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DHANALAKSHMI COLLEGE OF ENGINEERING

Tambaram, Chennai – 601 301

DEPARTMENT OF

COMPUTER SCIENCE AND ENGINEERING

CS6312 – DATABASE MANAGEMENT SYSTEM LABORATORY

III SEMESTER - R 2013

LABORATORY MANUAL

Name :

Register No. :

Section :

1 Format No.:DCE/Stud/LM/34/Issue:00/Revision:00

DHANALAKSHMI COLLEGE OF ENGINEERING

VISION

Dhanalakshmi College of Engineering is committed to provide highly disciplined, conscientious and

enterprising professionals conforming to global standards through value based quality education and training.

MISSION

● To provide competent technical manpower capable of meeting requirements of the industry

● To contribute to the promotion of Academic Excellence in pursuit of Technical Education at different levels

● To train the students to sell his brawn and brain to the highest bidder but to never put a price tag on heart

and soul

DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

VISION

To strive for acquiring, applying and imparting knowledge in Computer Science and Engineering through

quality education and to provide enthusiastic professionals with commitment

MISSION

● To educate the students with the state-of-art technologies to meet the growing challenges of the electronics

industry

● To carry out research through continuous interaction with research institutes and industry, on advances in

communication systems

● To provide the students with strong ground rules to facilitate them for systematic learning, innovation and

ethical practices


PROGRAMME EDUCATIONAL OBJECTIVES (PEOs)

1. Fundamentals

To impart students with fundamental knowledge in Mathematics, Science and fundamentals of engineering

that will would them to be successful professionals

2. Core Competence

To provide students with sound knowledge in engineering and experimental skills to identify complex

software problems in industry and to develop practical solution for them

3. Breadth

To provide relevant training and experience to bridge the gap between theory and practice this enables to

find solutions for real time problem in industry and organization and to design products requiring interdisciplinary

skills

4. Professionalism skills

To bestow students with adequate training and provide opportunities to work as team that will build up their

communication skills, individual leadership and supportive qualities and to develop them to adapt and work in

ever changing technologies

5. Lifelong Learning

To develop the ability of students to establish themselves as professionals in Computer Science and

Engineering and to create awareness about the need for lifelong learning and pursuing advanced degrees


PROGRAMME OUTCOMES (POs)

a) To apply basic knowledge of Mathematics, Science and Engineering fundamentals in Computer Science and

Engineering field

b) To design and conduct experiments as well as to analyze and interpret and apply the same in the career

c) To design and develop innovative and creative software applications

d) To understand a complex real world problems and develop an efficient practical solutions

e) To create, select and apply appropriate technique, resources, modern engineering and IT tools

f) To understand their roles as professionals and give the best to the society

g) To develop a system that will meet expected need with realistic constraints such as economic, environmental, social,

political, ethical, safe and sustainable

h) To communicate effectively and make others understand exactly what they are trying to convey in both verbal and

written forms

i) To work in a team as team member or a leader and make unique contributions and work with coordination

j) To engage lifelong learning and exhibit their technical skills

k) To develop and manage projects in multidisciplinary environments


CS6312 – DATABASE MANAGEMENT SYSTEMS LABORATORY

SYLLABUS

COURSE OBJECTIVES

1. Learn to create and use a database.

2. Be familiarized with a query language.

3. Have hands on experience on DDL Commands.

4. Have a good understanding of DML Commands and DCL commands.

5. Familiarize advanced SQL queries.

6. Be exposed to different applications.

LIST OF EXPERIMENTS:

1. Creation of a database and writing SQL queries to retrieve information from the database.

2. Performing Insertion, Deletion, Modifying, Altering, Updating and Viewing records.

3. Creation of Views, Synonyms, Sequence, Indexes, save point.

4. Creating an Employee database to set various constraints.

5. Creating relationship between the databases.

6. Study of PL/SQL block.

7. Write a PL/SQL block to satisfy some conditions by accepting input from the user.

8. Write a PL/SQL block that handles all types of exceptions.

9. Creation of Procedures.

10. Creation of database triggers and functions.

11. Mini project (Application Development using Oracle/ MySQL)

a) Inventory Control System

b) Material Requirement Processing

c) Hospital Management System

d) Railway Reservation System

e) Personal Information System

f) Web Based User Identification System

g) Timetable Management System

h) Hotel Management System


COURSE OUTCOMES

1. Apply data definition language.

2. Create table for real world scenarios.

3. Apply integrity constraints.

4. Manipulate the database using DML commands.

5. Frame Nested and Join Queries.

6. Develop a PL/SQL programs like procedures and functions.

7. Write a trigger.

8. Design and Develop a database.

9. Design a front end including forms and menus.

10. Develop a report using VB.

11. Design and implement a database schema for a given problem-domain.

12. Populate and query a database.

13. Create and maintain tables using PL/SQL.

14. Prepare reports.


CS6312 - DATABASE MANAGEMENT SYSTEMS LABORATORY

CONTENTS

Sl. No. Name of the Experiment Page No.

CYCLE 1 – EXPERIMENTS

1 DDL- Data Definition, Table Creation, Constraints 7

2 DML- Insert, Select Commands, Update & Delete Commands 10

3 TCL –Rollback, Commit commands 13

4 Writing and Practice of Simple Queries 15

5 Nested Queries & Queries Using Group By And Other Clauses 20

6 Problems Based On Views, Synonym, Sequence & Index 25

CYCLE 2 – EXPERIMENTS

7 Creating Database 29

8 High Level Programming Language Extension (PL/SQL Blocks) 34

9 PL/SQL Procedures 38

10 PL/SQL Functions 40

11 Triggers 44

12 Mini project – Payroll Processing System 49

CONTENT BEYOND SYLLABUS

13

SQL Join 52


Expt. No.: 1 DDL COMMANDS

Aim: To create, alter, truncate and drop a table using DDL commands

Software requirements: Front end: VB/VC ++/JAVA or Equivalent

Back end: Oracle / SQL / MySQL/ PostGress / DB2 or Equivalent

Hardware requirements:

Pentium Dual Core 1.9Ghz and 2GB RAM

DDL commands:

OVERVIEW:

CREATE - to create objects in the database

ALTER - alters the structure of the objects in the database.

DROP - delete objects from the database.

TRUNCATE - remove all records from a table, including all spaces allocated for the records are removed.

COMMENT - add comments to the data dictionary.

GRANT - gives user's access privileges to database.

REVOKE - withdraw access privileges given with the GRANT command.

SQL - CREATE TABLE

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

INPUT:

SQL> CREATE TABLE DEPARTMENT ( DEPTID VARCHAR(5) CONSTRAINT PKEY PRIMARY KEY,

DEPTNAME VARCHAR(15), DOS DATE CONSTRAINT NOTNULL );

OUTPUT: Table created.

SQL> CREATE TABLE EMPLOYE ( EMPID VARCHAR(5) CONSTRAINT PKEY PRIMARY KEY, EMPNAME

VARCHAR(15), JOB CHAR(10) CONSTRAINT UNIQ1 UNIQUE, MGRID VARCHAR(5)

CONSTRAINT FKEY1 REFERENCES EMP (EMPID), HIREDATE DATE, DEPTID VARCHAR(5) CONSTRAINT

FKEY2 REFERENCES DEPT(DEPTID), SALARY NUMBER(7,2));

OUTPUT: Table created.


SQL - DESC TABLE

SQL>desc Employee;

SQL - TRUNCATE TABLE

SQL>ALTER TABLE EMPLOYE ADD CONSTRAINT NTNLL NOT NULL (SALARY);

OUTPUT: Table Altered.

Similarly, ALTER TABLE EMPLOYE DROP CONSTRAINT UNIQ1;

SQL - TRUNCATE TABLE

TRUNCATE TRUNCATE TABLE <TABLE NAME>;

Result:

Thus the table was successfully created, altered, truncated and dropped successfully.

Outcome:

Thus the outcome of applying Data Definition Language has been attained.


Viva-voce

1. What is database?

2. What is DBMS?

3. What is a database system?

4. What are the advantages of DBMS?

5. What is the disadvantage in File Processing System?

6. Describe the three levels of data abstraction?

7. What is data independence?

8. What is an entity?

9. What is a key?

10. What are the categories of SQL command?

11. Define DDL.


Expt. No.: 2 DML COMMANDS

Aim:

To insert, select, update and delete records in a table using DML commands

Software requirements:

Front end: VB/VC ++/JAVA or Equivalent




DML Commands:

COMMANDS OVERVIEW:

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

SQL – SELECT FROM

Selecting all records from a table SELECT * FROM EMPLOYE

Listing the selected records which satisfies the given condition from a table

SELECT * FROM EMPLOYE WHERE DEPTID=’DPT01’

SQL - INSERT INTO

Syntax:

INSERT INTO tablename VALUES (value list)

Single-row insert

INSERT INTO DEPARTMENT VALUES (‘DPT02’,’TESTING’,’10-JAN-2012’)

Inserting one row, with values to the column in the given order

INSERT INTO DEPARTMENT (DEPTNAME, DEPTID) VALUES (‘DESIGN’, ‘DPT02’);

Inserting one or more rows with values from other tables.

INSERT INTO DEPARTMENT (DEPTID, DEPTNAME)

SELECT EMPID, EMPNAME FROM EMP WHERE MGRID=EMPID; OTHER


EXAMPLES:

INPUT:

SQL>INSERT INTO EMPLOYE VALUES (‘Emp121’,’Gayle’,’Testing’,’Emp01’,’01-Feb-2013’, ’Dpt01’, 32000);

OUTPUT: 1 row created.

INPUT:

SQL>INSERT INTO DEPARTMENT VALUES (‘&deptid,’&deptname’); Enter value for deptid: Dpt01

Enter value for deptname: Development

OUTPUT:

old 1: Insert into department values(‘&deptid,’&deptname’) new 1: Insert into department values(‘Dpt01’,'

Development’) 1 row created.

SQL - UPDATE

Syntax: UPDATE tablename SET COLUMN_NAME =value [ WHERE CONDITION]

INPUT::

SQL> UPDATE DEPT SET DOS=’01-JAN-2010’ WHERE DEPTID=’DPT01’;

1 row updated.

SQL - DELETE FROM

Syntax: DELETE FROM tablename WHERE condition

INPUT::

SQL> DELETE FROM DEPARTMENT;

OUTPUT: 1 row deleted.

Result:

Thus the table was successfully inserted, updated, select, and deleted.

Outcome:

Thus the outcome of applying DDL Commands has been attained.


