Re: Record Position in Ordered Selection
That looks like quite a useful reference in general. Thanks!
I don't fully understand the last query in the specified section, but this part of it seems to do what I need:
SELECT
s1.name, COUNT(s2.name) rank
FROM votes s1
JOIN votes s2 ON s1.votes < s2.votes or (s1.votes=s2.votes and s1.name = s2.name)
where s1.name = 'White';
I wonder about its efficiency, though. There are going to be thousands of listings, hopefully eventually tens or even hundreds of thousands. Am I going to be hammering the db server with this? What indices on the ranking table might help?
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Re: Record Position in Ordered Selection
June 07, 2009 07:05AM
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