MySQL中的Rank函数分配等级

Question

我需要为 TC5 而不是 id 分配等级。

SELECT a.id, a.user_id, a.tc1, a.tc4, min(a.tc5), a.tc2, b.avatar, c.username, @curRank := @curRank + 1 AS Rank
FROM 
treningove_casy a INNER JOIN  
sn_users b ON a.user_id=b.id INNER JOIN
users c ON a.user_id=c.id , (SELECT @curRank := 0) r
WHERE  a.tc2 LIKE  'Motokáry Modrice' 
GROUP BY a.user_id

那画面怎么样。请帮忙

修改后的代码

SELECT x.*, (@curRank := @curRank + 1) as Rank
FROM (SELECT a.id, a.user_id, a.tc1, a.tc4, min(a.tc5) as tc5,
             a.tc2, b.avatar, c.username, 
      FROM sbhgl_chronoengine_chronoforms_datatable_treningove_casy a INNER JOIN  
           sbhgl_jsn_users b
           ON a.user_id = b.id INNER JOIN
           sbhgl_users c
           ON a.user_id = c.id 
      WHERE  a.tc2 LIKE  'Motokáry Modřice' 
      GROUP BY a.user_id
     ) x CROSS JOIN
     (SELECT @curRank := 0) params
ORDER BY tc5 DESC;

Answer 1

SELECT a.id, a.user_id, a.tc1, a.tc4, a.tc2, b.avatar, c.username, 
    (select rank from (SELECT 
        IF (@score=s.tc5, @rank:=@rank, @rank:=@rank+1) rank,
        @score:=s.tc5 tc5s
        FROM treningove_casy s,
        (SELECT @score:=0, @rank:=0) r
        ORDER BY tc5 DESC) s ) as rank
FROM 
treningove_casy a INNER JOIN  
sn_users b ON a.user_id=b.id INNER JOIN
users c ON a.user_id=c.id , (SELECT @curRank := 0) r
WHERE  a.tc2 LIKE  'Motokáry Modrice' 
GROUP BY a.user_id

您可以尝试以上解决方案，希望对您有所帮助。

Answer 2

在 MySQL 中，您通常使用变量来表示排名。在您的情况下，您将使用子查询。我发现变量方法并不总是适用于GROUP BY：

SELECT x.*, (@curRank := @curRank + 1) as Rank
FROM (SELECT a.id, a.user_id, a.tc1, a.tc4, min(a.tc5) as tc5,
             a.tc2, b.avatar, c.username, 
      FROM treningove_casy a INNER JOIN  
           sn_users b
           ON a.user_id = b.id INNER JOIN
           users c
           ON a.user_id = c.id 
      WHERE  a.tc2 LIKE  'Motokáry Modrice' 
      GROUP BY a.user_id
     ) x CROSS JOIN
     (SELECT @curRank := 0) params
ORDER BY tc5 DESC;

注意：使用表别名是好的。但是，如果别名是表名的缩写，则更容易理解查询。