MySQL, Get users rank

有些话、适合烂在心里 提交于 2019-11-26 22:42:36
SELECT  uo.*, 
        (
        SELECT  COUNT(*)
        FROM    users ui
        WHERE   (ui.points, ui.id) >= (uo.points, uo.id)
        ) AS rank
FROM    users uo
WHERE   id = @id

Dense rank:

SELECT  uo.*, 
        (
        SELECT  COUNT(DISTINCT ui.points)
        FROM    users ui
        WHERE   ui.points >= uo.points
        ) AS rank
FROM    users uo
WHERE   id = @id

Solution by @Quassnoi will fail in case of ties. Here is the solution that will work in case of ties:

SELECT *,
IF (@score=ui.points, @rank:=@rank, @rank:=@rank+1) rank,
@score:=ui.points score
FROM users ui,
(SELECT @score:=0, @rank:=0) r
ORDER BY points DESC
 SET @rownum := 0;
 SELECT rank, score FROM ( SELECT @rownum := @rownum +1 AS rank, `score` ,`user_id`
 FROM leaderboard 
 ORDER BY  `score` DESC , `updated_timestamp` ) as result WHERE `user_id`=$user_id
 LIMIT 1
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