Question

我有一个论坛，我想找到最受欢迎的用户。最受欢迎的用户在帖子和评论上最受欢迎，也由最多的帖子和评论定义。 最喜欢（订单1），大多数帖子（订单2）和最多评论（订单3）的用户最受欢迎。同样的逻辑适用于下一个（第二个）最受欢迎的用户。

所以我有3张桌子：

发布表

id      user_id     likes
1       1           0
2       1           0
3       1           0
4       1           0
5       1           0
6       1           1
7       1           0
8       2           0
9       2           2
10      2           0
11      2           0
12      3           0
13      3           0
14      4           0
15      4           10

评论表

id      user_id     likes
1       1           0
2       1           1
3       1           1
4       1           0
5       2           0
6       2           2
7       2           1
8       4           1
9       4           0

用户表

id      name
1       John
2       Adam
3       Maggie
4       Steve

“赞”列包含其他用户在相应帖子（行）上提供的相似内容。我试过了：

 SELECT DISTINCT c.id, c.name,
    SUM(a.likes), SUM(b.likes), (SUM(a.likes) + SUM(b.likes)) as popular,
    COUNT(a.id) as mostp, COUNT(b.id) as mostc
    FROM posts as a, comments as b, users as c
    WHERE a.user_id=b.user_id AND a.user_id=c.id AND b.user_id=c.id
    GROUP BY a.user_id, b.user_id ORDER BY popular DESC, mostp DESC, mostc DESC

显然，这不起作用，因为如果你测试查询，它会提供比预期更多的喜欢（总和）。

这是实时查询 http://sqlfiddle.com/#!2/08900/3

Answer 1

您的查询的问题在于具有多个帖子和多个评论的用户，从而导致笛卡尔积并产生错误的总和。

以下查询（SQL Fiddle上的示例）应该有效，因为子查询已经按user_id分组：

SELECT
  u.name,
  COALESCE(p.likes,0) + COALESCE(c.likes,0) AS likes,
  COALESCE(p.cnt,0) AS post_count,
  COALESCE(c.cnt,0) AS comment_count
FROM users u
LEFT JOIN (
  SELECT user_id, COUNT(1) AS cnt, SUM(likes) AS likes
  FROM posts
  GROUP BY user_id
) p ON ( p.user_id = u.id )
LEFT JOIN (
  SELECT user_id, COUNT(1) AS cnt, SUM(likes) AS likes
  FROM comments
  GROUP BY user_id
) c ON ( c.user_id = u.id )
ORDER BY likes DESC, post_count DESC, comment_count DESC;

<强>结果：

|   NAME | LIKES | POST_COUNT | COMMENT_COUNT |
-----------------------------------------------
|  Steve |    11 |          2 |             2 |
|   Adam |     5 |          4 |             3 |
|   John |     3 |          7 |             4 |
| Maggie |     0 |          2 |             0 |

Answer 2

以下是如何做到这一点

SELECT
  u.id,
  u.name,
  (l.likes + r.likes) As TotalLikes,
  IFNULL(posts,0) AS TotalPosts,
  IFNULL(comments,0) AS TotalComments
FROM users AS u
  LEFT JOIN (SELECT
           user_id,
           IFNULL(SUM(likes),0) as likes,
           COUNT(likes) as posts
         FROM posts
         GROUP BY user_id) AS l
    on l.user_id = u.id
  LEFT JOIN (SELECT
           user_id,
           IFNULL(SUM(likes),0) as likes,
           COUNT(likes) AS comments
         FROM comments
         GROUP BY user_id) AS r
    on r.user_id = u.id
ORDER BY TotalLikes DESc

SQL Fiddle Demo

输出

| ID |   NAME | TOTALLIKES | TOTALPOSTS | TOTALCOMMENTS |
---------------------------------------------------------
|  4 |  Steve |         11 |          2 |             2 |
|  2 |   Adam |          5 |          4 |             3 |
|  1 |   John |          3 |          7 |             4 |
|  3 | Maggie |          0 |          2 |             0 |

选择喜欢，帖子和评论给出的最受欢迎的用户

2 个答案: