Question

我有一个名为“MyTable”的MySQL表，它基本上列出了用户名和点（两列，名称和点）。我想说的是“什么是joe1928的等级？”，这当然是基于他的观点。我怎么能在MySQL中执行此操作而无需下载所有数据并对其进行排序并自行确定排名？

得分最高的人将排名第一。

Answer 1

尝试获得分数高于用户的人数：

select count(*) from MyTable where score > (select score from MyTable where user = 'Joe');

这将为顶级用户返回0。

Answer 2

This page seems to describe and solve your problem.

该页面的注释：

SET @rownum := 0;
SELECT rank, correct FROM (
                SELECT @rownum := @rownum + 1 AS rank, correct, uid
                FROM quiz_user ORDER BY correct DESC
                ) as result WHERE uid=xxxxxxxx

Answer 3

SELECT @r AS Rank
FROM MyTable u, (SELECT @r := 0)
WHERE (@r := @r + 1) * (u.Username = 'joe1928')
ORDER BY u.Score DESC
LIMIT 1

Answer 4

select * from [TABLENAME] where [USERNAME] = blah order by [POINTS] desc limit 1;

Answer 5

根据@Dave发布的链接，您的查询将如下所示：

select Rank,name from
       (select @rownum:=@rownum+1 AS 'Rank', p.name
       from calls p, (select @rownum:=0) r
       order by p.points desc) as rankResults
where name = 'joe';

Answer 6

这是来自另一个堆栈溢出页面，似乎解决了你的问题。

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

在MySQL表中排名

6 个答案: