Time table for 2nd semester
09-April-2015 : ENGLISH
11-April-2015 : HINDI/KANNADA
15-April-2015 : BASIC of NETWORKING
17-April-2015 : CPP
22-April-2015 : DBMS
24-April-2015 : HR
Tuesday, 10 March 2015
TIME TABLE
Saturday, 7 March 2015
DBMS LAB Part-B 4
Ex : 4
Create two tables Production_detail and Purchased_detail with the following fields:
Production_detail (Product_no, Product_name, Qty_available, Price Profit(%))
Purchased_detail (Cust_name, Product_no, Qty_sold)
Create table product_detail
SQL> create table product_detail(
Prno varchar2(6) primary key,
prname varchar2(15) not null,
quantity number(3) not null,
price number(6) not null,
profit number(3) not null,
check(prno like 'p%'));
Table created.
Display table fields
SQL> desc product_detail;
Name Null? Type
-----------------------------------------------------
PRNO NOT NULL VARCHAR2(6)
PRNAME NOT NULL VARCHAR2(15)
QUANTITY NOT NULL NUMBER(3)
PRICE NOT NULL NUMBER(6)
PROFIT NOT NULL NUMBER(3)
Insert values into table
SQL> insert into product_detail values(
'p001','monitor',10,4500,20);
1 row created.
SQL> select * from product_detail;
PRNO PRNAME QUANTITY PRICE PROFIT
---------------------------------------------------------------------------------
p001 monitor 10 4500 20
p002 pendrive 50 250 5
p003 CDdrive 5 950 8
p004 keyboard 8 250 10
Create table purchased_detail
SQL> create table purchased_detail(
Custname varchar2(15) not null,
prno varchar2(6) references product_detail(prno),
qty_sold number(3) not null,
primary key(custname,prno));
Table created.
Display table fields
SQL> desc purchased_detail;
Name Null? Type
--------------------------------------------------------------------
CUSTNAME NOT NULL VARCHAR2(15)
PRNO NOT NULL VARCHAR2(6)
QTY_SOLD NOT NULL NUMBER(3)
Insert values into table
SQL> insert into purchased_detail values(
'raman','p003',2);
1 row created.
SQL> select * from purchased_detail;
CUSTNAME PRNO QTY_SOLD
------------------------------------------------------
raman p003 2
laxman p002 5
bharath p002 10
manish p001 3
amith p004 2
1. Query to display total amount spend by Mr. Laxman for purchasing any product.
SQL> select (p.price*c.qty_sold) "Total amount"
from product_detail p,purchased_detail c
where p.prno=c.prno and
c.custname='laxman';
Total amount
-----------------
1250
2. Query to display the name of product for which either quantity available is less than 10 or quantity sold is less than 4
SQL> select p.prname
from product_detail p,purchased_detail c
where p.prno=c.prno and
(p.quantity<10 or c.qty_sold<4);
PRNAME
--------------
CDdrive
monitor
keyboard
3. Query to display the name of product and quantity taken by Mr. Bharath.
SQL> select p.prname,c.qty_sold
from product_detail p,purchased_detail c
where p.prno=c.prno and
c.custname='bharath';
PRNAME QTY_SOLD
-------------------------------------
pendrive 10
4. How much the profit does the shop keeper gets on Mr. Manish purchase?
SQL> select (p.price*c.qty_sold*profit/100) "Profit"
from product_detail p,purchased_detail c
where p.prno=c.prno and
c.custname='manish';
Profit
--------
2700
5. How many “Pen drives” have sold?
SQL> select sum(c.qty_sold) "Pendrives sold"
from product_detail p,purchased_detail c
where p.prno=c.prno and
p.prname='pendrive';
Pendrives sold
------------------
15
DBMS LAB Part-B 3
Ex : 3
Create two tables Department_detail and Employee_detail with the following fields:
Department_detail (DEPTNO, DEPTNAME, TOTEMP, CHARGE/HOUR, BONUS%PER_HOUR)
Employee_detail (EMPNAME, DEPTNO, TOTAL_HOURS)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Create table department_detail
SQL> create table department_detail(
deptno varchar2(6) primary key,
deptname varchar2(15) not null,
totemp number(3) not null,
charge_hour number(5) not null,
bonus_per_hr number(3) not null,
check (deptno like ‘d%’));
Table created.