Viva-voce

1. What is the role of DML compiler in the database architecture?

2. Is it possible to delete a column using delete command?

3. Can update command can be used to change the structure of the table.

4. How to add a new constraint to a table using update command?

5. What is DML (Data Manipulation Language)?

6. What is DML Compiler?

7. Define TCL

8. Define Cursor

9. What are the privileges that can be granted on a table by a user to others?

10. What command is used to get back the privileges offered by the GRANT command?

11. What command is used to create a table by copying the structure of another table?


Expt. No.: 3 TCL COMMANDS

Aim:

To create save point, commit and rollback while using a transaction






TCL commands:

COMMANDS OVERVIEW

COMMIT - save work done & it is visible to other users

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

SAVEPOINT

Syntax: savepoint username;

INPUT:

SQL> savepoint emp;

Output:

savepoint created;

ROLLBACK

Syntax:

rollback to savepoint_text_identifier;

INPUT:

SQL> rollback to emp;

Output:

Rollback complete.


COMMIT Syntax: commit; INPUT:

SQL> commit;

Output:

Commit completed.

Result:

Thus the table was successfully saved, committed and rollback.

Outcome:

Thus the outcome of applying TCL Commands has been attained.

Viva-voce

1. What is savepoint in Oracle?

2. Write the syntax for savepoint.

3. What are the TCL commands?

4. What is VDL (View Definition Language)?

5. What is SDL (Storage Definition Language)?

6. What is Data Storage?

7. What is Functional dependency?

8. What is DCL?

9. What is transaction control language?

10. What do you mean by oracle language?


ENO

7369

ENAME

SMITH

JOB

CLERK

MGR

7902

HIREDATE

17-DEC-10

SAL

8000

COMM

0

DEPTNO

20

AGE

25

7499 ALLEN SALESMAN 7698 20-FEB-11 16000 300 30 25

7521 WARD SALESMAN 7698 22-FEB-11 12500 500 30 25

7566 JONES MANAGER 7839 02-APR-11 29705 500 20 25

7698 BLAKE MANAGER 7839 01-MAY-12 28500 1400 30 25

Expt. No.: 4 WRITING AND PRACTICE OF SIMPLE QUERIES

Aim:

To write simple SQL queries and to practice them

Software Requirements:



Hardware Requirements:

Standalone desktops 30 Numbers

Problem Statements:

List all employee details SQL>SELECT * FROM EMP;

OUTPUT:

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;

OUTPUT: ENAME

CLARK

rows selected.

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);

OUTPUT:

ENAME ------- SCOTT


FORD

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

INPUT: SQL> SELECT ENAME FROM EMP WHERE ENAME LIKE ‘J%’ OR ENAME LIKE ‘T%’; OUTPUT:

ENAME

--------- JONES

TURNER

JAMES

2. 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%’;

OUTPUT:

ENAME JOB

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

JONE MANAGER

BLAKE MANAGER

CLARK MANAGER

KING PRESIDENT

3. List all jobs available in employee table. INPUT:

SQL> SELECT DISTINCT JOB FROM EMP;

OUTPUT:

JOB

---------

ANALYST

CLERK

MANAGER

PRESIDENT

SALESMAN

ASSISTANT

CLERK

7 rows selected.

4. What is the difference between maximum and minimum salaries of employees in the organization? INPUT: SQL>

SELECT MAX(SAL)-MIN(SAL) FROM EMP;


OUTPUT:

MAX(SAL)-MIN(SAL)

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

4997

5. 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);

OUTPUT:

ENAME SAL

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

JONES 2975

BLAKE 2850

CLARK 2450

6. Find how much amount the company is spending towards salaries. INPUT: SQL> SELECT SUM (SAL) FROM

EMP;

OUTPUT: SUM(SAL)

---------

32928

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

INPUT: SQL>SELECT ENAME FROM EMP WHERE DEPTNO = 20;

OUTPUT :

ENAME

--------- SMITH JONES

SCOTT ADAMS

8. List ename whose commission is NULL.

INPUT: SQL> SELECT ENAME FROM EMP WHERE COMM IS NULL; OUTPUT:

ENAME

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

CLARK SCOTT KING ADAMS JAMES FORD

6 rows selected.

9. Find no.of dept in employee table.


INPUT: SQL> SELECT COUNT (DISTINCT ENAME) FROM EMP;

OUTPUT:

COUNT(DISTINCT ENAME)

--------------------------------------- 17

10. List ename whose manager is not NULL.

INPUT:

SQL> SELECT ENAME FROM EMP WHERE MGR IS NOT NULL; OUTPUT: ENAME

---------- SMITH ALLEN WARD JONES MARTIN

5 rows selected.

11. Calculate the experience in years of each programmer and display along with their 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;

12. In which month most number of programmers has 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’);

Result:

Thus the given problem statements were solved using simple queries.

Outcome:

Thus the outcome for creating table for real world scenarios has been attained


Viva-voce

1. Which command used to display all details from employee?

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

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

4. Which date function is used to find the difference between two dates?

5. What is the use of DESC in SQL?

6. What do you mean by where clause?

7. Which of the SQL command can be used to modify existing data in a database table?

8. What is E-R model?

9. What is a checkpoint and when does it occur?

10. List any two advantages of view.


Expt. No.: 5 NESTED QUERIES & QUERIES USING GROUP BY

AND OTHER CLAUSES

Aim:

To write queries using Set operations, to write nested queries and also to write queries using clauses such as

GROUP BY, ORDER BY, etc. and retrieving information by joining tables






Set Operations & Other Clauses:

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

UNION - OR

INTERSECT - AND

EXCEPT - NOT

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.

Problem Statements:

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);

OUTPUT:

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));

OUTPUT:

PNAME SAL SPLACE

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

MARY 4500S BHARI

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;

7. Display total salary spent for each job category.

INPUT: SQL>SELECT JOB,SUM (SAL) FROM EMP GROUP BY JOB; OUTPUT:

JOB SUM(SAL)

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

CLERK 23050

MANAGER 8275

PRESIDENT 5000

SALESMAN 5600

7 rows selected.


8. 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);

OUTPUT:

ENAME SAL

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

CHAI 3

JAMES 950

MILLER 1000

ADAMS 1100

russel 2200

5 rows selected.

9. 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;

OUTPUT:

DNAME COUNT(ENAME)

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

ACCOUNTING 3

RESEARCH 5

SALES 9

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

INPUT: SQL> SELECT PNAME, SUM(SCOST) FROM SOFTWARE GROUP BY PNAME; OUTPUT:

PNAMESUM(SCOST)

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

john 12000

kamala 12000

raju 12333

3 rows selected.

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

INPUT: SQL>SELECT PNAME, COUNT(TITLE) FROM SOFTWARE GROUP BY PNAME;

12. 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;

13. Display each institute name with number of students.

INPUT: SQL> SELECT SPLACE, COUNT(PNAME) FROM STUDY GROUP BY SPLACE;

OUTPUT:

SPLACE COUNT(PNAME)

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

BDPS 2

BITS 1

BNRILLIANI 1

COIT 1

HYD 1

5 rows selected.

14. 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);

999

OUTPUT: SOLD

--------- 11

15.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’);

OUTPUT:

no rows selected

Result:

Thus the given problem statements were solved using nested queries and group by & other clause.

Outcome:

Thus the outcome of Framing and Joining Nested Queries has been attained.


Viva-voce

1. What is the purpose of Having clause?

2. What is format of Date?

3. What is the purpose of ROWNUM?

4. Differentiate Nested Loop from Merge Join.

5. How are the OUTER JOIN operations different from JOIN operations?

6. How is the OUTER UNION operation different from UNION?

7. List the types of privileges available in SQL.

8. What are the commands available for modifying the database in SQL?

9. What is view in SQL? How is it defined?

10. List the SQL domain Types.

11. What are aggregate functions?

12. What is the use of group by clause?

13. What is the use of sub queries?


Expt. No.: 6 PROBLEMS BASED ON VIEWS, SYNONYM,

SEQUENCE & INDEX

Aim:

To solve the problem statements based on views, synonym, sequence and Indexes

Software Requirements:



Hardware Requirements: Pentium Dual Core 1.9Ghz and 2GB RAM

Overview:

VIEW:

A view is simply the representation of a SQL statement that is stored in memory so that it can easily be re-used.

It’s also referred as logical table.

SYNTAX:

CREATE OR REPLACE VIEW <view name > AS < select statement >

SYNONYM:

A synonym is an alias or alternate name for a table, view, sequence, or other schema object. They are used

mainly to make it easy for users to access database objects owned by other users.

SYNTAX:

CREATE OR REPLACE SYNONYM <synonym_name> FOR <object_name>

SEQUENCE:

A sequence is a database object from which multiple users may generate unique integers. User can use

sequences to automatically generate primary key values.

INDEXES:

Database system uses indexes to avoid the need for large-table, full-table scans and disk sorts, which are

required when the SQL optimizer cannot find an efficient way to service the SQL query.


Problem Statements:

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);

OUTPUT:

1 row created;

INPUT: SQL>UPDATE VIEW1 SET SALARY =50000 WHERE PNAME LIKE ‘SARVAN’; OUTPUT:

1 row updated.

Note: update command does not works for all queries on views. INPUT: SQL>DELETE FROM VIEW1 WHERE

PNAME LIKE ‘SMITH’;

OUTPUT:

1 row deleted.

6. Drop views which you generated. INPUT: SQL>DROP VIEW VIEW1;

OUTPUT: View dropped.

7. Create a synonym for a table created by other user.

SQL> CREATE SYNONYM EMP FOR SP.EMPLOYE;

8. Try to perform all DML operations through the created synonym.

SQL> INSERT INTO EMP VALUES (‘EMP10’,’LARA’,’MAINTENANCE’,’27-OCT-1998’);

OUTPUT:

1 row created.

SQL> UPDATE EMP SET EMPNAME=’BRAIN LARA’ WHERE EMPID=’EMP10’; OUTPUT:

1 row Updated.

9. Create a sequence to generate unique value for empid field in the employee table while inserting. SQL> CREATE

SEQUENCE EMP_SEQ START WITH 1000;


INCREMENT BY 1 NOCACHE NOCYCLE;

Sequence Created.

SQL> INSERT INTO EMP VALUES (EMP_SEQ.NEXTVAL,’SAHA’,’TESTING’,’27-DEC-1999’); OUTPUT:

1. Declarations: This section starts with the keyword DECLARE. It is an optional section and defines all variables,

cursors, subprograms, and other elements to be used in the program.

2. Executable Commands: This section is enclosed between the keywords BEGIN and END and it is a mandatory

section. It consists of the executable PL/SQL statements of the program. It should have at least one executable line

of code, which may be just a NULL command to indicate that nothing should be executed.

3. Exception Handling: This section starts with the keyword EXCEPTION. This section is again optional and

contains exception(s) that handle errors in the program.Every PL/SQL statement ends with a semicolon (;). PL/SQL

blocks can be nested within other PL/SQL blocks using BEGIN and END.

