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Structure Query Language
Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables
TABLE: BOOK
|
Code |
BNAME |
TYPE |
|
F101 |
The priest |
Fiction |
|
L102 |
German easy |
Literature |
|
C101 |
Tarzan in the lost world |
Comic |
|
F102 |
Untold Story |
Fiction |
|
C102 |
War Heroes |
Comic |
TABLE: MEMBER
|
MNO |
MNAME |
CODE |
ISSUEDATE |
|
M101 |
RAGHAV SINHA |
L102 |
2016-10-13 |
|
M103 |
SARTHAK JOHN |
F102 |
2017-02-23 |
|
M102 |
ANISHA KHAN |
C101 |
2016-06-12 |
(i) To display all details from table MEMBER in descending order of ISSUEDATE.
(ii) To display the BNO and BNAME of all Fiction Type books from the table BOOK
(iii) To display the TYPE and number of books in each TYPE from the table BOOK
(iv) To display all MNAME and ISSUEDATE of those members from table MEMBER who have books issued (i.e ISSUEDATE) in the year 2017.
(v)SELECT MAX(ISSUEDATE) FROM MEMBER;
(vi) SELECT DISTINCT TYPE FROM BOOK;
(vii) SELECT A.CODE, BNAME, MNO, MNAME FROM BOOK A, MEMBER B WHERE A.CODE=B.CODE ;
(viii) SELECT BNAME FROM BOOK
WHERE TYPE NOT IN ('FICTION', 'COMIC');
(i)SELECT * FROM MEMBER ORDER BY ISSUEDATE DESC;
(ii) SELECT Code, BNAME FROM BOOK WHERE TYPE=’Fiction’;
(iii) SELECT COUNT(*), TYPE FROM BOOK GROUP BY TYPE;
(iv) SELECT MNAME, ISSUEDATE FROM MEMBER WHERE ISSUEDATE>=’2017-01-01’ AND ISSUEDATE<=’2017-12-31’;
(v)MAX(ISSUEDATE)
2017-02-23
(vi) DISTINCT TYPE
Fiction
Literature
Comic
(vii)
|
CODE |
BNAME |
MNO |
MNAME |
|
L102 |
The priest |
M101 |
RAGHAV SINHA |
|
F102 |
Untold Story |
M103 |
SARTHAK JOHN |
|
C101 |
Tarzan in the lost world |
M102 |
ANISHA KHAN |
(viii) BNAME
German Easy
Some More Questions From Structure Query Language Chapter
Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables
TABLE: BOOK
Code
BNAME
TYPE
F101
The priest
Fiction
L102
German easy
Literature
C101
Tarzan in the lost world
Comic
F102
Untold Story
Fiction
C102
War Heroes
Comic
TABLE: MEMBER
MNO
MNAME
CODE
ISSUEDATE
M101
RAGHAV SINHA
L102
2016-10-13
M103
SARTHAK JOHN
F102
2017-02-23
M102
ANISHA KHAN
C101
2016-06-12
(i) To display all details from table MEMBER in descending order of ISSUEDATE.
(ii) To display the BNO and BNAME of all Fiction Type books from the table BOOK
(iii) To display the TYPE and number of books in each TYPE from the table BOOK
(iv) To display all MNAME and ISSUEDATE of those members from table MEMBER who have books issued (i.e ISSUEDATE) in the year 2017.
