How to use query dense_Rank() and rank()?

Question

3.67/5 (2 votes)

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SQL

SELECT Name, rank() OVER (ORDER BY Score ASC) AS RankScore
FROM Rank

But it still can't work. Anyone, please help. Thanks.

Posted 18-Jun-14 5:03am

Ananta Pratama

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4 solutions

Solution 1

Try like this..

select score, DENSE_RANK() over(order by score) as rank from Rank1

Posted 18-Jun-14 5:44am

priya dharshan

Comments

Ananta Pratama 18-Jun-14 12:27pm

Sorry, but it's still can't work.. Here's the warning from my SQL : #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(order by score) as rank from Rank LIMIT 0, 30' at line 1
Why it can happen?

Solution 2

Check this out[^]

Posted 18-Jun-14 6:41am

Peter Leow

Solution 4

i thought you are try to work in mysql look at this link[^]

Posted 18-Jun-14 22:52pm

Thava Rajan

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King Fisher · Accepted Answer · 2014-06-18T08:01:00

Solution 3

ascending:

SQL

SELECT Name, rank() OVER (ORDER BY Score Asc) AS rankscore
FROM Rank

Descending:

SQL

SELECT Name, rank() OVER (ORDER BY Score DESC ) AS rankscore
FROM Rank

Posted 18-Jun-14 8:01am

King Fisher

Updated 18-Jun-14 8:04am

v2