop;

end;

Output :

30 sales chicago

40 operating boston

20 research dallas


Explicit cursor

10.Write a program to fetch all the date from a table and display it on the screen using

%row type attribute.

declare


drec c1%rowtype;

begin

open c1;

dbms_output.put_line('no'||' '||'Name'||' '||'LOC');

loop

fetch c1 into drec;

exit when c1%notfound;

dbms_output.put_line(drec.deptno||' '||drec.dname||' '||drec.loc);

end loop;

close c1;

end;

Output :

no Name LOCATION

30 sales chicago

40 operating boston

20 research dallas


11. Write a program to display the details of employees in a specified deptno using

parameterized cursor.

declare

cursor c1(no in number) is select * from emp where deptno=no;

erec c1%rowtype;

begin

open c1(&no);

loop

fetch c1 into erec;


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dbms_output.put_line('no='||erec.empno);

dbms_output.put_line('name='||erec.ename);

dbms_output.put_line('sal='||erec.sal);

dbms_output.put_line('job='||erec.job);

end loop;

close c1;

end;

Output :

no=7893

name=Tim

salary=5000

job=President

no=7934

name=miller

salary=1300

job=clerk

no=7782

name=clerk

salary=2450

job=manager

12. Write a program to display deptno, empno, ename from emp table using %Found

attribute.

declare

cursor c1 is select empno,ename,deptno from emp;

no emp.empno%type;

name emp.ename%type;

dno emp.deptno%type;

begin

open c1;

loop

fetch c1 into no,name,dno;

if(c1%found) then

dbms_output.put_line('no='||NO||' '||'name='||NAME||' '||'dno='||DNO);

else

exit;

end if;

end loop;

close c1;

end;

Output :

no=7893 name=Tim dno=10

no=7369 name=smith dno=20

no=7499 name=allen dno=30

no=7876 name=adams dno=20

no=7900 name=james dno=30

no=7934 name=miller dno=10

no=7698 name=blake dno=30

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no=7782 name=clerk dno=10

no=7788 name=scott dno=20

no=7654 name=martin dno=30

no=7566 name=jones dno=20

no=7521 name=ward dno=30

no=7844 name=turter dno=30

13. Write a program to find the number of records of any given table using %rowcount.

declare

cursor c1 is select * from emp;

erec c1%rowtype;

begin

open c1;

loop

fetch c1 into erec;


end loop;

dbms_output.put_line('no of rows='||c1%rowcount);

close c1;

end;

Output :

no of rows=13


14. Write a program to check whether the cursor is opened or not. If cursor is opened

“Display cursor already opened” else open the cursor and display the message “Just

Opened the cursor.”

declare


begin

open c1;

if(c1%isopen) then

dbms_output.put_line('cursor is already open');

else

open c1;

dbms_output.put_line('cursor is just opened');

end if;

close c1;

end;

Ouput :

cursor is already open


Cursor loop

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15. Write a program to fetch all the data from salgrade table using cursor for loop.

declare

cursor c1 is select * from salgrade;

begin

for rec in c1 loop

dbms_output.put_line('grade'||' '||'hisal'||' '||'Losal');

dbms_output.put_line(rec.grade||' '||rec.hisal||' '||rec.losal);

end loop;

end;

Output :

grade highsalary Lowsalary

3 2000 1401


1 1200 700


4 3000 2001


2 1400 1201

Viva Questions:

1) What is a Cursor?

A database Cursor is a control which enables traversal over the rows or records in the table.

This can be viewed as a pointer to one row in a set of rows. Cursor is very much useful for

traversing such as retrieval, addition and removal of database records.

2) What is PL/SQL?

PL/SQL (Procedural Language/SQL) is basically a procedural extension of Oracle – SQL.

PL/SQL helps the user to develop complex database applications using control structures,

procedures, functions, modules, etc.

3) Explain the purpose of %TYPE and %ROWTYPE data types with the example?

PL/SQL uses the %TYPE declaration attribute for anchoring. This attribute provides the

datatype of a variable, constant or column. %TYPE attribute is useful while declaring a

variable that has the same datatype as a table column.

