Question

我有评论表，其中存储了所有内容，我必须对所有内容进行SUM并添加BEST ANSWER * 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

Answer 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

Answer 2

如果您的MySQL版本支持，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：

 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：围绕这个有很多未解决的问题，但上述两种技术是确定排名的简单方法。你需要改变上面的内容以满足你的确切需要，因为我对member_id的排名构成了一点模糊。

Answer 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，即使总数相同，等级也总是在增加。

如何根据SUM获得排名？

3 个答案: