Ip Project File-12th Class

*EMP_ ID SHOULD BE DECLARED AS PRIMARY KEY.

*USER_ID SHOULD BE UNIQUE.

*SALARY MUST BE GREATER THAN 5000.

*FIRST_NAME AND LAST_NAME MUST NOT REMAIN IN BLANK.

Project Question No. 1

Q1. CREAT TABLE ‘EMPLOYEE’ WITH THE FOLLOWING STRUCTURE. TABLE : EMPLOYEENAME OF COLUMN TYPEEMP_ID INTEGERFIRST_NAME VARCHAR2(30)LAST_NAME VARCHAR2(30)USER_ID VARCHAR2(10)SALARY INTEGER

CREAT TABLE EMP

(

EMP_ID PRIMARY KEY, INTEGER,

FIRST_NAME VARCHAR2(30), NOT NULL,

LAST_NAME VARCHAR2(30), NOT NULL,

USER_ID UNIQUE, VARCHAR2(10),

SALARY INTEGER CHECK(SALARY>5000);

Project Question No. 2

Q2. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’STUDENT’’

TABLE : STUDENT

STUDENT NO.

CLASS NAME GAME GRADE1 SUPW GRADE2

10 7 VAIBHAV CRICKET B PHOTOGRAPHY A11 8 PRABHJO

TTENNIS A GARDENING C

12 7 VIPIN SWIMMING B PHOTOGRAPHY B13 7 VIJAY TENNIS C COOKING A14 9 DANISH CRICKET A LITERATURE A15 10 PANKAJ SKATTING A GARDENING C

*DISPLAY ALL THE DETAILS FROM TABLE ‘’STUDENT’’.

*DISPLAY THE NAME OF THE STUDENT WHO ARE GETTING ‘’C’’ IN EITHER GAME OR SUPW.

*DISPLAY THE DIFFERENT GAME OFFERED IN THE SCHOOL.

*DISPLAY THE SUPW TAKEN UP BY THE STUDENT, WHOSE NAME STARTS WITH ‘’V’’.

SQL>SELECT*FROM STUDENT;

SQL>SELECT NAME FROM STUDENT WHERE GRADE 1 = ‘’C’’ OR GRADE 2= ‘’C’’;

SQL>SELECT DISTINCT GAME FROM STUDENT;

SQL>SELECT SUPW FROM STUDENT EHRE NAME LIKE ’’V’’;

Project Question No.3

Q3. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “HOSPITAL”.

NO. NAME AGE DEPARTMENT D.O.ADM CHARGES SEX

1 SANDEEP 65 SURGERY 23-2-98 300 M2 RAVINA 24 ORTHOPAEDIC 20-1-98 200 F3 KARAN 45 ORTHOPAEDIC 15-2-98 500 M4 TARUN 12 SURGERY 01-1-98 300 M5 ZUBIN 36 ENT 12-1-98 250 M6 KETAKI 16 ENT 24-2-98 300 F

7 ANKITA 29 CARDIOLOGY 20-2-98 800 F

8 ZAREEN 45 GYNAECOLOGY 22-2-98 300 F

8 KUSH 19 CARDIOLOGY 13-1-98 800 M

10 SHAILYA 31 NUCLEAR 19-1-98 400 F

TABLE : HOSPITAL

*DISPLAY ALL THE INFORMATION ABOUT THE PATIENTS OF CARDIOLOGY DEPARTMENT.

*TO LIST THE NAMES OF FEMALE PATIENTS WHO ARE IN ORTHOPAEDIC DEPARTMENT.

*TO LIST NAMES OF ALL PATIENTS WITH THEIR DATE OF ADMISSION IN ASCENDING ORDER.

*DISPLAY PATIEN’S NAME, CHARGE, AGE OF MALE PATIENS ONLY.

*TO COUNT THE NUMBER OF PATIENTS WITH AGE>20.

*TO INSERT A NEW ROW IN THE “HOSPITAL” TABLES WITH THE FOLLOWING DATA:

(11, ‘MUSTAFA’,37,’ENT’,’25-2-98’,’250’M’).

SQL>SELECT*FROM HOSPITAL WHERE DEPARTMENT=’CARDIOLOGY’;

SQL>SELECT NAME FROM HOSPITAL WHERE DEPARTMENT=’ORTHOPEADIC’ AND SEX=’F’;

SQL>SELECT NAME, DOADM. FROM HOSPITAL ORDER BY DOADM ASC;

SQL>SELECT NAME, CHARGES, AGE, FROM HOSPITAL WHERE SEX=’M’; SQL>SELECT COUNT (*) FROM HOSPITAL WHERE AGE>20;

SQL>INSERT INTO HOPITAL VALUES

(11, ‘MUSTAFA’,37,’ENT’,’25-2-98’,’250’M’);

PROJECT QUESTION NO. 4

Q4. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’SUPPLIER’’.

TABLE : SUPPLIER

S.NO. P NAME SNAME QTY CITY51 BREAD BRITANNIA 8.00 DELHI52 CAKE BRITANNIA 20.00 MUMBAI53 COFFEE NESCAFE 45.00 DELHI54 CHOCOLATE AMUL 10.00 MUMBAI55 SAUCE KISAN 36.00 JAIPUR

*DISPLAY DATA FOR ALL PRODUCT WHOSE QUANTITY IS BETWEEN 170 AND 370.

*DISPLAY DATA FOR ALL PRODUCT SORTED BY THEIR QUANTITY.

*FIND ALL PRODUCTS THAT HAVE NO SUPPLIER.

*GIVE SNAME, PNAME, PRICE FOR ALL THE PRODUCT WHOSE QUANTITY IS<200.

*TO DISPLAY S.NO., PNAME, SNAME, QTY IN DESCENDING ORDER OF QUANTITY FROM THE ‘SUPPLIER’ TABLE.