For example, the variable m_empno has the same data type and size as the column empno in

table emp.

m_empno emp.empno%TYPE;

%ROWTYPE attribute is used to declare a variable to be a record having the same structure

as a row in a table. The row is defined as a record and its fields have the same names and data

types as the columns in the table or view.

For example: dept_rec dept%ROWTYPE;

This declares a record that can store an entire row for the DEPT table.

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4 ) What do you understand by PL/SQL packages?

PL/SQL packages are schema objects that group functions, stored procedures, cursors and

variables at one place.

Packages have 2 mandatory parts:

Package Specifications

Package body

5 ) What do you understand by PL/SQL cursors?

PL/SQL requires a special capability to retrieve and process more than one row and that

resource is known as Cursors. A cursor is a pointer to the context area, which is an area of

memory containing SQL statements and information for processing the statements.

PL/SQL Cursor is basically a mechanism under which multiple rows of the data from the

database are selected and then each row is individually processed inside a program.

6) Explain cursor types.

There are two types of cursors.

They are explained as follows:

a) Explicit Cursors: For queries that return more than one row, an explicit cursor is declared

and named by a programmer. In order to use explicit cursor in PL/SQL, 4 steps are followed

Declare the cursor Syntax: CURSOR<cursor_name>is SELECT statement;

Here, <cursor_name> is the name assigned to the cursor and SELECT statement is the query

that returns rows to the cursor active set.

Open the cursor Syntax: OPEN <cursor_nam>;

Where, <cursor_name> is the name of the previously defined cursor.

Fetch rows from the cursor

Syntax: FETCH <cursor_name> INTO <record_list>;

Here, <cursor_name> refers to the name of the previously defined cursor from which rows

are being fetched.

<record_list> represents the list of variables that will receive the data being fetched.

Closing the cursor Syntax: CLOSE <cursor_name>;

Here, <cursor_name> is the name of the cursor being closed.

b) Implicit cursors: When any SQL statement is executed, PL/SQL automatically creates a

cursor without defining such cursors are known as implicit cursors.

For the following statements, PL/SQL employs implicit cursors

INSERT

UPDATE

DELETE

SELECT ( queries that return exactly one row)

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7) What are actual parameters and formal parameters?

The variables or an expression referred to as parameters that appear in the procedure call

statement is known as Actual parameters.

For example: raise_sal(emp_num, merit+ amount);

Here in the above example, emp_num and amount are the two actual parameters.

The variables that are declared in the procedure header and are referenced in the procedure

body are called as Formal parameters.

For example:

PROCEDURE raise_sal( emp_id INTEGER) IS

curr_sal REAL:

………..

BEGIN

SELECT sal INTO cur_sal FROM emp WHERE empno = emp_id;

…….

END raise_sal;

Here in the above example, emp_id acts as a formal parameter.

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EXPERIMENT-8

Aim: PL/SQL Programming II :Creating stored procedures, functions

Programs:

1. Write a procedure to display ename,sal when user inputs empno using in/out

parameters.

create or replace procedure empdisp(no emp.empno%type) is name char(10);

salary number;

begin

select ename,sal into name,salary from emp where empno=no;

dbms_output.put_line('ename='||name||' '||'esal='||salary);

exception


dbms_output.put_line('Invalid empno');

end empdisp;

Output:

Procedure created.

SQL> exec empdisp(7369);

ename=SMITH esal=800

PL/SQL procedure successfully completed.

2. Write a procedure to update sal of Emp table taking empno as in parameter

create or replace procedure empset(no emp.empno%type,fsal emp.sal%type)is

isal emp.sal%type;

usal emp.sal%type;

begin

select sal into isal from emp where empno=no;

dbms_output.put_line('isal='||isal);

update emp set sal=fsal where empno=no;

select sal into usal from emp where empno=no;

dbms_output.put_line('usal='||usal);

end empset;

Output:

Procedure created.

SQL> exec empset(7369,1000);

isal=1000

usal=1000


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3. Write a function to check the validity of emp number from the Emp table.

create or replace function empvalid(no emp.empno%type)

return varchar2 is

emprec emp%rowtype;

begin

select * into emprec from emp where empno=no;

loop

if(emprec.empno=no) then

return('valid number');

end if;

end loop;

exception


return('invalid number');

end empvalid;

Output:

Function created. SQL>select empvalid(7369) from dual;

EMPVALID(7369) -----------------

VALID NUMBER

SQL>select empvalid(7777) from dual;

EMPVALID(7777) -----------------

INVALID NUMBER

4. Write a function, which returns Deptno when user inputs dname.

create or replace function dno(name dept.dname%type)

return

number is

no number;

begin

select deptno into no from dept where dname=name;

return(no);

exception


return(-999);

end dno;

Output:

Function created. SQL> select dno('SALES') from dual;

DNO('SALES')

------------ 30

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5. Create a function with a name insert_emp which is used to insert a record into the

emp table.

create or replace function insert_emp(no emp.empno%type,name emp.ename%type,salary

emp.sal%type,des emp.job%type)

begin

insert into emp(empno,ename,sal,job)

values(no,name,salary,des);

end insert_emp;

Output:

Function created

Exceptions

6. Write a program using exception whenever given deptno is not exist then create a

new department in the dept table.

declare

dept_no number;

begin

select deptno into dept_no from dept where dname='Delhi';

exception


insert into dept values(80,'Principal','Delhi');

end;

/

Output:

7. Write a program to get the salary of an employee

declare

vempno emp.empno%type;

vsal emp.sal%type;

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begin

vempno := 7499;

select sal into vsal from emp where empno = vempno;

dbms_output.put_line(‘salary = ‘ || vsal);

exception


dbms_output.put_line(‘employee not found…’);

end;

Output:

salary = 1600


8. a)Check what happened when we input a 5 digit number.

b)Check what happens when we input employee number in single quotes.

c)Check what happens when we input alpha-numeric information.

d)Check what happens when there is a duplicate empno.

declare

vempno emp.empno%type;

vsal emp.sal%type;

begin

vempno := 7900;

select sal into vsal from emp where empno = vempno;

dbms_output.put_line(‘salary = ‘ || vsal);

exception


dbms_output.put_line(‘employee not found…’);

when others then

dbms_output.put_line(sqlcode || sqlcode);

end;

Output:

Salary = 950


9. Creation of Packages

create or replace package samplepak

is

procedure proc1( n number, n1 out number);

function fun1(n number) return number;

end;

/

Output:

Package created.

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9.1. Creation of package body

create or replace package body samplepak

is

procedure proc1(n number, n1 out number)

is

begin

n1 := n * 5;

end proc1;

function fun1(n number) return number

is

n1 number;

begin

n1 := n * 2;

return n1;

end fun1;

end samplepak;

Output:

Package Body created

SQL> VARIABLE N NUMBER

SQL> EXECUTE SAMPLEPAK.PROC1(5,:N)


SQL> PRINT N

N

----------

25

SQL> EXECUTE :N := SAMPLEPAK.FUN1(4)


SQL> PRINT N

N

----------

8

9.2.Program to test Polymorphism

create or replace package body polypack

is

procedure proc1(n number, n1 out number)

is

DBMS LAB

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begin

n1 := n * 5;

end proc1;

procedure proc1(x varchar2,y varchar2, z out varchar2)

is

begin

z := concat(x,y);

end proc1;

end polypack;

/

Output:

SQL> VARIABLE N NUMBER

SQL> VARIABLE ST VARCHAR2(100)

SQL> EXECUTE POLYPACK.PROC1(5,:N)


SQL> PRINT N

N

----------

25

SQL> EXECUTE POLYPACK.PROC1('RAVI',' KIRAN',:ST)


SQL> PRINT ST

ST

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

RAVI KIRAN

Viva Questions:

1. What is a stored procedure?

Stored Procedure is a function consists of many SQL statement to access the database system.

Several SQL statements are consolidated into a stored procedure and execute them whenever

and wherever required.

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2. Advantages and Disadvantages of Stored Procedure?

Stored procedure can be used as a modular programming – means create once, store and call

for several times whenever required. This supports faster execution instead of executing

multiple queries. This reduces network traffic and provides better security to the data.

Disadvantage is that it can be executed only in the Database and utilizes more memory in the

database server.

3. What is recursive stored procedure?

A stored procedure which calls by itself until it reaches some boundary condition. This

recursive function or procedure helps programmers to use the same set of code any number of

times.

4. What do you understand by Exception handling in PL/SQL?

When an error occurs in PL/SQL, the exception is raised. In other words, to handle undesired

situations where PL/SQL scripts terminated unexpectedly, an error handling code is included

in the program. In PL/SQL, all exception handling code is placed in an EXCEPTION section.

There are 3 types of EXCEPTION:

Predefined Exceptions: Common errors with predefined names.

Undefined Exceptions: Less common errors with no predefined names.

User-defined Exceptions: Do not cause runtime error but violate business rules.

5. Enlist some predefined exceptions?

Some of the predefined exceptions are: NO_DATA_FOUND: Single row SELECT statement where no data is returned.

TOO_MANY_ROWS: Single row SELECT statement where more than one rows

are returned.

INVALID_CURSOR: Illegal cursor operation occurred.

ZERO_DIVIDE: Attempted to divide by zero.

6. What are the parameters that we can pass through a stored procedure?

We can pass IN, OUT & INOUT parameters through a stored procedure and they should be

defined while declaring the procedure itself.

7. What are the packages in PL SQL?

A package is a group of related database objects like stored procs, functions, types, triggers,

cursors, etc. that are stored in the Oracle database. It is a kind of library of related objects

which can be accessed by multiple applications if permitted.

PL/SQL Package structure consists of 2 parts: package specification & package body.

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EXPERIMENT-9

Aim: Triggers and auditing triggers

Program:

1. To create table level or statement triggers

create table client_master(clientno number(10),name varchar2(10),bal_due number(10,2),

address varchar2(10),city varchar2(10));

Output:

Table created

insert into client_master values(1000,'abc',10000,'hhh','kkk');

Output:

1 row(s) inserted

insert into client_master values(2000,'def',20000,'iii','lll');

Output:

1 row(s) inserted

insert into client_master values(3000,'ghi',30000,'jjj','mmm');

Output:

1 row(s) inserted

create table audit_client(clientno number(10),name varchar2(10),bal_due

number(10,2),operation varchar2(10),userid varchar2(10),odate date);

Output:

Table created

BEFORE UPDATE:

1 create or replace trigger audit_Trail

2 before update on client_master

3 for each row

4 declare

5 oper varchar2(10);

6 clientno client_master.clientno%type;

7 name client_master.name%type;

8 bal_due client_master.bal_due%type;

9 begin

10 if updating then

11 oper:='update';

12 end if;

13 if deleting then

14 oper:='delete';

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15 end if;

16 clientno:=:old.clientno;

17 name:=:old.name;

18 bal_due:=:old.bal_due;

19 insert into audit_client values(clientno,name,bal_due,oper,user,sysdate);

20 end;

Output :

Trigger created.

select * from client_master;

Output:

CLIENTNO NAME BAL_DUE ADDRESS CITY

1000 abc 10000 hhh kkk

3000 ghi 30000 Jjj mmm

2000 def 20000 Iii lll

update client_master

set bal_due=bal_due+100

where clientno=1000;

Output:

1 row(s) updated.




Output:

1 row(s) updated.




Output:

1 row(s) updated.

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Output:



3000 ghi 30100 jjj mmm

2000 def 20100 iii lll

select * from audit_client;

Output:

CLIENTNO NAME BAL_DUE OPERATION USERID ODATE

1000 abc 10100 update ANONYMOUS 03-

MAR-21

2000 def 20100 update ANONYMOUS 03-

MAR-21

3000 ghi 30100 update ANONYMOUS 03-

MAR-21

AFTER DELETING :

create or replace trigger audit_Trail

after delete on client_master

for each row

declare

oper varchar2(10

clientno client_master.clientno%type;

name client_master.name%type;

bal_due client_master.bal_due%type;

begin

if updating then

oper:='update';

end if;

if deleting then

oper:='delete';

end if;

clientno:=:old.clientno;

name:=:old.name;

bal_due:=:old.bal_due;

insert into audit_client values(clientno,name,bal_due,oper,user,sysdate);

end;

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Output :

Trigger created.


Output :



3000 ghi 30100 Jjj mmm


delete client_master


Output:

1 row(s) deleted.

delete client_master


output :

1 row(s) deleted.

select * from audit_client;

outptut:

CLIENTNO NAME BAL_DUE OPERATION USERID ODATE

1000 abc 10000 update ANONYMOUS 23-FEB-

21

2000 def 20000 update ANONYMOUS 23-FEB-

21

3000 ghi 30000 update ANONYMOUS 23-FEB-

21

1000 abc 10100 delete ANONYMOUS 23-FEB-

21

3000 ghi 30300 delete ANONYMOUS 23-FEB-

21


Output:

DBMS LAB

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Viva Questions:

1. What is a Trigger?

The Trigger allows us to execute a batch of SQL code when table event occurs (INSERT,

UPDATE or DELETE commands are executed against a specific table).

2. Explain the difference in the execution of triggers and stored procedures?

A stored procedure is executed explicitly by issuing a procedure call statement from another

block via a procedure call with arguments.

The trigger is executed implicitly whenever any triggering event like the occurrence of DML

statements happens.

3. What is a trigger and what are its types?

A trigger is a stored program which is written in such a way that it gets executed

automatically when some event occurs. This event can be any DML or a DDL operation.

PL/SQL supports two types of triggers:

Row Level

Statement Level

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ADDITIONAL EXPERIMENTS

1.AIM: To Check whether a number is Armstrong or not using functions.

FUNCTION BODY:

set serveroutput on;

create or replace function armstrong(n number) return number is

r number(10);

a number(10);

b number(10);

c number(10);

begin

b:=0;

c:=n;

while(c>0)

loop

r:=c mod 10;

b:=b+(r*r*r);

c:=floor(c/10);

end loop;

return b;

end armstrong;

PROGRAM:

declare

n number(10);

m number(10);

o number(10);

begin

n:=&n;

o:=n;

m:=armstrong(n);

if(m=o) then

dbms_output.put_line('Given no is armstrong number');

else

dbms_output.put_line('Given no is not an armstrong

number');

end if;

end;

OUTPUT:

Enter value for n: 153

Given no is Armstrong number


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2. AIM: To Perform Banking Operations Using Procedures

TABLE:

create table acc(ac_no number(5), name varchar(20), balance number(6));

Table created.

insert into acc values(&ac_no,'&name',&balance);

select * from acc;

AC_NO NAME BALANCE

101 aaa 10000

102 bbb 6000

103 ccc 15000

104 ddd 5500

PROCEDURE DEFINITION:


create or replace procedure withdraww (ac_no1 in number, amount1 in number) is

begin

update acc set balance=balance - amount1 where ac_no=ac_no1;

end;

create or replace procedure depositt (ac_no1 in number, amount1 in number) is

begin

update acc set balance=balance + amount1 where ac_no=ac_no1;

end;

PROGRAM:


declare

choice number;

ac_no1 number(5);

amount number(5);

begin

ac_no1:= &ac_no1;

choice:= &choice;

amount:= &amount;

if choice=1 then

depositt(ac_no1,amount);

else

withdraww(ac_no1,amount);

end if;

end;

OUTPUT:

Enter value for ac_no1: 103

Enter value for choice: 1

Enter value for amount: 2000


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Enter value for ac_no1: 103

Enter value for choice: 2

Enter value for amount: 1000


select * from acc;

AC_NO NAME BALANCE

101 aaa 10000

102 bbb 6000

103 ccc 16000

104 ddd 5500

DATABASE MANAGEMENT SYSTEMS LABchalapathiengg.ac.in/cse lab manual/dbms2021.pdfDBMS LAB 3 8. ETHICS:...

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