(v)SELECT MAX(ISSUEDATE) FROM MEMBER;
(vi) SELECT DISTINCT TYPE FROM BOOK;
(vii) SELECT A.CODE, BNAME, MNO, MNAME FROM BOOK A, MEMBER B WHERE A.CODE=B.CODE ;
(viii) SELECT BNAME FROM BOOK
WHERE TYPE NOT IN ('FICTION', 'COMIC');
TABLE: BOOK
Code
BNAME
TYPE
F101
The priest
Fiction
L102
German easy
Literature
C101
Tarzan in the lost world
Comic
F102
Untold Story
Fiction
C102
War Heroes
Comic
MNO
MNAME
CODE
ISSUEDATE
M101
RAGHAV SINHA
L102
2016-10-13
M103
SARTHAK JOHN
F102
2017-02-23
M102
ANISHA KHAN
C101
2016-06-12
(ii) To display the BNO and BNAME of all Fiction Type books from the table BOOK
(iii) To display the TYPE and number of books in each TYPE from the table BOOK
(iv) To display all MNAME and ISSUEDATE of those members from table MEMBER who have books issued (i.e ISSUEDATE) in the year 2017.
(v)SELECT MAX(ISSUEDATE) FROM MEMBER;
(vi) SELECT DISTINCT TYPE FROM BOOK;
(vii) SELECT A.CODE, BNAME, MNO, MNAME FROM BOOK A, MEMBER B WHERE A.CODE=B.CODE ;
(viii) SELECT BNAME FROM BOOK
WHERE TYPE NOT IN ('FICTION', 'COMIC');
Observe the following STUDENTS and EVENTS tables carefully and write the name of the RDBMS operation which will be used to produce the output as shown in LIST? Also, find the Degree and Cardinality of the LIST.
STUDENTS
No
Name
1
Tara Mani
2
Jaya Sarkar
3
Tarini Trikha
EVENTS
EVENTCODE
EVENTNAME
1001
Programming
1002
IT Quiz
LIST
No
Name
EVENTCODE
EVENTNAME
1
Tara Mani
1001
Programming
1
Tara Mani
1002
IT Quiz
2
Jaya Sarkar
1001
Programming
2
Jaya Sarkar
1002
IT Quiz
3
Tarini Trikha
1001
Programming
3
Tarini Trikha
1002
IT Quiz
Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables
Table: VEHICLE
CODE
VTYPE
PERKM
101
VOLVO BUS
160
102
AC DELUXE BUS
150
103
ORDINARY BUS
90
105
SUV
40
104
CAR
20
Note:
- PERKM is Freight Charges per kilometre
- VTYPE is Vehicle Type
Table: TRAVEL
NO
NAME
TDATE
KM
CODE
NOP
101
Janish
2015-11-13
200
101
32
103
Vedika Sahai
2016-04-21
100
103
45
105
Tarun Ram
2016-03-23
350
102
42
102
John Fen
2016-02-13
90
102
40
107
Ahmed Khan
2015-01-10
75
104
2
104
Raveena
2016-05-28
80
105
4
106
Kripal Anya
2016-02-06
200
101
25
Note:
• NO is Traveller Number
• KM is Kilometer travelled
• NOP is number of travellers travelled in vehicle
• TDATE is Travel Date
- To display NO, NAME, TDATE from the table TRAVEL in descending order of NO.
- To display the NAME of all the travellers from the table TRAVEL who are travelling by vehicle with code 101 or 102.
- To display the NO and NAME of those travellers from the table TRAVEL who travelled between ‘2015â12â31’ and ‘2015â04â01’.
- To display all the details from table TRAVEL for the travellers, who have travelled distance more than 100 KM in ascending order of NOP.
- SELECT COUNT(*), CODE FROM TRAVEL
GROUP BY CODE HAVING COUNT(*)>1;
- SELECT DISTINCT CODE FROM TRAVEL;
- SELECT A.CODE,NAME,VTYPE
FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B.CODE AND KM<90;
- SELECT NAME,KM*PERKM
FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B.CODE AND A.CODE=’105’;
- PERKM is Freight Charges per kilometre
- VTYPE is Vehicle Type
• NO is Traveller Number
• KM is Kilometer travelled
• NOP is number of travellers travelled in vehicle
• TDATE is Travel Date
GROUP BY CODE HAVING COUNT(*)>1;
FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B.CODE AND KM<90;
FROM TRAVEL A,VEHICLE B
WHERE A.CODE=B.CODE AND A.CODE=’105’;
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