如何根据 SUM 获得排名？答案

【问题标题】：How to get rank based on SUM's?如何根据 SUM 获得排名？
【发布时间】：2010-05-25 20:55:16
【问题描述】：

我有存储所有内容的 cmets 表，我必须对所有内容求和并添加最佳答案 *10。我需要整个列表的排名，以及如何显示指定用户/ID 的排名。

这里是 SQL：

   SELECT m.member_id AS member_id, 
          (SUM(c.vote_value) + SUM(c.best)*10) AS total
     FROM comments c
     LEFT JOIN members m ON c.author_id = m.member_id
     GROUP BY c.author_id
     ORDER BY total DESC
    LIMIT {$sql_start}, 20

【问题讨论】：

我不明白 - 查询看起来不错。也许一些示例数据和预期输出会有所帮助？
排名，如何显示排名，以及指定的UID

标签： mysql rank

【解决方案1】：

这样的事情怎么样：

SET @rank=0;
SELECT * FROM (
   SELECT @rank:=@rank+1 AS rank, m.member_id AS member_id, 
      (SUM(c.vote_value) + SUM(c.best)*10) AS total
   FROM comments c
   LEFT JOIN members m ON c.author_id = m.member_id
   GROUP BY c.author_id
   ORDER BY total DESC
) as sub
LIMIT {$sql_start}, 20

【讨论】：

其实我不能用SET，它的ExpressionEngine和SQL模块不允许SET @rank=0;一切都必须在一行中，没有“;”
在那种情况下，我不知道如何在 MySQL 中做到这一点。但是你可以从你的极限中得出排名，对吧？你只需要在你的程序而不是 MySQL 中计算排名。

【解决方案2】：

如果您的 MySQL 版本支持它们，您可能需要查看 windowing functions...

 SELECT m.member_id AS member_id, 
          (SUM(c.vote_value) + SUM(c.best)*10) AS total,
          RANK() OVER (ORDER BY (SUM(c.vote_value) + SUM(c.best)*10)) as ranking
     FROM comments c
     LEFT JOIN members m ON c.author_id = m.member_id
     GROUP BY c.author_id
     ORDER BY total DESC;

另一种可能性是这样的：

 SELECT m.member_id AS member_id, 
          (SUM(c.vote_value) + SUM(c.best)*10) AS total,
          (SELECT count(distinct <column you want to rank by>)
           FROM comments c1
           WHERE c1.author_id = m.member_id) as ranking
     FROM comments c
     LEFT JOIN members m ON c.author_id = m.member_id
     GROUP BY c.author_id
     ORDER BY total DESC;

注意：这方面有很多悬而未决的问题，但上述两种技术通常是确定排名的简单方法。您需要更改以上内容以满足您的确切需求，因为我对 member_id 排名的构成有点模糊。

【讨论】：

【解决方案3】：

SELECT
    @rank:=@rank+1 as rank,
    m.member_id AS member_id, 
    (SUM(c.vote_value) + SUM(c.best)*10) AS total
FROM comments c,
(SELECT @rank:=0) as init
LEFT JOIN members m ON c.author_id = m.member_id
GROUP BY c.author_id
ORDER BY total DESC
LIMIT {$sql_start}, 20

在解决方案中，排名总是在增加，即使总数相同。

【讨论】：