如何获得基于SUM的排名？

我有评论表，其中的一切都存储，我必须汇总一切，并添加BEST ANSWER * 10。我需要排列整个列表，以及如何显示指定用户/ ID的排名。如何获得基于SUM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

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2010-05-25 Kenan

我不明白 - 查询看起来很好。也许一些样本数据和预期产出会有所帮助？ – 2010-05-25 20:59:02

Rank，如何显示排名，以及指定的UID – Kenan 2010-05-25 20:59:46

怎么是这样的：

SET @rank=0; 
SELECT * FROM (
    SELECT @rank:[email protected]+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

来源

2010-05-25 21:01:51 Wolph

其实我不能使用SET，它的ExpressionEngine和SQL模块不允许SET @ rank = 0;一切都必须在一条线上，不要“;” – Kenan 2010-05-25 21:04:09

在这种情况下，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获得等级，对吧？你只需要在程序中而不是在MySQL中计算排名。 – Wolph 2010-05-26 00:28:22

你可能想看看windowing functions如果你的MySQL版本支持他们......

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：

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：围绕此问题有很多开放性问题，但上述两种技术是确定排名的一般方法。你需要改变上面的内容以适合你的确切需要，因为我对member_id的排名构成了一点模糊。

来源

2010-05-25 21:16:34

SELECT 
    @rank:[email protected]+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

在该解决方案中，即使总数相同，排名也始终在增加。

来源

2010-05-25 21:36:24 newtover

如何获得基于SUM的排名？

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