Display table fields
SQL> desc department_detail;
Name Null? Type
---------------------------------------------------------------------
DEPTNO NOT NULL VARCHAR2(6)
DEPTNAME NOT NULL VARCHAR2(15)
TOTEMP NOT NULL NUMBER(3)
CHARGE_HOUR NOT NULL NUMBER(5)
BONUS_PER_HR NOT NULL NUMBER(3)
Insert values into table
SQL> insert into department_detail values(
‘d001’,’accounts’,5,750,10);
1 row created.
SQL> select * from department_detail;
DEPTNO DEPTNAME TOTEMP CHARGE_HOUR BONUS_PER_HR
------------------------------------------------------------------------------------------
d001 accounts 5 750 10
d002 sales 10 2000 10
d003 service 5 500 10
d004 purchase 4 600 10
Create table employee_detail
SQL> create table employee_detail(
Empname varchar2(15) not null,
deptno varchar2(6)references department_detail(deptno),
total_hours number(3) not null,
primary key(empname, deptno));
Table created.
Display table fields
SQL> desc employee_detail;
Name Null? Type
--------------------------------------------------------------------------
EMPNAME NOT NULL VARCHAR2(15)
DEPTNO NOT NULL VARCHAR2(6)
TOTAL_HOURS NOT NULL NUMBER(3)
Insert values into table
SQL> insert into employee_detail values(
'raman','d002',8);
1 row created.
SQL> select * from employee_detail;
EMPNAME DEPTNO TOTAL_HOURS
---------------------------------------------------------------
raman d002 8
deekshith d001 6
ramnath d003 4
naveen d002 7
shyam d004 5
1. Query to display employee name, department number and department name of all employees.
SQL> select e.empname,e.deptno,d.deptname
from department_detail d,employee_detail e
where d.deptno=e.deptno;
EMPNAME DEPTNO DEPTNAME
----------------------------------------------------------
raman d002 sales
deekshith d001 accounts
ramnath d003 service
naveen d002 sales
shyam d004 purchase
2. Query to display the department name to which Mr. Ramnath belongs to.
SQL> select d.deptname from department_detail d,
employee_detail e
where d.deptno=e.deptno and
e.empname='ramnath';
DEPTNAME
-----------------
Service
3. Query to display the bonus got by Mr. Raman.
SQL> select(d.charge_hour*e.total_hours
*d.bonus_per_hr/100) "bonus"
from department_detail d,employee_detail e
where d.deptno=e.deptno and
e.empname='raman';
Bonus
--------
1600
Query to display the total number of hours taken by accounts department.
SQL> select sum(e.total_hours) "total hours"
from department_detail d,employee_detail e
where d.deptno=e.deptno and
d.deptname='accounts';
total hours
--------------
6
DBMS LAB Part-B 2
Ex : 6
Create two table’s employee_master and attendance with the following fields:
Employee_master(emp_id, emp_name, emp_add, phone, email, boj, dor).
Attendance (emp_id, wom, mhrs, tuhrs, whrs, thhrs, frhrs, sahrs, suhrs).
Identify primary and foreign keys. Specify the not null property for mandatory keys (doj<dor and valid values of wom<=5)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Create table employee_master
SQL> create table employee_master(
emp_id number(6) primary key,
emp_name varchar2(20) not null,
emp_add varchar2(30) not null,
phone number(10),
email varchar2(30),
doj date not null,
dor date not null,
check (doj<dor));
Table created.
Display table fields
SQL> desc employee_master;
Name Null? Type
----------------------------------------------------
EMP_ID NOT NULL NUMBER(6)
EMP_NAME NOT NULL VARCHAR2(20)
EMP_ADD NOT NULL VARCHAR2(30)
PHONE NOT NULL NUMBER(10)
EMAIL NOT NULL VARCHAR2(30)
DOJ NOT NULL DATE
DOR NOT NULL DATE
Insert values into table
SQL> insert into employee_master
values(2001,'harsha','kavoor',224560,
'harsha@yahoo.com','30-jun-1985','15-jan-1997');
1 row created.
SQL> select * from employee_master;
EMP_ID EMP_NAME EMP_ADD PHONE EMAIL DOJ DOR
-------------------------------------------------------------------------------------------------
2001 harsha kavoor 224560 harsha@yahoo.com 30-JUN-85 15-JAN-97
2002 mahesh bejai 254608 mahesh@yahoo.com 25-MAR-96 10-MAY-10
2003 swarna balmatta 268754 23-JUL-00 12-AUG-16
2004 priya kulur 268945 21-DEC-02 29-DEC-15
2005 hema pvs 258743 15-NOV-96 31-JAN-13
2006 preetham marakada 268745 preethu@yahoo.com 21-OCT-90 25-SEP-14
2007 shilpa madikeri 268475 24-APR-10 30-OCT-20
2008 manasa Bangalore man@www.com 05-JUL-94 05-JUL-19
2009 rashmi kavoor ras@www.com 15-JUL-90 18-SEP-16
Create table attendance
SQL> create table attendance (
emp_id number(6) references employee_master(emp_id)
on delete cascade,
wom number(1) not null,
mhrs number(3) not null,
tuhrs number(3) not null,
whrs number(3) not null,
thhrs number(3) not null,
frhrs number(3) not null,
sahrs number(3) not null,
suhrs number(3) not null,
check (wom<=5));
Table created.
Display table fields
SQL> desc attendance;
Name Null? Type
-----------------------------------------------------
EMP_ID NOT NULL NUMBER(6)
WOM NOT NULL NUMBER(1)
MHRS NOT NULL NUMBER(3)
TUHRS NOT NULL NUMBER(3)
WHRS NOT NULL NUMBER(3)
THHRS NOT NULL NUMBER(3)
FRHRS NOT NULL NUMBER(3)
SAHRS NOT NULL NUMBER(3)
SUHRS NOT NULL NUMBER(3)
Insert values into table
SQL> insert into attendance values(
2002,2,4,6,7,5,8,0,0);
1 row created.
SQL> select * from attendance;
EMP_ID WOM MHRS TUHRS WHRS THHRS FRHRS SAHRS SUHRS
----------------------------------------------------------------------------------------
2002 2 4 6 7 5 8 0 0
2001 1 6 0 11 3 4 0 0
2005 3 6 8 9 0 1 0 0
2001 5 0 0 5 3 0 0 0
2009 2 5 3 0 0 4 0 4
2005 3 0 0 0 4 6 0 4
2001 1 5 6 0 4 2 0 7
Query to display employee name and email_id of all the employees who are working on Sunday.
SQL> select distinct emp_name,email
from employee_master
where emp_id in( select emp_id from attendance
where suhrs>0);
EMP_NAME EMAIL
----------------------------------------
harsha harsha@yahoo.com
hema
rashmi ras@www.com
Query to display the total hours worked by the employee ‘mahesh’.
SQL> select sum(a.mhrs+a.tuhrs+a.whrs+a.thhrs+a.frhrs
+a.sahrs+a.suhrs) "Total hours worked by mahesh"
from employee_master e, attendance a
where e.emp_id=a.emp_id and
e.emp_name in('mahesh');
Total hours worked by mahesh
----------------------------
30
Query to display the names of the employee who never attended the duty so far.
SQL> select distinct emp_id,emp_name
from employee_master
where emp_id not in(select distinct emp_id from
attendance);
EMP_ID EMP_NAME
-------------------------------
2003 swarna
2004 priya
2006 preetham
2007 shilpa
2008 manasa
4. Query to Delete the records of the employees who are already retiered.
SQL> delete from employee_master
where dor<sysdate;
2 rows deleted.
5. Query to Display the names of employee who have total number of hours more than 20 hours a week.
SQL> select distinct e.emp_name from employee_master
e,attendance a
where e.emp_id=a.emp_id and
(a.mhrs+a.tuhrs+a.whrs+a.thhrs+a.frhrs
+a.sahrs+a.suhrs)>20;
EMP_NAME
--------------
harsha
hema
mahesh
DBMS LAB Part-B 1
Create the following tables by identifying primary and foreign keys. Specify the not null property for mandatory keys.
Suppliers (supplier_no, sname, saddress, scity)
Computer_items (item_no, supplier_no,item_name,iquantity)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Create table Supplier
SQL> create table suppliers
(supplier_no number(6) primary key,
sname varchar2(20) not null,
saddress varchar2(30) not null,
scity varchar2(15) not null);
Table created.