SQL>SELECT*SUPPLIER WHERE QTY BETWEEN 170 AND 370;

SQL>SELECT*FROM SUPPLIER ORDER BY QTY;

SQL>SELECT*FROM SUPPLIER WHERE SNAME IS NULL;

SQL>SELECT SNAME FROM SUPPLIER WHERE PNAME IS LIKE C%;

SQL>SELECT PNAME, SNAME, QTY FROM SUPPLIER ORDER BY QTY DESC;


Q5. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’TEACHER’’

TABLE : TEACHER

S.NO. NAME AGE DEPARTMENT D.O.ADM SALARRY SEX1 JUGAL 34 COMPUTER 10-01-97 12000 M2 SHAMILA 31 HISTORY 24-03-98 20000 F3 SANDEEP 32 MATHS 12-12-96 30000 M4 RAKESH 42 MATHS 05-09-99 40000 F5 SHYAM 50 HISTORY 27-06-98 30000 M6 SHIV OM 44 COMPUTER 25-02-97 21000 M7 SANGEETA 35 HISTORY 01-07-99 40000 F8 SHALAKH

A33 MATHS 31-07-97 20000 F

*TO SHOW ALL INFORMATION ABOUT THE TEACHER OF HISTORY DEPARTMENT.

*TO LIST THE NAME OF FEMALE TEACHER WHO ARE IN MATHS DEPARTMENT.

*TO LIST NAME OF ALL TEACHER WITH THEIR DATE OF ADMISSION IN ASCENDING ORDER.

*DISPLAYTEACHER’S NAME, SALARY, AGE FOR MALE TEACHER ONLY.

*TO COUNT TNE NUMBER OF TEACHER WITH AGE > 23.

*TO INSERT A NEW ROW IN THE TEACHER TABLE WITH THE FOLLOWING DATA :

(9,’RAJA’,26,’COMPUTER’,’13-5-95,23000,’M’);

SQL>SELECT*FROM TEACHER WHERE DEPARTMENT = ‘’HISTORY’’;

SQL>SELECT NAME FROM TEACHER WHERE DEPARTMENT = ‘’MATHS’’ AND SEX = ‘’F’’;

SQL>SELECT NAME FROM TEACHER ORDER BY D.O.ADM;

SQL>SELECT NAME, SALARY, AGE FROM TEACHER WHERE SEX = ‘’M’’;

SQL>SELECT COUNT(*) FROM TEACHER WHERE AGE >23;

SQL>INSERT INTO TEACHER VALUE

(9,’RAJA’,26,’COMPUTER’,’13-5-95,23000,’M’);


Q6. . WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’CLUB’’.

TABLE : CLUB

COACH_ID COACH_NAME AGE SPORTS D.O.APP PAY SEX1 KUKREJA 35 KARATE 27-03-96 1000 M2 RAVINA 34 KARATE 20-01-98 1200 F3 KARAN 34 SQUASH 19-02-98 2000 M4 TARUN 33 BASKETBALL 01-01-98 1500 M5 ZUBIN 36 SWIMMING 12-01-98 750 M6 KETAKI 36 SWIMMING 24-02-98 800 F7 ANKITA 39 SQUASH 20-02-98 2200 F8 ZAREEN 37 KARATE 22-02-98 1100 F9 KUSH 41 SWIMMING 13-01-98 900 M10 SHAILYA 37 BASKETBALL 19-02-98 1700 M

*TO SHOW ALL THE INFORMATION ABOUT SWIMMING COACHES IN THE CLUB.

*TO LIST NAMES OF ALL COACHES WITH THEIR D.O.ADM IN APP. IN DESCENDING ORDER.

*TO DISPLAY A REPORT, SHOWING COACH_NAME, PAY, AGE AND BONUS (15%OF PAY) FOR ALL THE COACHES.

*TO INSERT A NEW ROW IN THE CLUB TABLE WITH THE FOLLOWING DATA:

(11,’RAJIV’,40,’HOCKEY’,’27-05-2000,’M’);

SQL>SELECT*FROM CLUB WHERE SPORTS = ‘SWIMMING’;

SQL>SELECT COACH_NAME FROM CLUB ORDER BY D.O.APP DESC;

SQL>SELECT COACH_NAME, PAY, AGE, PAY*15 AS BONUS FROM CLUB;

SQL>INSERT INTO CLUB VALUE

(11,’RAJIV’,40,’HOCKEY’,’27-05-2000,’M’);

PROJECT QUESTION NO.7

Q7. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’GRADUATE’’.

TABLE : GRATUATE

S. NO. NAME STIPEND SUBJECT AVERAGE DIV1 KARAN 400 PHYSICS 68 12 DIVAKAR 450 COMPUTER SC 68 13 DIVYA 300 CHEMISRTY 62 24 ARUN 350 PHYSICS 63 15 SABINA 500 MATHS 70 16 JOHN 400 CHEMISTRY 55 27 ROBERT 250 PHYSICS 64 18 RUBINA 450 MATHS 64 19 VIKAS 500 COMPUTER SC 62 110 MOHAN 300 MATHS 57 2

*LIST THE NAMES OF THOSE STUDENTS WHO HAVE OBTAINED DIV = 1 SORTED BY NAME.

*DISPLAY THE REPORT, LISTING NAME, STIPEND, SUBJECT AND AMOUNT OF STIPEND RECEIVED INA YEAR ASSUMING THAT THE STIPNED IS PAID EVER MONTH.

*TO COUNT THE NUMBER OF STUDENTS WHO ARE EITHER PHYSICS OR COMPUTER SC GRADUATES.

*TO INSERT AN NEW ROW IN THE GRADUATE TABLE:

(11,’KAJOL’,300,’COMPUTER SC’,75,1);

SQL>SELECT NAME FROM GRADUATE WHERE DIV = ‘’1’’ AND ORDER BY NAMEL

SQL>SELECT NAME, STIPEND, SUBJECT, STIPEND*12 FROM GRADUATES;

SQL>SELECT COUNT (*) FROM GRADUATES WHERE SUBJECT = ‘’PHYSICS’’ OR ‘’COMPUTER SC’’;

SQL>INSERT INTO GRADUATES

(11,’KAJOL’,300,’COMPUTER SC’,75,1);


Q8. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’LAB’’.

TABLE : LAB S. NO.

ITEM NAME

COST /ITEM

QTY. D.O.PURCHASE WARRANTY OPERATIONAL

1 COMPUTER

6 000 9 21-05-96 2 7

2 PRINTER 15000 3 21-05-97 4 23 SCANNER 18000 1 29-08-98 3 14 CAMERA 21000 2 13-06-98 1 25 HUB 8000 1 31-10-99 2 16 UPS 5000 5 21-05-96 1 47 PLOTTER 25000 2 11-01-00 2 2

*TO SELECT THE ITEM NAME PURCHASE AFTER 31-10-97.

*TO LIST THE ITEM NAME, WHICH ARE WITHIN THE WARRANTY PERIOD TILL PRESENT DATE.

*TO LIST THE ITEM NAME IN ASCENING ORDER OF THE DATE OF PURCHASE WHERE QTY. IS MORE THAN 3.

*TO DISPLAY ITEM NAME, COST PER ITEM AND QTY. WHOSE WARRANTY IS OVER.

*TO COUNT THE NUMBER OF ITEMS WHOSE COST IS MORE THAN 10000.

*TO INSERT A NEW ROW IN THE LAB TABLE WITH FOLLOWING DATA:

(8,’VCR’,40,10000,2,’2-02-00’,1,2);

SQL>SELECT ITEM NAME FROM LAB WHERE DATE OF PURCHASE > 30-10-97;

SQL>SELECT ITEM NAME FROM LAB WHERE WARRANTY = ‘’1’’;

SQL>SELECT ITEM NAME FROM LAB ORDER BY QTY > 3;

SQL>SELECT ITEM NAME, COST PER ITEM, QUANTITY FROM LAB WHERE QUANTITY = ‘’NUL’’;

SQL>SELECT COUNT(*) FROM LAB WHERE COST PER ITEM > 10000;

SQL> INSERT INTO LAB VALUE

(8,’VCR’,40,10000,2,’2-02-00’,1,2);


Q9. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’FAMILY’’

TABLE : FAMILY

S. NO. NAME FEMALE MEMBER

MALE MEMBER

INCOME OCCUPATION

1 MISHRA 3 2 7000 SERVICE2 GUPTA 4 1 50000 BUSINESS3 KHAN 6 3 8000 MIXED4 CHADDHA 2 2 25000 BUSINESS5 YADAV 7 2 20000 MIXED6 JOSHI 3 2 14000 SERVICE7 MAURA 6 3 5000 FARMING8 RAO 5 2 10000 SERVICE

*TO SELECT ALL THE INFORMATION OF FAMILY WHOSE OCCUPATION IS SERVICE.

*TO LIST THE NAME OF FAMILY WHERE FEMALE NUMBER IS MORE THAN 3.

*TO LIST ALL NAME OF FAMILY WITH INCOME IN ASCENDING ORDER.

*TO DISPLAY FAMILY’S NAME, MALE MEMBERS AND OCCUPATION OF BUSINESS FAMILY.

*TO COUNT THE NUMBER OF FAMILIES WHOSE INCOME IS LESS THAN 1000.

*TO INSERT A NEW ROW IN THE FAMILY TABLE WITH FOLLOWING DATA:

(9,’D’SOUZA’,2,1,15000,’SERVICE’);

SQL>SELECT*FROM FAMILY WHERE OCCUPATION = ‘SERVICE’;

SQL>SELECT NAME FROM FAMILY WHERE FEMALE MEMBER > 3;

SQL>SELECT NAME, INCOME FROM FAMILY ORDER BY;

SQL>SELECT NAME, MALE MEMBER FROM FAMILY WHERE OCCUPATION = ‘BUSINESS’;

SQL>SELECT COUNT (*) FROM FAMILY INCOME < 1000;

SQL>INSERT INTO FAMILY VALUE

(9,’D’SOUZA’,2,1,15000,’SERVICE’);


Q10. WRITE SQL COMMAND FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’TRAIN’’

TABLE : TRAIN

S. NO. TRAI N NO. NAME CLASS SEAT NO. AGE FARE1 50001 ANU 1 11 16 7002 50002 SAGAR AC-CHAIR 34 43 13003 50003 SAMIR AC-1 2 78 90004 50004 GAURAV 11 23 17 6005 50005 KAPIL 1 3 25 17006 50006 ANKUR 11 6 20 5007 50007 PRIYA AC-11 1 11 1200

*TO DISPLAY THE NAME AND SEAT NO.WHERE CLASS IS 11.

*TO DISPLAY TRAIN NO. AND SEAT NO. WHERE THE FARE IS > 1300.

*TO DISPLAY LIST OF PASSENGER HAVING NAME STARTING WITH ‘A’ IN ASCENDING ORDER OF TRAIN NO.

*TO DISPLAY THE HIGHEST AMOUNT PAID AS FARE.

*TO DISPLAY THE TRAIN NO. AND THE NUMBER OF PASSENGER OF EACH TRAIN.

*TO INSERT A NEW ROW IN THE TRAIN TABLE WITH FOLLOWING DATA:

(8,’S0002’,’RAMANLAL’,’11’,2,66,3000);

SQL>SELECT NAME, SEAT NO.FROM TRAIN WHERE CLASS = ‘11’;

SQL>SELECT TRAIN NO, AND SEAT NO. FROM TRAIN WHERE FARE > 1300;

SQL>SELECT NAME FROM TRAIN WHERE NAME LIKE A% ORDER BY NAME;

SQL>SELECT MAX(FARE) FROM TRAIN;

SQL>SELECT DISTINCT TRAIN NO. AND NO. FROM TRAIN;

SQL>INSERT INTO TARIN VALUE

(8,’S0002’,’RAMANLAL’,’11’,2,66,3000);


Q11. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’SHOP’’

TABLE : SHOP

S. NO. SHOP NAME SALE AREA COST_% RATING CITY1 S.M. SONS 250000 WEST 68.6 C DELHI2 DHAROHAR 500000 SOUTH 81.8 A MUMBAI3 KRITI ART 300000 NORTH 79.8 A KOLKATA4 RIPPLE 380000 NORTH 88.0 A MUMBAI5 BISWAS 456000 EAST 92.0 A DELHI6 CRYSTA; 29000 SOUTH 66.7 A KOLKATA

*TO DISPLAY THE NAME OF ALL SHOPS WHICH ARE IN THE SOUTH.*TO DISPLAY NAME AND COUSTOMER PERCENTAGE OF ALL THE SHOPS HAVING COST_PERCENTAGE > 80.*TO DISPLAY LIST OF ALL THE SHOPS WITH SALE > 300000 IN ASCENDING ORDER OF SHOP_NAME, AND RATING FOR EACH SHOP IN THE TABLE, FOR ONLY THOSE SHOPS WHOSE

SALE IS BETWEEN 350000 AND 400000 (INCLUDING BOTH 35000 AND 400000).

*TO DISPLAY THE CITY AND THE NUMBER OF SHOPS IN EACH CITY IN THE TABLE SHOP.

*TO INSERT DETAIL OF A NEW SHOP IN THE TABLE SHOP WITH THE FOLLOWING DATA.

(7,’THE SHOP’,55000,’SOUTH’,90.8,’A’,’AHMDABAD’);

SQL>SELECT SHOP_NAME FROM SHOP WHERE AREA = ‘SOUTH’;SQL>SELECT SHOP_NAME, COST_PERCENT FROM SHOP WHERE COST_PERCENT > 80;

SQL>SELECT*FROM SHOP WHERE SALE > 300000 AND ORDER BY SHOP_NAME ASC;

SQL>SELECT SHOP_NAME, AREA, RATING FROM SHOP WHERE SALE BETWEEN 35000 AND 400000;

SQL>SELECT CITY, COUNT(*) FROM SHOP GROUP BY CITY;

SQL>INSERT INTO SHOP VALUE

(7,’THE SHOP’,55000,’SOUTH’,90.8,’A’,’AHMDABAD’);


Q12. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “FLIGHT”

TABLE : FLIGHT

S.NO FLIGHT NO. FROM CITY TO CITY DISTANCE SEATS FREQUENCY1 T1412 BLUE ISLAND YOYO TOWN 1500 210 32 A3242 PIKSWA YOYO TOWN 900 170 73 T1412 PERIMILA YOYO TOWN 1000 210 74 C3120 KETTY ISLAND BLUE ISLAND 1200 120 75 T1412 JASWAKA PERMILA 500 120 46 C3120 PIKSWAKA ARCHIC LAND 1000 170 47 A3242 HARAPAN ISLAND KETTY ISLAND 900 210 18 T1412 BLUE ISLAND HARAPAN VALLEY 1200 120 29 T1618 MUMSUI BLUE ISLAND 1200 210 4

*TO DISPLAY THE FLIGHT NO.OF THOSE FLIGHT WHICH HAVE FREQUENCY MORE THAN 4 TIMES IN A WEEK.

*DISPLAY FLIGHT NO.FROM CITY AND TO CITY OF ALL THOSE FLIGHT WHICH ARE STARTING FROM CITY BLUE ISLAND OR ARE HAVING DESTINATION AS BLUE ISLANDS.

*TO DISPLAY FLIGHT NO. FREQUENCY AND DISTANCE OF ALL THOSE FLIGHTS, WHICH ARE COVERING DISTANCE MORE THAN 900 KILOMETERS.

*TO DISPLAY FLIGHT NO.AND SEATING CAPACITY OF ALL THE THOSE FLIGHT IN WHICH SEATING CAPACITY IS MORE THAN 140.

*TO DISPLAY FLIGHT NO. OF ALL THOSE FLIGHTS WHICH ARE WEEKLY FLIGHTS AND HAVING SEATING CAPACITY LESS THAN 180.

*TO INSERT A NEW ROW IN THE FLIGHT TABLE WITH FOLLOWING DATA.

(9,’T1618’,’MUMSUI’,’BLUE ISLAND’,1200,210,4);

SQL>SELECT FLIGHT NO. FROM FLIGHT WHERE FREQUENCY > 4;

SQL>SELECT FLIGHT NO. FROM CITY FROM FLIGHT WHERE FROM CITY = ‘BLUE ISLAND’ OR TO CITY = ‘BLUE CITY’;

SQL>SELECT FLIGHT NO., SEAT FROM FLIGHT WHERE SEATS > 140;

SQL>SELECT FLIGHT NO., FREQUENCY, DISTANCE FROM FLIGHT WHERE DISTANCE > 900;

SQL>SELECT FLIGHT NO. FROM FLIGHT WHERE FREQUENCY = ‘1’ AND SEAT < 180;

SQL>INSERT INTO VLUE

(9,’T1618’,’MUMSUI’,’BLUE ISLAND’,1200,210,4);


Q13. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “LIBRARY”

TABL7E : LIBRARY

S.NO TITILE AUTHOR SUBJECT PUBLISHER QTY PRICE1 DATA STUCTURE LIPSSCHUTE DS MC GRAW 4 2172 DOS GUIDE NORTON OS PHI 3 1753 TURBOO C++ ROBOT LAFORE PROG GALGOTIA 5 2704 DBASE DUMIES PALMER DBMS PUSTAK M 7 1305 MAST.WINDOW COWART OS BPB 1 2256 COMP.STUDIES FRENCH FDN GALGOTIA 2 757 COBOL STERN PROG JOHN W 4 100008 GUIDE NETWOK FREED NET ZPRESS 3 2009 BACE FOR BEGINNNERS NORTON PROG BPB 3 4010 ADV. PASCAL SCHILTD PROG MCGRAW 4 350

*TO DISPLAY THE TITLE OF ALL THE BOOKS WITH PRICE BETWEEN 100 AND 300.

*TO DISPLAY TITLE AND AUTHOR OF ALL THE BOOKS HAVING TYPE PROG AND PUBLISHER BY BPB.

*TO DISPLAY LIST OF ALL THE BOOKS WITH THE PRICE MORE THAN 130 IN ASCENDING ORDER OF QTY.

*TO DISPLAY A REPORT WITH TITLE, MAISPLACEMENT CHARGES FOR EACH BOOK IN THE TABLE.

*TO DISPLAY THE PUBLISHER AND THE NUMBER OF BOOKS OF EACH PUBLISHER IN THE TABLE LIBRARY.

TO INSERT A NEW ROW IN THE TABLE LIBRARY WITH THE FOLLOWING DATA

(11,’COMPUTER SC.C++’,’RITA’,’KHANNA’,1225);

SQL>SELECT TITLE FROM LABRARY WHERE PRICE BETWEEN 100 AND 300;

SQL>SELECT TITLE, AUTHOR FROM LIBRARY WHERE SUBJECT = ‘PROG’ AND PUBLISHER = ‘BPB’;

SQL>SELECT TITLE FROM LIBRARY WHERE PRICE > 130 AND ORDER BY QTY;

SQL>SELECT TITLE, PRICE*1.5 AS ‘MISPLACEMENT CHARGES’ FROM LIBRARY;

MYDQL>SELECT DISTINCT PUBLISHER, COUNT(*) FROM LIBRARY GROUP BY PUBLISHER;

SQL>INSERT INTO LIBRARY VALUE

(11,’COMPUTER SC.C++’,’RITA’,’KHANNA’,1225);


Q14. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “BANK”

TABLE : BANK

ACC_NO. CNAME BNAME AMOUNT D.O.OPEN T_TRANSACTION1 KARAN B. OF BARODA 15000 12-01-98 102 PUNEET SATE BANK 2500 01-02-97 093 ANIRBAN ORIENTAL B. 17000 15-07-99 054 YATIN STND. CHAR. 38000 10-02-99 115 SUNNY STATE BANK 47000 06-02-98 156 JAYANT UCO BANK 34000 10-08-98 077 NIKHIL B. OF BARODA 56000 02-02-99 128 TARUN ORIENTAL B. 22000 04-04-99 089 JISHA UCO BANK 34500 05-01-98 11

*DISPLAY DATA FOR ALL CUSTOMER WHOSE TRANSACTION IS BETWEEN 8 AND 11.

*DISPLAY DATA FOR ALL CUSTOMER SORTED BY THEIR D.O.OPEN.

*TO COUNT THE NUMBER OF CUSTOMER WITH AMOUNT < 30000.

*LIST THE MINIMUM AND MAXIMUM AMOUNT FROM THE BANK.

*TO LIST CNAME, BNAME, AMOUNT FOR ALL THE CLIENTS WHOSE AMOUNT IS < 20000.

*TO DISPLAY ACC_NO., CNAME, BNAME, TOTAL TRANSACTION IN DESCENDING ORDER OF AMOUNT.

SQL>SELECT*FROM BANK WHERE T_TRANSACTION BETWEEN 8 AND 11;

SQL>SELECT*FROM BANK ORDER BY D.O.OPEN;

SQL>SELECT COUNT (*) FROM BANK WHERE AMOUNT < 30000;

SQL>SELECT MIN (AMOUNT) FROM BANK;

SQL>SELECT CNAME, BNAME, AMOUNT FROM BANK WHERE AMOUNT < 20000;

SQL>SELECT ACC_NO., BNAME, AMOUNT, T_TRANSACTION FROM BANK ORDER BY AMOUNT DESC;


Q15. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “VOTER”

TABLE : VOTER

V. NO V. NAME AGE ADDRESS PHONE1 DIVAKAR 22 ROHINI 270452492 RAJIV 23 SAROJINI NAGAR 255678923 SMITH 24 SAROJINI NAGAR 254566974 ARPIT 19 PASCHIM VIHAR 254578785 SUNNY 26 DEV NAGAR 254873046 SUNIL 23 VIKAS PURI 254787847 RAJIV 27 ROHINI 251463658 ROHIT 24 ROHINI 251469879 VIDHI 34 PITAM PURA 2544565610 ANAND 26 BANK VIHAR 25466566

*TO LIST V.NO., V. NAME, AGE FOR ALL THE VOTERS. THIS INFORMATION SHOULD BE SORTED ON V. NAME.

*TO LIST ALL THOSE VOTERS WHO ARE EITHER RESIDING IN ROHINI OR WHOSE AGE < 25.

*DISPLAY ALL THE VOTER WITH AGE > 27.

*LIST DIFFERENT VOTERS WITH UNIQUE AGE.

*COUNT THE NUMBER OF VOTER OF VOTERS WHER ADDRESS IS “ROHINI”.

*INSERT A NEW ROW IN THE VOTER TABLE. FILL THE ENTIRE COLUMN WITH VALUES

(11,’VAIBHAV’,18,’PASCHIM VIHAR’,’25255186’);

SQL>SELECT V. NO, AGE FROM VOTERS ORDER BY V.NAME;

SQL>SELECT*FROM VOTERS WHERE ADDRESS=’ROHINI’ OR AGE > 27;

SQL>SELECT DISTINCT AGE FROM VOTERS;

SQL>SELECT COUNT (*) FROM VOTERS WHERE ADDRESS = ‘ROHINI’;

SQL>INSERT VOTER VALUE

(11,’VAIBHAV’,18,’PASCHIM VIHAR’,’25255186’);


Q16. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “DIRECTORY”

TABLE : DIRECTORY

S. NO. FNAME LNAME PHONE ADDRESS1 ARPIT THUKRAL 43590465 PASCHIM VIHAR2 RAM SHARMA 27894563 VIKAS PURI3 VIKAS MALHOTRA 27123200 PITAM PURA4 ROHIT ARORA 27541552 PREET VIHAR5 KISAN KAUSHIK 27456566 ROHINI6 RAHUL VERMA 27889443 ROHINI7 RAKESH GULATI 27141145 PITAN PURA8 PARUL ARORA 27136464 ROHINI

*TO SELECT ALL THE INFORMATION FROM DIRECTORY OF ROHINI AREA.

*UPDATE THE DATABASE SET THE PHONE NO. AS 27047645 WHERE PHONE NO. IS 27123200.

*TO CREAT A VIEW CALLED DIRECTORY WITH THE FOLLOWING FIELDS FNAME, PHONE, ADDRESS.