Result:

Thus the given problem statements based on views, synonyms, sequences and indexes were solved.

Outcome:

Thus the outcome for developing a PL/SQL programs based on views, synonyms, sequences and indexes has

been attained.


Viva-voce

1. What is view?

2. List out the types of views.

3. How many types of indexes were available in Oracle?

4. What are the four broad categories of constraints?

5. What is meant by Cost Estimation?

6. List the SQL statements used for transaction control.

7. With an example explain referential integrity.

8. What are the advantages of referential integrity constraints?

9. What is sequence?

10. What is a synonym in Oracle?


Expt. No.: 7 CREATING DATABASE

Aim:

To create database and also to create relationship among database






Command Overview:

SYNTAX:

CREATE DATABASE [ database

{ USER SYS IDENTIFIED BY password

| USER SYSTEM IDENTIFIED BY password

| CONTROLFILE REUSE

| MAXDATAFILES integer

| MAXINSTANCES integer

| CHARACTER SET charset

| NATIONAL CHARACTER SET charset

| SET DEFAULT

{ BIGFILE | SMALLFILE } TABLESPACE

| database_logging_clauses

| tablespace_clauses

| set_time_zone_clause

};


Problem Statement:

1. Create a new database by the name student_db with required features.

SQL> CREATE DATABASE STUDENT_DB USER SYS IDENTIFIED BY ORACLE USER SYSTEM IDENTIFIED

BY ORACLE

DATAFILE ‘C:\oracle\oradata\ORA11\SYSTEM01.DBF’ SIZE 325M REUSE AUTOEXTEND ON NEXT 10240K

MAXSIZE UNLIMITED

SYSAUX DATAFILE ‘C:\oracle\oradata\ORA11\SYSAUX01.DAT’ SIZE 120M REUSE AUTOEXTEND ON NEXT

5M MAXSIZE 2048M

DEFAULT TABLESPACE USERS DATAFILE ‘C:\oracle\oradata\ORA11\USERS01.DBF’ SIZE 50M REUSE

AUTOEXTEND ON MAXSIZE UNLIMITED;

Result:

Thus the database was created with the specified parameters.

Outcome:

Thus the outcome for developing and designing a database has been attained.


Viva-voce

1. What is Data?

2. How to know the current database we are working with?

3. How to create table space?

4. What are the types of storage devices?

5. Write the syntax for database creation.

6. What is the "integrity rules"?

7. What is extension and intension?

8. What is an Entity?


Expt. No.: 8 HIGH LEVEL PROGRAMMING LANGUAGE

EXTENSION (PL/SQL BLOCKS)

Aim:

To solve the given problem statements using PL/SQL blocks






PL/SQL block:




section. It consists of the executable PL/SQL statements of the program. It should have at least one executable line

of code, which may be just a NULL command to indicate that nothing should be executed.


contains exception(s) that handle errors in the program.Every PL/SQL statement ends with a semicolon (;). PL/SQL

blocks can be nested within other PL/SQL blocks using BEGIN and END.

Problem Statements:

REFERED TABLES:

Employe (emp_id,emp_name,dept_id,basic,hra,da,pf,net);

1. Create a PL/SQL block for inserting values in the Employee table. Only emp_id, emp_name, department & basic

should be received as input while executing the block and for the rest of the fields the values need to be calculated

as given below.

Calculations: HRA=50% OF BASIC DA=20% OF BASIC PF=7% OF BASIC

NETPAY=BASIC+DA+HRA-PF

INPUT:


DECLARE

ENO1 employe.emp_id%type; ENAME1 employe.emp_name%type; DEPTNO1 employe.dept_id%type; BASIC1

employe.basic%type;

HRA1 employe.HRA%type; DA1 employe.DA%type; PF1 employe.pf%type; NETPAY1

employe.net%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 EMPLOYE VALUES (ENO1, ENAME1, DEPTNO1, BASIC1, HRA1, DA1, PF1,NETPAY1); END;

OUTPUT:

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;

OUTPUT:

EMP_ID

EMP_ NAME

DEPT_ID

BASIC

HRA

DA

PF

NETPAY

101 GAYLE 10 5000 2500 1000 350 8150

102 SARVAN 20 5000 2500 1000 350 8150

103 POINTING 20 5500 2750 1100 385 8965

104 SMITH 10 6000 3000 1200 420 9780

105 ALIS 10 6000 3000 1200 420 9780

2. Create a PL/SQL block for updating records in Employe table where the user should provide the emp_id and the

new basic salary and thus the HRA,DA, PF and NETPAY should get calculated and updated accordingly.

INPUT: DECLARE

ENO1 employe.emp_id%type;

BASIC1 employe.basic%type; HRA1 employe.HRA%type; DA1 employe.DA%type;

PF1 employe.pf%type; NETPAY1 employe.net%type;

BEGIN ENO1:=&ENO1; BASIC1:=&BASIC1;

HRA1:=(BASIC1*50)/100; DA1:=(BASIC1*20)/100; PF1:=(BASIC1*7)/100; NETPAY1:=BASIC1+HRA1+DA1-PF1;

UPDATE EMPLOYE SET BASIC=BASIC1, HRA=HRA1, DA=DA1, PF=PF1, NETPAY=NETPAY1

WHERE EMP_ID=ENO1;

END; OUTPUT: SQL>/

Enter value for eno1: 105 old 11: ENO1:=&ENO1;

new 11: ENO1:=105;

Enter value for basic1: 8000 old 14: BASIC1:=&BASIC1;

new 14: BASIC1:=8000;


Result:

Thus PL/SQL statements were written for inserting & updating the rows in the EMPLOYE table.

