HI,
I actually posted similar (or same?) question yesterday, but I thought I need to post a new question since I have short, but clear question.
I have the following table.
id point
1 30
2 30
3 29
4 27
5 28
6 26
what I want:
get all the users order by rank. user #1 and #2 should have 1 as their rank value because they both have 30 points
I want to query a rank by user id. I like to get 1 as the result of my rank when I query user #1 and #2 because both of them have 30 points
Added on: 3/18
I tried Logan's query, but got the following result
id point rank
1 30 1
2 30 1
3 29 3
4 27 5
5 28 4
6 26 6
The subquery approach that you have seen recommended will scale quadratically. http://www.xaprb.com/blog/2006/12/02/how-to-number-rows-in-mysql/ shows a much more efficient approach with user variables. Here is an untested adaptation to your problem:
@points := -1; // Should be an impossible value.
@num := 0;
SELECT id
, points
, @num := if(@points = points, @num, @num + 1) as point_rank
, @points := points as dummy
FROM `users`
ORDER BY points desc, id asc;
Just count how many people have more points then them.
select count(1) from users
where point > (select point from users where id = 2) group by point
This will give you the number of people that have more points for the given user. So for user 1 and user 2 the result will be 0 (zero) meaning they are first.
When I needed to do something similar, I created a view that looked like this:
CREATE VIEW rankings_view
AS
SELECT id
, point
, (select count(1)
from points b
where b.point > a.point) +1 as rank
FROM points as a;
This assumes that the original table was named points, obviously. Then you can get the rank of any id, or the id corresponding to any rank, by querying the view.
EDIT
If you want to count the number of distinct point values above each point value instead of the number of entries with point values above the current point value, you can do something like:
CREATE VIEW rankings_view2
AS
SELECT id
, point
, (SELECT COUNT(1) +1 AS rank
FROM ( SELECT DISTINCT point
FROM points b
WHERE b.point >a.point ))
FROM points AS a;
NOTE
Some of the other solutions presented definitely perform better than this one. They're mysql specific, so I can't really use them for what I'm doing. My application has, at most, 128 entities to rank, so this works well enough for me. If you might have tons of rows, though, you might want to look at using one of the other solutions presented here or limiting the scope of the ranking.
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