*TO DISPLAY THE DATA FOR ARPIT, RAHUL, KISAN.

*TO DELETE THE ROW WHERE THE ADDRESS IS ROHINI.

*TO DELETE THE DATA PHYSICALLY.

SQL>SELECT* FROM DIRECTORY WHERE ADDRESS = “ROHINI”;

SQL>UPDATE DIRECTORY SET PHONE = “27047645” WHERE PHONE = “27123200”;

SQL>INSERT INTO DIRECTORY VALUE (FNAME VARCHAR(30), PHONE INTEGER, ADDRESS VARCHAR(30));

SQL>SELECT* FROM DIRECTORY WHERE FNAME = “ARPIT” AND “RAHUL” AND “KISAN”;

SQL>ALTER TABLE DIRECTORY DROP ADDRESS = “ROHINI”;

SQL>DROP TABLE DIRECTORY;


Q17. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “MOVIE”

TABLE : MOVIE

S.NO. TITLE TYPE RATING STARS QTY PRICE1 GONE WITH WIND DRAMA G GABLE 4 39.952 FRIDAY THE 13TH HORROR R JASON 2 69.953 TOP GUNS DRAMA PG CRUISE 7 49.954 SPLASH COMEDY PG 13 HANKS 3 29.955 INDEPENDENCE DAY DRAMA R TURNER 3 19.956 RISKY BUSINESS COMEDY R CRUISE 2 44.957 CANOON SCIFI PG AMECHE 2 31.958 CROCODILE DUNDEE COMEDY PG 13 HARIS 2 69.959 101 DALMATIANS COMEDY G HOFFMAN 3 59.9510 TOOTSIE COMEDY PG HOFFMAN 1 29.95

*FIND THE TOTAL VALUE OF THE MOVIE CASSETTES AVAILABLE IN THE LIBERARY.

*DIPLAY A LIST OF ALL THE MOVIE WITH THE PRICE OVER 20 AND SORTED BY PRICE.

*DIPLAY ALL THE MOVIE SORTED BY QTY. IN DESCENDING ORDER.

*DIPLAY A REPORT LISTING A MOVIE NUMBER, CURRENT VALUE AND REPLACEMENT VALUE FOR EACH MOVIE IN THE ABOVE TABLE. CALCULATE THE REPLACEMENT VALUE FOR ALL THE MOVIES AS QTY*PRICE*1.15.

*COUNT THE NUMBER OF MOVIE WHERE RATING IS NOT “G”.

*INSERT A NEW ROW IN THE MOVIE TABLE FILL ENTIRE COLUMN WITH VALUE (11,’FIMALDESTINATION’,’MYSTREOUS’,AE,’DANIEL’,4,’99.00’);

SQL>SELECT COUNT(*)FROM MOVIE.

SQL>SELECT TITLE FROM MOVIE WHERE PRICE > 20 AND ORDER BY PRICE;

SQL>SELECT TITLE FROM MOVIE ORDER BY DESC;

SQL>SELECT NO. TITLE FROM MOVIE WHERE QTY*PRICE*1.15;

SQL>SELECT COUNT(*) FROM MOVIE WHERE RATING = ‘G’;

SQL>INSERT INTO MOVIE VALUE

(11,’FIMALDESTINATION’,’MYSTREOUS’,AE,’DANIEL’,4,’99.00’);


Q18. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “PRODUCT”.

TABLE : PRODUCT

S. NO. NAME PRICE SUPPLIER STOCK1 MOTHER BOARD 7000 INTEL 202 KEYBOARD 1000 TVSE 703 MOUSE 500 LOGITECH 604 SOUND CARD 600 SAMSUNG 505 SPEAKER 600 SAMSUNG 256 MONITOR 3000 PHILIPS 227 CD-ROM 2800 CREATIVE 328 PRINTER 79000 HP 10

*TO DISPLAY DATA FOR THE ENTIRE ITEM SORTED BY THEIR NAME.

*TO DISPLAY THE NAME AND PRICE FROM THE TABLE ITEM IN DESCENDING ORDER OF THEIR STOCK.

*LIST ALL THE NAMES AND PRICE WITH PRICE BETWEEN 3000 AND 7000.

*WRITE A SQL COMMAND TO DELETE ROWS WITH STOCK BETWEEN 20 TO 40.

*TO COUNT THE NUMBER OF PRODUCTS WITH THE STOCK < 5.

SQL>SELECT*FROM PRODUCT ORDER BY NAME;

SQL>SELECT NAME, PRICE FROM PRODUCT ORDER BY STOCK DESC;

MYDQL>SELECT NAME, PRICE FROM PRODUCT WHERE PRICE BETWEEN 3000 AND 7000.

SQL>DELETE FROM PRODUCT WHERE STOCK BETWEEN ‘20’ OR ‘40’;

SQL>SELETC COUNT (*) FROM PRODUCT WHERE STOCK < 5;


Q19. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “EMP_SALARY”.

TABLE : EMP_SALARY

EMP_ID FNAME LNAME COUNTRY

SALARY BENEFITS DESIGNATION

010 GEORGE SMITH US 75000 15000 MANAGER105 MARY JONES UK 65000 15000 MANAGER152 SAM TONES FRANCE 80000 25000 DIRECTOR215 SARAH ACKER US 75000 12500 NAMAGER244 MANILA SEN USA 50000 12000 CLERK300 RABERT SAMUEL US 45000 10000 CLERK335 HENRY WILLIAM US 40000 10000 CLERK400 RACHEL LEE UK 32000 7500 SALESMAN441 PETER THOMAS FRANCE 28000 7500 SALESMAN

*TO DISPLAY FNAME, LNAME, COUNTRY AND SALARY OF ALL THE EMPLOYEES LIVING IN US FROM THE TABLE EMP_SALARY.

*TO DISPLAY THE CONTENTS OF EMP_SALARY TABLE IN DESCENDING ORDER OF FNAME.

*TO DISPLAY THE FNAME, LNAME, SALARY ALL THE MANAGER FROM THE TABLE.

*TO DISPLAY THE MAXIMUM SALARY AMONG MANAGER AND CLERK FROM THE TABLE.

*TO DELETE THE TABLE PHYSICALLY.

SQL>SELECT FNAME, LNAME, COUNTRY, SALARY FROM EMP_SALARY WHERE COUNTRY = “US”;

SQL>SELECT * FROM EMP_SALARY ORDER BY FNAME DESC;

SQL>SELECT FNAME, LNAME, SALARY FROM EMP_SALARY WHERE DESIGNATION = ‘MANAGER’;

SQL>SELECT MAX SALARY FROM EMP_SALARY DESIGNATION = ‘MANAGER’ OR ‘CLERK’;

SQL>DROP TABLE EMP_SALARY;


Q20. WRITE SQL COMMANDS FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE “FURNITURE”.

TABLE : FURNITURE S. NO. ITEM NAME TYPE D.O.STOCK PRICE DISCOUNT1 WHITE LOTUS DOUBLE BED 23-03-2002 30000 252 PINK FEATURE BABY COT 20-01-2002 7000 253 DOLPHIN BABY COT 29-02-2002 9500 204 DECENT OFFICE TABLE 02-01-2002 25000 305 COMFORT ZONE DOUBLE BED 12-02-2002 25000 236 DONALD BABY COT 23-03-2002 6500 157 ROYAL FINISH OFFICE TABLE 24-03-2002 18000 308 ROYAL TIGER SOFA 24-02-2002 31000 309 ECONO SETTING SOFA 28-01-2002 95-- 2510 EATING PARADISE DINNING TABLE 1-02-2002 11500 25

*TO SHOW ALL THE INFORMATION ABOUT THE BABY COTS FROM THE FURNITURE TABLE.

*TP LIST THE ITEM NAME WHICH ARE PRICED AT MORE THAN 15000 FROM THE FURNITURE TABLE.

*TO LIST ITEM NAME AND TYPE OF THOSE ITEM, IN WHICH DATE OF STOCK IS BEFORE 01-02-2002

FROM THE FURNITURE TABLE IN DESCENDING ORDER OF ITEM NAME.

*TO DISPLAY ITEM NAME AND DATE OF STOCK OF THOSE ITEMS, IN WHICH THE DISCOUNT PRECENTAGE IS MORE THAN 25 FROM FURNITURE TABLE.

*TO COUNT THE NUMBER OF ITEMS, WHOSE TYPE IS “SOFA” FROM FURNITURE TABLE.

SQL>SELECT*FROM FURNITURE WHERE TYPE = ‘BABY COT’;

SQL>SELECT ITEM NAME FROM FURNITURE WHERE PRICE > 15000;

SQL>SELECT ITEM NAME, TYPE FROM FURNITURE WHERE D.O.STOCK < 01-02-2002 ORDER BY ITEM NAME DESC;

SQL>SELECT ITEM NAME, D.O.STOCK FROM FURNITURE WHERE DISCOUNT > 25;

SQL>SELECT COUNT (*) FROM FURNITURE WHERE TYPE = “SOFA”;


Q21. WRITE SQL COMMAND FOR THE FOLLOWING ON TH BASIS OF GIVEN TABLE BOOKS”

TABLE : BOOKS

BOOK ID BOO K NAME AUTHOR NAME PUBLISHER PRICE TYPE QTY.

1 FAST COOK LATA KAPOOR EPB 355 COOKERY 52 THE TEARS WILLIAM SMITH FIRST PUBL. 650 FICTION 203 MY FIRST C++ BRAIN & BROOKE EPB 350 TEXT 104 C++ BRAINWRKS A.W. ROSSAINE TDH 350 TEXT 155 THUNDERBOLTS ANNA ROBERT FIRST PUBL. 750 FICTION 50

*TO SHOW BOOK NAME, AUTHOR NAME AND PRICE OF BOOKS OF FIRST PUBL. PUBLISHER.

*TO LIST ALL THE NAME FROM BOOKS OF TEXT TYPE.

*TO DISPLAY THE NAMES AND PRICE FROM BOOKS IN ASCENDING ORDER OF THEIR PRICE.

*TO INCREASE THEIR PRICE OF ALL THE BOOKS OF EPB PUBLISHER BY 50.

SQL>SELECT BOOK NAME, AUTHOR NAME, PRICE FROM BOOKS WHERE PUBLISHER = “FIRST PUBL”;

SQL>SELECT BOOK NAME FROM BOOKS WHERE TYPE = “TEXT”;

SQL>SELECT BOOK NAME, PRICE FROM BOOKS ORDER BY PRICE ASC;

SQL>UPDATE BOOKS SET PRICE WHERE PUBLISHER = “EPB” AND PRICE = “PRICE+50”;


Q22. WRITE SQL COMMAND FOR THE FOLLOWING ON TH BASIS OF GIVEN TABLE GARMENTS

TABLE : GARMENTS

G.CODE DESCRIPTION PRICE F.CODE READY DATE10023 PENCIL SKIRT 1150 F03 19-02-200810001 FROMAL SHIRT 1250 F01 12-01-200810012 INFORMAL SHIRT 1550 F02 06-01-200810024 BABY COT 750 F03 07-04-200710090 TULIP SKIRT 850 F02 31-03-200710019 EVENING GOWN 850 F03 06-01-200810009 INFORMAL PANTS 1500 F02 20-10-200810007 FORMAL PANTS 1350 F01 09-03-200810020 FROCK 850 F04 09-09-200710089 SLACKS 750 F03 20-10-2007

*TO DISPLAY G.CODE AND DESCRIPTION OF EACH GARMENT IN DESCINDING ORDER OF G.CODE.

*TO DISPLAY THE DETAILS OF ALL THE GARMENTS, WHICH HAVE READY DATE IN BETWEEN 08-12-2008 AND 16-06-2008.

*TO DISPLAY THE AVERAGE PRICE OF ALL THE GARMENTS, WHICH ARE MADE UP OF FABRIC WITH F.CODE AS F03.

*TO DISPLAY FABRIC WISE HIGHEST AND LOWEST PRICE OF GARMENTS FROM GARMENTS TABLE. (DISPLAY F.CODE OF EACH GARMENT ALONG WITH HIGHEST AND LOWEST PRICE.)