Outcome:

Thus the outcome for developing a PL / SQL programs has been attained.


Viva-voce

1. What is PL SQL?

2. What are the uses of cursor?

3. What are the two types of exceptions?

4. What is Raise_application_error?

5. What is function in PL/SQL block?

6. Explain two virtual tables available at the time of database trigger execution.

7. What are the rules to be applied to NULLs whilst doing comparisons?

8. How is a process of PL /SQL compiled?

9. Define Implicit and Explicit Cursors

10. When is a declare statement required?

11. What is the importance of SQLCODE and SQLERRM?


Expt. No.: 9 PROCEDURES

Aim:

To solve the given problem statements using procedures






PL/SQL procedures:

An Oracle stored procedure is a program stored in an Oracle database.

The following are the advantages of using procedures.

Better performance: Oracle stored procedures load once into the shared pool and remain there unless they

become paged out. Subsequent executions of the Oracle stored procedure are far faster than executions of

external code.

Coupling of data with behavior: DBAs can use naming conventions to couple relational tables with the behaviors

associated with a table by using Oracle stored procedures as "methods". If all behaviors associated with the

employee table are prefixed with the table name--employee. Hire, employee.give_raise, for example--the data

dictionary can be queries to list all behaviors associated with a table (select * from dba_objects where owner =

'EMPLOYEE'), and it's easy to identify and reuse code via stored procedures.

Isolation of code: Since all SQL is moved out of the external programs and into the Oracle stored procedures, the

application programs become nothing more than calls to Oracle stored procedures. As such, it becomes v e r y

simple to swap out one database and swap in another one.

A view is simply the representation of a SQL statement that is stored in memory so that it can easily be re-used.

Its also referred as logical table.





section. It consists of the executable PL/SQL statements of the program. It should have at least one executable

line of code, which may be just a NULL command to indicate that nothing should be executed.


contains exception(s) that handle errors in the program.Every PL/SQL statement ends with a semicolon (;).

PL/SQL blocks can be nested within other PL/SQL blocks using BEGIN and END.

SYNTAX:

CREATE OR REPLACE VIEW <view name > AS < select statement >

SYNONYM:

A synonym is an alias or alternate name for a table, view, sequence, or other schema object. They are

used mainly to make it easy for users to access database objects owned by other users.

SYNTAX:

CREATE OR REPLACE SYNONYM <synonym_name> FOR <object_name>

SEQUENCE:

A sequence is a database object from which multiple users may generate unique integers. User can

use sequences to automatically generate primary key values.

INDEXES:

Database system uses indexes to avoid the need for large-table, full-table scans and disk sorts, which

are required when the SQL optimizer cannot find an efficient way to service the SQL query.

SYNTAX:

CREATE [OR REPLACE] PROCEDURE procedure_name (parameters list) IS

<declaration_section>

BEGIN

<executable_section>

EXCEPTION

<exception_section

> END;


Problem Statements:

REFERRED TABLES:

Borrow(acc_no , rollno, date_issue);

1.Write a procedure to insert a record in borrower relation. Before inserting check whether the book is available or

not.

CREATE OR REPLACE PROCEDURE PROC_BORROW(ACCNO NUMBER, ROLL VARCHAR, DOI DATE) IS

CNT NUMBER(5); BEGIN

SELECT COUNT(*) INTO CNT FROM BORROW WHERE ACC_NO=ACCNO;

IF(CNT=0) THEN

INSERT INTO BORROW VALUES (ACCNO,ROLL,DOI); ELSE

DBMS_OUTPUT.PUT_LINE(‘BOOK NOT AVAILABLE’); END IF;

END;

OUTPUT:

SQL> @ e:\proc.sql; Procedure Created.

SQL> exec pro1(‘123’,’CS01’,’27-OCT-2013’);

Procedure successfully completed. SQL> SELECT * FROM BORROW;

Result:

Thus PL/SQL Procedures were created to solve the given problem statements.

Outcome:

Thus the outcome for developing a PL / SQL program using Procedures has been attained.


Viva-voce

1. What is a stored procedure?

2. What is a stand-alone procedure?

3. What are the modes of parameters that can be passed to a procedure?

4. What are advantages of Stored Procedures?

5. What operators deal with NULL?

6. Does SQL*Plus also have a PL/SQL Engine?

7. What packages are available to PL SQL developers?

8. Explain 3 basic parts of a trigger.

9. What are character functions?

10. Show the cursor attributes of PL/SQL.


Expt. No.: 10 FUNCTIONS

Aim:

To solve the given problem statements using PL/SQL functions






PL/SQL function:

A PL/SQL function is same as a procedure except that it returns a value. A standalone function is created using

the CREATE FUNCTION statement.

SYNTAX:

CREATE [OR REPLACE] FUNCTION function_name (parameter_name [IN | OUT | IN OUT] type [, ...]) RETURN

return_datatype

{IS | AS} BEGIN

< function_body > END;

Problem Statement:

REFERRED TABLES:

Transaction (accno number(5), amount number(7,2), trans_type varchar(5),dot date);

1. Create a function to insert the records into the transaction table, after performing each transaction in the

transaction table show the net balance of the particular account.