Display table fields
SQL> desc suppliers;
Name Null? Type
----------------------------------------------------------------
SUPPLIER_NO NOT NULL NUMBER(6)
SNAME NOT NULL VARCHAR2(20)
SADDRESS NOT NULL VARCHAR2(30)
SCITY NOT NULL VARCHAR2(15)
Insert values into table
SQL> insert into suppliers
values(101,'microtech','kulur','mangalore');
1 row created.
SQL> select * from suppliers;
SUPPLIER_NO SNAME SADDRESS SCITY
------------------------------------------------------------------------
101 microtech kulur mangalore
102 cats kavoor mangalore
103 electrotech lalbag bangalore
104 onida bejai mangalore
105 samsung PVS bangalore
Create table computer_items
SQL> create table computer_items(
item_no number(6) primary key,
supplier_no number(6)references suppliers(supplier_no),
item_name varchar2(15) not null,
iquantity number(6) not null);
Table created.
Display table fields
SQL> desc computer_items;
Name Null? Type
-----------------------------------------------------
ITEM_NO NOT NULL NUMBER(6)
SUPPLIER_NO NOT NULL NUMBER(6)
ITEM_NAME NOT NULL VARCHAR2(15)
IQUANTITY NOT NULL NUMBER(6)
Insert values into table
SQL> insert into computer_items values(
501,101,'monitor',3);
1 row created.
SQL> select * from computer_items;
ITEM_NO SUPPLIER_NO ITEM_NAME IQUANTITY
------------------------------------------------------------------------
501 101 monitor 3
502 103 keyboard 5
503 105 mouse 7
504 102 pendrive 6
506 104 floppy_disk 8
505 103 hard disk 5
507 102 joystick 3
508 104 printer 9
509 105 scanner 6
510 104 ups 9
1. Query to list item and suppliers details:
SQL> select s.supplier_no,s.sname,s.saddress,s.scity,
c.item_name
from suppliers s,computer_items c
where s.supplier_no=c.supplier_no;
SUPPLIER_NO SNAME SADDRESS SCITY ITEM_NAME
--------------------------------------------------------------------------------
101 microtech kulur mangalore monitor
103 electrotech lalbag bangalore keyboard
105 samsung PVS bangalore mouse
102 cats kavoor mangalore pendrive
104 onida bejai mangalore floppy_disk
103 electrotech lalbag bangalore hard disk
102 cats kavoor mangalore joystick
104 onida bejai mangalore printer
105 samsung PVS bangalore scanner
104 onida bejai mangalore ups
2. Query to list the names of the supplier who are supplying keyboard:
SQL> select s.supplier_no,s.sname
from suppliers s,computer_items c
where s.supplier_no=c.supplier_no and
c.item_name='keyboard';
SUPPLIER_NO SNAME
--------------------------------
103 electrotech
3. Query to display the items supplied by ‘microtech’:
SQL> select c.item_no,c.item_name,c.iquantity
from suppliers s,computer_items c
where s.supplier_no=c.supplier_no and
s.sname='microtech';
ITEM_NO ITEM_NAME IQUANTITY
---------------------------------------------
501 monitor 3
4. Query to list the items supplied by the suppliers ‘cats’ and ‘electrotech’:
SQL> select c.item_no,c.item_name,c.iquantity,s.sname
from suppliers s,computer_items c
where s.supplier_no=c.supplier_no and
s.sname in('cats','electrotech');
ITEM_NO ITEM_NAME IQUANTITY SNAME
-------------------------------------------------------------------------
502 keyboard 5 electrotech
504 pendrive 6 cats
505 hard disk 5 electrotech
507 joystick 3 cats
DBMS LAB Part-A 4
Ex 4:
To create table book using SQL command to store ISBN, TITLE, PUBLISHER, CATEGORY, YEAR, PRICE.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SQL> CREATE TABLE BOOK(
ISBN VARCHAR2(10) PRIMARY KEY,
TITLE VARCHAR2(20) NOT NULL,
PUBLISHER VARCHAR2(15) NOT NULL,
CATEGORY VARCHAR2(10) NOT NULL,
YEAR NUMBER(04) NOT NULL,
PRICE NUMBER(8,2) NOT NULL);
Insert values into table:
INSERT INTO BOOK
VALUES(‘C01001’,’JAVA’,’MICROSOFTPRESS’,
’COMP SC’,2009,2000);
SQL> SELECT * FROM BOOK;
ISBN TITLE PUBLISHER CATEGORY YEAR PRICE
-------------------------------------------------------------------------------------------------------
C01001 JAVA MICROSOFT PRESS COMP SC 2009 2000
C01002 C++ MICRO COMP SC 2009 1500
CO1003 INDIAN CIVIL SUMUKHA HISTORY 2011 999
C01004 ELECTRONICS RAMAN PHYSICS 2006 500
C01005 LIVING ORGANISMS STROUPTOS BIOLOGY 2010 450
C01006 INDIAN CIVIL C.V CIVICS 2003 200
C01007 COA MICRO COMP SC 2009 150
C01008 FORMULAS MATHEMATICS MATHS 2000 100
C01009 JAVA MICROSOFT PRESS COMP SC 2004 156
C01010 FORMULAS MATHEMATICS MATHS 2001 250
10 rows selected.
