Question

以下是我所说的话。

  userid score 
    1       8    
    2       5    
    3       4    
    4       4    
    5      10    
    6       3

我想做的是:

userid score position
    5      10     1
    1       8     2
    2       5     3
    3       4     4
    4       4     4
    6       3     5

<>光>

我的代码如下:

userid score position
    5      10     1
    1       8     2
    2       5     3
    3       4     4
    4       4     4
    6       3     6

法典

SELECT userid, score, 
(SELECT COUNT(*) FROM fschema.mytab3 u2 
WHERE 
u2.score > u1.score) + 1 AS position FROM fschema.mytab3 u1
ORDER BY position

我希望用户6有<条码><> > > > 代码>,作为<条码>5而不是<条码>。

Answer 1

这一点是什么?

SELECT userid, score, 
(SELECT COUNT(distinct u2.score) FROM fschema.mytab3 u2 
WHERE 
u2.score > u1.score) + 1 AS position FROM fschema.mytab3 u1
ORDER BY position

Answer 2

引证:

SELECT
  *,
  @r:=IF(@score IS NULL OR @score <> score, @r+1, @r) position, @score:=score
FROM
  fschema.mytab3,
  (SELECT @r:=0, @score:=NULL) t
ORDER BY
  score DESC, userid

Answer 3

I think that you could try it using variables, which looks easier. Something like this:

SELECT userid, score, @rownum:=@rownum+1 as position
FROM fschema.mytab3 u1, (SELECT @rownum:=0) r
ORDER BY score;

(Currently I m unable to check MySql queries, please, excuse me if there is some error)

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