CREATE OR REPLACE FUNCTION FUNC_TRANS (ACC_ID NUMBER,AMNT NUMBER,TYPE VARCHAR)

RETURN NUMBER

IS BALANCE NUMBER; BEGIN

INSERT INTO TRANSACTION VALUES(ACC_ID,AMNT,TYPE);

SELECT SUM(AMNT) INTO BALANCE FROM TRANSACTION WHERE ACC_NO=ACC_ID; RETURN

BALANCE;


END;

PL/SQL FUNCTION:

DECLARE

BALANCE_AMNT NUMBER(6); ACC_NO VARCHAR(5);

AMNT NUMBER(5); TYPE VARCHAR(2); RESULT NUMBER(5); BEGIN

ACC_NO:=’&ACC_NO’; AMNT:=’&AMNT’; TYPE:=’&TYPE’;

BALANCE_AMNT:=FUNC1(ACC_ID,AMNT,TYPE); DBMS_OUTPUT.PUT_LINE(‘TOTAL AMOUNT ‘||

TOTAL_AMNT);

END;

OUTPUT:

SQL> @E:\SQL\FUNCPLSQL.SQL;

Enter value for accno: 001 Enter value for amt: 25000 Enter value for type=’CRDT’ RESULT 26000


Result:

Thus PL/SQL functions were created to solve the given problem statements.

Outcome:

Thus the outcome for developing a PL / SQL programs using functions has been attained.


Viva-voce

1. What is the difference between FUNCTION, PROCEDURE AND PACKAGE in PL/SQL?

2. Is the function return value?

3. What are the basic functions for master, msdb, model, tempdb databases?

4. Write the syntax for function.

5. How to display return value the PL/SQL function?


Expt. No.: 11 TRIGGERS

Aim:

To solve the given problem statements using triggers






PL/SQL Triggers:

Triggers are stored programs, which are automatically executed or fired when some events occur. Triggers are, in

fact, written to be executed in response to any of the following events: A database manipulation (DML) statement

(DELETE, INSERT, or UPDATE).

A database definition (DDL) statement (CREATE, ALTER, or DROP).

A database operation (SERVERERROR, LOGON, LOGOFF, STARTUP, or SHUTDOWN).

Triggers could be defined on the table, view, schema, or database with which the event is associated.

SYNTAX:

CREATE OR REPLACE TRIGGER < trigger_name >

{BEFORE | AFTER | INSTEAD OF }

{INSERT [OR] | UPDATE [OR] | DELETE} [OF col_name]

ON < table_name >

[REFERENCING OLD AS o NEW AS n] [FOR EACH ROW]

WHEN < condition > DECLARE

< Declaration-statements > BEGIN

< Executable-statements > EXCEPTION

< Exception-handling-statements > END;


Problem Statement:

REFERRED TABLES:

Account ( accnt_no,cst_id,acnt_type,last_trans_date,balance ) Account_bckup(accnt_no,last_trans_date,balance)

Loan (ln_id,cst_id,ln_amount,ln_date);

1. Create a trigger for Account relation such that whenever a record is inserted in the Account table the same record

also gets inserted in the backup table.

CREATE OR REPLACE TRIGGER TRIG_ACNT_BCKUP AFTER INSERT ON ACCOUNT FOR EACH ROW

DECLAR BEGIN

INSERT INTO ACCOUNT_BCKUP VALUES (:NEW.ACCNT_NO, :NEW.LAST_TRANS_DATE,

NEW.BALANCE); END;

OUTPUT:

SQL> @e:/plsql/accnt_trig.sql Trigger Created.

SQL> INSERT INTO ACCOUNT VALUES (‘AC010’,’CST011’,’SVNG’,’27-AUG-2013’,12000);

1 Row inserted.

SQL> SELECT * FROM ACCOUNT_BCKUP;

2. Create a trigger for account relation such that whenever account record is inserted in account relation with

negative relation then that record should also be inserted in the loan relation with positive balance.

CREATE OR REPLACE TRIGGER TRIG_LOAN AFTER INSERT ON ACCOUNT

FOR EACH ROW DECLARE BEGIN

IF(:NEW.BALANCE<0)

THEN

INSERT INTO LOAN VALUES (:NEW.ACCNT_NO, :NEW.CST_ID, -(:NEW.BALANCE), SYSDATE); END IF;

END;

OUTPUT:

SQL> @e:/plsql/loan_trig.sql Trigger Created.

SQL> INSERT INTO ACCOUNT VALUES (‘AC011’,’CST011’,’SVNG’,’27-DEC-2013’,-8000);

1 Row inserted.

SQL> SELECT * FROM LOAN;

LN_ID CST_ID LN_AMOUNT LN_DATE

AC010 CST011 8000 30-DEC-2013

3. Create a trigger for Account relation such that whenever a record is inserted in the Account table the same record

also gets inserted in the backup table.


CREATE OR REPLACE TRIGGER TRIG_ACNT_BCKUP AFTER INSERT ON ACCOUNT FOR EACH ROW

DECLAR BEGIN

INSERT INTO ACCOUNT_BCKUP VALUES (:NEW.ACCNT_NO, :NEW.LAST_TRANS_DATE,

:NEW.BALANCE); END;

OUTPUT:

SQL> @e:/plsql/accnt_trig.sql Trigger Created.