1. List the details of the book whose publishers name start with m.
SQL> SELECT * FROM BOOK
WHERE PUBLISHER LIKE 'M%';
ISBN TITLE PUBLISHER CATEGORY YEAR PRICE
-----------------------------------------------------------------------------------------------
C01001 JAVA MICROSOFT PRESS COMP SC 2009 2000
C01002 C++ MICRO COMP SC 2009 1500
C01007 COA MICRO COMP SC 2009 150
C01008 FORMULAS MATHEMATICS MATHS 2000 100
C01009 JAVA MICROSOFT PRESS COMP SC 2004 156
C01010 FORMULAS MATHEMATICS MATHS 2001 250
6 rows selected.
2. List the details of the publisher having A as second character in their name.
SQL> SELECT * FROM BOOK
WHERE PUBLISHER LIKE '_A%';
ISBN TITLE PUBLISHER CATEGORY YEAR PRICE
-------------------------------------------------------------------------------------------------------
C01004 ELECTRONICS RAMAN PHYSICS 2006 500
C01008 FORMULAS MATHEMATICS MATHS 2000 100
C01010 FORMULAS MATHEMATICS MATHS 2001 250
3. Find the books published in 2010,2011 and 2012.
SQL> SELECT * FROM BOOK
WHERE YEAR IN(2010,2011,2012);
ISBN TITLE PUBLISHER CATEGORY YEAR PRICE
-------------------------------------------------------------------------------------------------------
C01001 JAVA MICROSOFT PRESS COMP SC 2009 2000
C01002 C++ MICRO COMP SC 2009 1500
CO1003 INDIAN CIVIL SUMUKHA HISTORY 2011 999
C01005 LIVING ORGANISMS STROUPTOS BIOLOGY 2010 450
C01007 COA MICRO COMP SC 2009 150
4. Display the book_id, title, publisher of all the books in the descending order in the year.
SQL> SELECT ISBN,TITLE,PUBLISHER FROM BOOK
ORDER BY YEAR DESC;
ISBN TITLE PUBLISHER
--------------------------------------------------------------------------
CO1003 INDIAN CIVIL SUMUKHA
C01005 LIVING ORGANISMS STROUPTOS
C01001 JAVA MICROSOFT PRESS
C01002 C++ MICRO
C01007 COA MICRO
C01004 ELECTRONICS RAMAN
C01009 JAVA MICROSOFT PRESS
C01006 INDIAN CIVILIZ C.V
C01010 FORMULAS MATHEMATICS
C01008 FORMULAS MATHEMATICS
10 rows selected.
5. Display the details of all the books other than ‘Microsoft press’ publishers.
SQL> SELECT * FROM BOOK
WHERE PUBLISHER NOT IN('MICROSOFT PRESS');
ISBN TITLE PUBLISHER CATEGORY YEAR PRICE
-------------------------------------------------------------------------------------------------------
C01002 C++ MICRO COMP SC 2009 1500
CO1003 INDIAN CIVIL SUMUKHA HISTORY 2011 999
C01004 ELECTRONICS RAMAN PHYSICS 2006 500
C01005 LIVING ORGANISMS STROUPTOS BIOLOGY 2010 450
C01006 INDIAN CIVIL C.V CIVICS 2003 200
C01007 COA MICRO COMP SC 2009 150
C01008 FORMULAS MATHEMATICS MATHS 2000 100
C01010 FORMULAS MATHEMATICS MATHS 2001 250
8 rows selected.
