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 kilometer
•VTYPE is Vehicle Type
| NO |
NAME |
TDATE |
KM |
CODE |
NOP |
| 101 |
Janish Kin |
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 |
AhmedKhan |
2015-01-10 |
75 |
104 |
2 |
| 104 |
Raveena |
2015-05-28 |
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
(i) To display NO, NAME, TDATE from the table TRAVEL in descending order of NO.
(ii) To display the NAME of all the travellers from the table TRAVEL who are travelling by vehicle with code 101 or 102.
(iii) To display the NO and NAME of those travellers from the table TRAVEL who travelled between ‘2015-12-31 ’and ‘2015-04-01 ’.
(iv) To display all the details from table TRAVEL for the travellers, who have travelled distance more than 100 KM in ascending order of NOP.
(v) SELECT COUNT (*), CODE FROM TRAVEL GROUP BY CODE HAVING COUNT(*)>1;
(vi) SELECT DISTINCT CODE FROM TRAVEL;
(vii) SELECT A.CODE,NAME,VTYPE FROM TRAVEL A,VEHICLE B WHERE A.CODE=B. CODE AND KM <90;
(viii) SELECT NAME,KM*PERKM FROM TRAVEL A, VEHICLE B WHERE A.CODE=B.CODE AND A.CODE=‘105 ’;