SQL> INSERT INTO ACCOUNT VALUES (‘AC010’,’CST011’,’SVNG’,’27-AUG-2013’,12000);

1 Row inserted.

SQL> SELECT * FROM ACCOUNT_BCKUP;

4. Create a trigger for account relation such that whenever account record is inserted in account relation with

negative relation then that record should also be inserted in the loan relation with positive balance.

CREATE OR REPLACE TRIGGER TRIG_LOAN AFTER INSERT ON ACCOUNT

FOR EACH ROW DECLARE BEGIN

IF(:NEW.BALANCE<0)

THEN

INSERT INTO LOAN VALUES (:NEW.ACCNT_NO, :NEW.CST_ID, -(:NEW.BALANCE), SYSDATE); END IF;

END;

OUTPUT:

SQL> @e:/plsql/loan_trig.sql Trigger Created.

SQL> INSERT INTO ACCOUNT VALUES (‘AC011’,’CST011’,’SVNG’,’27-DEC-2013’,-8000);

1 Row inserted.

SQL> SELECT * FROM LOAN;

LN_ID CST_ID LN_AMOUNT LN_DATE

AC010 CST011 8000 30-DEC-2013


Result:

Thus the triggers are created accordingly to solve the given problem statements.

Outcome:

Thus the outcome for developing a Trigger has been attained.


Viva-voce

1. What is a Trigger?

2. What is a stored procedure?

3. What is the difference between Trigger and Stored Procedure?

4. What is a transaction? What are ACID properties?

5. What are indexes?

6. Compare trigger, procedure and cursors.

7. What are the various types of triggers?

8. What is a PL/SQL trigger?

9. Describe triggers features and limitations.

10. What are the variables used in triggers?


Expt. No.: 12 Payroll Processing System

Aim: To develop a mini project using vb/oracle






Procedure:

• Create a database for payroll processing which request the using SQL

• Establish ODBC connection

• In the administrator tools open data source ODBC

• Click add button and select oracle in ORA home 90

• click finish

• A window will appear given the data source home as oracle and select TNS source name as lion and give the

used id as SWTT

• ADODC CONTROL FOR SALARY FORM:-

• The above procedure must be follow except the table ,A select the table as salary

• Write appropriate Program in form each from created in VB from each from created in VB form project.

SQL>create table emp(eno number primary key,enamr varchar(20),age number,addr varchar(20),DOB date,phno

number(10));

Table created.

SQL>create table salary(eno number,edesig varchar(10),basic number,da number,hra number,pf number,mc

number,met number,foreign key(eno) references emp);

Table created.

TRIGGER to calculate DA,HRA,PF,MC

SQL> create or replace trigger employ


2 after insert on salary

3 declare

4 cursor cur is select eno,basic from salary;

5 begin

6 for cur1 in cur loop 7

update salary set

8 hra=basic*0.1,da=basic*0.07,pf=basic*0.05,mc=basic*0.03 where hra=0;

9 end loop;

10 end;

11 /

Trigger created

Output


Result:

Thus the design and implementation of payroll processing system using oracle, Vb was successfully done.


Exp. No. 13 SQL Join

Aim:

To solve the given problem statements using triggers





Standalone desktops 30 Numbers

SQL JOIN

An SQL JOIN clause is used to combine rows from two or more tables, based on a common field between them.

Different SQL JOINs

• LEFT JOIN: Return all rows from the left table, and the matched rows from the right table

• RIGHT JOIN: Return all rows from the right table, and the matched rows from the left table

• FULL JOIN: Return all rows when there is a match in ONE of the tables

SQL LEFT JOIN Keyword

The LEFT JOIN keyword returns all rows from the left table (table1), with the matching rows in the right table (table2). The result is NULL in the right side when there is no match.

SQL LEFT JOIN Syntax

SELECT column_name(s) FROM table1 LEFT JOIN table2 ON table1.column_name=table2.column_name;

SQL LEFT JOIN Example

SELECT Customers.CustomerName, Orders.OrderID

FROM Customers

LEFT JOIN Orders


ON Customers.CustomerID=Orders.CustomerID

ORDER BY Customers.CustomerName;

SQL RIGHT JOIN Keyword

The RIGHT JOIN keyword returns all rows from the right table (table2), with the matching rows in the left table (table1). The result is NULL in the left side when there is no match.

SQL RIGHT JOIN Syntax

SELECT column_name(s)

FROM table1

RIGHT JOIN table2

ON table1.column_name=table2.column_name;

SQL RIGHT JOIN Example

SELECT Orders.OrderID, Employees.FirstName

FROM Orders

RIGHT JOIN Employees

ON Orders.EmployeeID=Employees.EmployeeID

ORDER BY Orders.OrderID;

SQL FULL OUTER JOIN Keyword

The FULL OUTER JOIN keyword returns all rows from the left table (table1) and from the right table (table2).

The FULL OUTER JOIN keyword combines the result of both LEFT and RIGHT joins.

SQL FULL OUTER JOIN Syntax

SELECT column_name(s)

FROM table1

FULL OUTER JOIN table2

ON table1.column_name=table2.column_name;

SQL FULL OUTER JOIN Example

SELECT Customers.CustomerName, Orders.OrderID

FROM Customers

FULL OUTER JOIN Orders

ON Customers.CustomerID=Orders.CustomerID

ORDER BY Customers.CustomerName;

Result:

Thus the sqljoin are created and executed successfully.

DHANALAKSHMI COLLEGE OF ENGINEERING · dhanalakshmi college of engineering tambaram, chennai –...

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