6. Display title, price of all the books with the price more than 2000 and less than 3000 (using BETWEEN) operator.
SQL> SELECT TITLE, PRICE FROM BOOK
WHERE PRICE BETWEEN 2000 AND 3000;
TITLE PRICE
---------------------------------
JAVA 2000
DBMS LAB Part-A 3
Ex 3:
To create table client_master using SQL command to store CLIENT_NO, NAME, ADDRESS1, ADDRESS2, CITY, PINCODE, STATE, BAL_DUE.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. For every client in client_master table decrease bal_due by 10%.
SQL> UPDATE CLIENT_MASTER
SET BAL_DUE = BAL_DUE + (BAL_DUE*0.1);
10 rows updated.
2. Update the table client_master, change the contents of the field name to ‘vijay kadam’ and the contents of the field address to ‘SCT Jay Apartment’ for the record with client_no=’C01002’.
UPDATE CLIENT_MASTER SET
NAME='VIJAY KADAM',
ADDRESS1='SCT JAY APT'
WHERE CLIENT_NO='C01002';
1 row updated.
SQL> SELECT * FROM CLIENT_MASTER;
CLIENT NAME ADDRESS1 ADDRESS2 CITY PINCODE STATE BAL_DUE
------------------------------------------------------------------------------------------------------------------------------
C01001 ANIS MADI BEJAI UDUPI 574144 KARNARTAKA 9000
C01002 VIJAY KADAM SCT JAY APT NAGAR UDUPI 671125 KERALA 13500
C01003 GURU SERE KOTTARA UPUPI 562232 KARNARTAKA 9900
C01004 DHANRAJ KS BEJAI UDUPI 575003 KARNARTAKA 13500
C01005 MANOJ PANDESHWAR YAKKAR UDUPI 567651 KARNARTAKA 12150
C01006 RAKSHITH PERMUDE BCROAD UDUPI 575223 KARNARTAKA 16200
C01007 LIKHIN MAROLI BEJAI UDUPI 567876 KARNARTAKA 11160
C01008 RAKSHAN MALALI SULYA UDUPI 574144 KARNARTAKA 5000
C01009 NISHANK POLALI PADAV UDUPI 564333 KARNARTAKA 17100
C01010 BHARATH GANDHINAG BAJPE UDUPI 475767 GOA 9900
10 rows selected.
3. Add a new column name penalty number(10,2)to the table client_master.
SQL> ALTER TABLE CLIENT_MASTER ADD PENALTY NUMBER(10,2);
Table altered.
SQL> DESC CLIENT_MASTER;
Name Null? Type
------------------------------------------------------
CLIENT_NO NOT NULL VARCHAR2(6)
NAME NOT NULL VARCHAR2(20)
ADDRESS1 NOT NULL VARCHAR2(30)
ADDRESS2 NOT NULL VARCHAR2(30)
CITY NOT NULL VARCHAR2(15)
PINCODE NOT NULL NUMBER(8)
STATE NOT NULL VARCHAR2(15)
BAL_DUE NOT NULL NUMBER(10,2)
PENALTY NOT NULL NUMBER(10,2)
4. Change the size of the column penalty to(8,2) in the client_master.
SQL> ALTER TABLE CLIENT_MASTER MODIFY PENALTY NUMBER(8,2);
Table altered.
SQL> DESC CLIENT_MASTER;
Name Null? Type
-----------------------------------------------------
CLIENT_NO NOT NULL VARCHAR2(6)
NAME NOT NULL VARCHAR2(20)
ADDRESS1 NOT NULL VARCHAR2(30)
ADDRESS2 NOT NULL VARCHAR2(30)
CITY NOT NULL VARCHAR2(15)
PINCODE NOT NULL NUMBER(8)
STATE NOT NULL VARCHAR2(15)
BAL_DUE NOT NULL NUMBER(10,2)
PENALTY NOT NULL NUMBER(8,2)
5. Change the name of the table Client_master to client_master1.
SQL> RENAME CLIENT_MASTER TO CLIENT_MASTER1;
Table renamed.
6. Delete all the newly created table.
SQL> DROP TABLE CLIENT_MASTER1;
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