SQL>SELECT DISTINCT G.CODE, DESCRIPTION FROM GARMENT ORDER BY G. CODE DESC;

SQL>SELECT * FROM GARMENT WHERE READY DATE BETWEEN 08-12-1007 AND 16-06-2008;

SQL>SELECT AVG PRICE FROM GARMENT WHERE FCODE = “F03”;

SQL>SELECT MAX FCODE FROM GARMENT ORDER BY PRICE;


Q23. WRITE SQL COMMAND FOR THE FOLLOWING ON TH BASIS OF GIVEN TABLE “STORE”.

TABLE : STORE

ITEM NO. ITEM SCODE QTY RATE LAST BUY2005 SHARPENER CLASSIC 23 60 6 31-06-20092003 BALL PEN NO.25 22 50 25 01-02-20102002 GEL PEN PREMIUM 21 150 12 24-02-20102006 GEL PEN CLASSIC 21 250 20 11-03-20092001 ERASER SMALL 22 220 6 19-01-20092004 ERASER BIG 22 110 8 02-12-10092009 BALL PEN 0.5 21 180 18 03-11-2009

*TO DISPLAY DETAILS OF ALL THE ITEMS IN THE STORETABLE IN ASCENDING ORDER OF LAST BUY.

*TO DISPLAY ITEM NO.AND ITEM NAME OF ALL THOSE ITEMS FROM STORE TABLE WHOSE RATE IS MORE THAN 15 RUPEES.

*TO DISPLAY THE DETAILS OF THOSE ITEMS WHOSE SUPPLIER CODE (SCODE) IS 22 OR QTY IN STORE (QTY) IS MORE THAN 110 FROM THE TABLE STORE.

*TO DISPLAY MINIMUM RATE OF ITEMS FOR EACH SUPPLIER INDIVIDUALLY AS PER SCODE FROM THE TABLE STORE.

SQL>SELECT * FROM STORE ORDER BY LAST BUY;

SQL>SELECT ITEM NO., ITEM NAME FROM STORE WHERE RATE > 15;

SQL>SELECT * FROM STORE WHERE SCODE = “22” AND QTY > 110;

SQL>SELECT DISTINCT MIN RATE FROM STORE ORDER BY SCODE;


Q24. WRITE SQL COMMAND FOR THE FOLLOWING ON TH BASIS OF GIVEN TABLE “DRESS’’

TABLE : DRESS

DCODE DESCRIPTION PRICE MCODE LAUNCH DATE10001 FORMAL SHIRT 1250 M001 2008-01-1210020 FROCK 750 M004 2007-09-0910012 INFORMAL SHIRT 1450 M002 2008-06-0610019 EVENING GOWN 850 M003 2007-06-0610090 TULIP SKIRT 850 M002 2008-03-3110023 PENCIL SKIRT 1250 M003 2008-12-1910089 SLACKS 850 M003 2008-10-2010007 FORMAL PANT 1450 M001 2008-03-0910009 IMFORMAL

PANTS14050 M002 2008-10-20

10024 BABY TOP 650 M003 2007-04-07

*TO DISPLAY DCODE AND DESCRIPTION OF EACH DRESS IN DESCENDING ORDER OF CODE.

*TO DISPLAY THE DETAILS OF ALL THE DRESS WHICH HAVE LAUNCH DATE IN BETWEEN 2007-12-05 AND 2008-06-08.

*TO DISPLAY THE AVERAGE PRICE OF ALL THE DRESS WHICH ARE MADE UP OF MATERIAL WITH MCODE AS M003.

*TO DISPLAY MATERIAL WISE HIGHEST AND LOWEST PRICE OF DRESS FROM DRESS TABLE.

SQL>SELECT DCODE, DESCCRIPTION FROM DRESS ORDER BY DCODE DECS;

SQL>SELECT*FROM DRESS WHERE LAUNCH DATE BETWEEN 2007-12-05 AND 2008-06-08;

SQL>SELECT AVG PRICE FROM DRESS WHERE MCODE = ‘M003’;

SQL>SELECT MAX PRICE, MIN PRICE FROM DRESS;


Q25. WRITE SQL COMMAND FOR THE FOLLOWING ON THE BASIS OF GIVEN TABLE ‘’COACHING’’.

TABLE : COACHING

NAME AGE DEPARTMENT

D.O.ADM FEE SEX

RAHUL 24 HINDI 2009-01-10 2000 MSHALU 21 COMPUTER 2008-03-24 1200 FABHA 23 HISTORY 2010-12-09 3000 FSUMAN 25 HISTORY 2001-09-09 2500 FKULKARNI 22 COMPUTER 2002-05-18 4000 MROHIT 25 HINDI 2009-15-17 3500 M

*TO SHOW THE INFORMATION ABOUT THE STUDENT OF COMPUTER DEPARTMENT.

*TO LIST NAMES OF MALE STUDENT WHO ARE IN HINDI DEPARTMENT.

*TO LIST THE NAMES OF ALL THE STUDENT IN ASCENDING ORDER OF THEIR AGE.

*TO DISPLAY AGE, FEE OF FEMALE STUDENT WHO PAY MORE THAN 350.

*TO DISPLAY ALL THE NAMES THAT END WITH ‘’NI’’.

*TO DISPLAY THE DISTINCT DEPARTMENT.

SQL>SELECT * FROM COACHING WHERE DEPARTMENT = ‘’COMPUTER’’;

SQL>SELECT NAME FROM COACHING WHERE SEX = ‘M’ AND DEPARTMENT = ‘HINDI’;

SQL>SELECT NAME FROM COACHING ORDER BY AGE ASC;

SQL>SELECT AGE, FEE FROM COACHING WHERE SEX = ‘F’ AND FEE > 350;

SQL>SELECT NAME FROM COACHING WHERE NAME LIKE ‘NI%’;

SQL>SELECT COUNT (*) FROM COACHING WHERE AGE > 22;

SQL> SELECT DISTINCT DEPARTMENT FROM COACHING;

Ip Project File-12th Class

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