Reputation: 540
sorry to ask so many questions about SQL but I am doing exercises that help me revise for my test.
I am trying to display the names of all people who have 2 Ls in their name and are in depno = 30 or their super is "7782". I have written some code for it but its giving an error. I have used the Column name "SUPER" for the manager.
SELECT ENAME
FROM emp
WHERE ENAME LIKE 'L%', DEPTNO = 30;
OR SUPER = '7782';
Thanks alot again!
-Jay
Found the answer: The answer should actually be:
SELECT ENAME
FROM emp
WHERE ENAME LIKE '%L%L%' AND DEPTNO = 30
OR SUPER = '7782';
because the other codes given here give me an error as the end of third line is not supposed to have a semicolon.
Upvotes: 1
Views: 2380
Reputation:
Select *
from yourtable
where name like '%L%L%'
or depno = 30
or super = 7782
Upvotes: 2
Reputation: 3318
This should do it:
SELECT ENAME
FROM emp
WHERE ENAME LIKE '%L%L%', DEPTNO = 30;
OR SUPER = '7782';
Be aware that the LIKE '%L' might prevent the usage of an index.
Upvotes: 3