Relational Database and SQL — Question 35
Back to all questionsWrite SQL Commands/output for the following on the basis of the 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 |
(i) List the names of those students who have obtained rank 1 sorted by NAME.
(ii) Display a list of all those names whose AVERAGE is greater than 65.
(iii) Display the names of those students who have opted COMPUTER as a SUBJECT with an AVERAGE of more than 60.
(iv) List the names of all the students in alphabetical order.
(v) SELECT * FROM GRADUATE WHERE NAME LIKE "%I%";
(vi) SELECT DISTINCT RANK FROM GRADUATE;
(i)
SELECT NAME
FROM GRADUATE
WHERE `RANK` = 1
ORDER BY NAME;+--------+
| NAME |
+--------+
| DIVYA |
| GAURAV |
| KARAN |
| LIZA |
| PUJA |
| RAJ |
| VARUN |
+--------+
(ii)
SELECT NAME
FROM GRADUATE
WHERE AVERAGE > 65;+--------+
| NAME |
+--------+
| KARAN |
| RAJ |
| GAURAV |
| LIZA |
+--------+
(iii)
SELECT NAME
FROM GRADUATE
WHERE SUBJECT = 'COMPUTER' AND AVERAGE > 60;+------+
| NAME |
+------+
| LIZA |
+------+
(iv)
SELECT NAME
FROM GRADUATE
ORDER BY NAME;+--------+
| NAME |
+--------+
| DEEP |
| DIVYA |
| GAURAV |
| KARAN |
| LIZA |
| MANAV |
| NISHA |
| PUJA |
| RAJ |
| VARUN |
+--------+
(v) SELECT * FROM GRADUATE WHERE NAME LIKE "%I%";
+-------+-------+---------+-----------+---------+------+
| S.No. | name | stipend | subject | average | rank |
+-------+-------+---------+-----------+---------+------+
| 4 | DIVYA | 350 | CHEMISTRY | 63 | 1 |
| 8 | LIZA | 450 | COMPUTER | 68 | 1 |
| 10 | NISHA | 300 | COMPUTER | 57 | 2 |
+-------+-------+---------+-----------+---------+------+
(vi) 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;
+------+
| RANK |
+------+
| 1 |
| 2 |
+------+