Write SQL commands for (a) to (d) and write the output for (e) and (f) on the basis of given table GRADUATE:

Table: GRADUATE

S NO	NAME	STIPEND	SUBJECT	AVERAGE	RANK
1	KARAN	400	PHYSICS	68	1
2	RAJ	450	CHEMISTRY	68	1
3	DEEP	300	MATHS	62	2
4	DIVYA	350	CHEMISTRY	63	1
5	GAURAV	500	PHYSICS	70	1
6	MANAV	400	CHEMISTRY	55	2
7	VARUN	250	MATHS	64	1
8	LIZA	450	COMPUTER	68	1
9	PUJA	500	PHYSICS	62	1
10	NISHA	300	COMPUTER	57	2

(a) List the names of those students who have obtained rank 1.

(b) Display a list of all those names whose average is greater than 65.

(c) Display the names of those students who have opted for computer as a subject with average of more than 60.

(d) List the names of all students whose name ends with 'a'.

(e) SELECT * FROM GRADUATE WHERE Subject = "Physics";

(f) SELECT DISTINCT RANK FROM GRADUATE;

(a)

SELECT NAME FROM GRADUATE 
WHERE `RANK` = 1;

Output

+--------+
| NAME   |
+--------+
| KARAN  |
| RAJ    |
| DIVYA  |
| GAURAV |
| VARUN  |
| LIZA   |
| PUJA   |
+--------+

(b)

SELECT NAME
FROM GRADUATE
WHERE AVERAGE > 65;

Output

+--------+
| NAME   |
+--------+
| KARAN  |
| RAJ    |
| GAURAV |
| LIZA   |
+--------+

(c)

SELECT NAME
FROM GRADUATE
WHERE SUBJECT = 'COMPUTER' AND AVERAGE > 60;

Output

+------+
| NAME |
+------+
| LIZA |
+------+

(d)

SELECT NAME FROM GRADUATE 
WHERE NAME LIKE "%a";

Output

+-------+
| NAME  |
+-------+
| DIVYA |
| LIZA  |
| PUJA  |
| NISHA |
+-------+

(e)

SELECT * FROM GRADUATE WHERE Subject = "Physics";

Output

+-------+--------+---------+---------+---------+------+
| S.No. | name   | stipend | subject | average | RANK |
+-------+--------+---------+---------+---------+------+
|     1 | KARAN  |     400 | PHYSICS |      68 |    1 |
|     5 | GAURAV |     500 | PHYSICS |      70 |    1 |
|     9 | PUJA   |     500 | PHYSICS |      62 |    1 |
+-------+--------+---------+---------+---------+------+

(f) Since 'RANK' is a reserved keyword in SQL, we encounter an error while running this query. To avoid such errors, we can enclose the column name 'RANK' in backticks to treat it as a literal identifier.

The corrected query is :

SELECT DISTINCT `RANK` FROM GRADUATE;

Output

+------+
| RANK |
+------+
|    1 |
|    2 |
+------+