Question

以下是我的表格结构：

-- users
+----+--------+
| id |  name  |
+----+--------+
| 1  | Jack   |
| 2  | Peter  |
| 3  | Ali    |
+----+--------+

-- reputations
+----+-------------+---------+-------+------------+------------+
| id | post_id     | user_id | score | reputation | date_time  |
+----+-------------+---------+-------+------------+------------+ -- Suppose:
| 1  | 1           | 1       | 1     | 5          | 1500489844 | -- out of last week
| 2  | 4           | 3       | -1    | -2         | 1500499815 | -- out of last week
| 3  | 2           | 3       | 1     | 5          | 1500584821 |
| 4  | 3           | 1       | 1     | 5          | 1501389166 |
| 5  | 2           | 4       | 1     | 5          | 1501399142 |
| 6  | 2           | 1       | -1    | -2         | 1501399142 |
| 7  | 4           | 1       | 0     | 15         | 1501481186 |
+----+-------------+---------+-------+------------+------------+
-- Note: the last row came from an accepted-answer, that's why its score is 0

-- post_tag
+---------+--------+
| post_id | tag_id |
+---------+--------+
| 1       | 2      |
| 1       | 4      |
| 2       | 2      |
| 3       | 1      |
| 3       | 4      |
| 4       | 3      |
+---------+--------+

-- tags
+----+--------+
| id |  name  |
+----+--------+
| 1  | php    |
| 2  | html   |
| 3  | css    |
| 4  | mysql  |
+----+--------+

现在我想获得上周声明，分数，前两个标签的用户列表。我怎么能这样做？

注1： ＆＃34; top tow标签＆＃34; 表示用户在其中获得更多回购的两个标签。

注2：每个问题至少有1个标记。

这是预期输出：

+----+--------+-------+------------+----------+
| id |  name  | score | reputation |   tags   |
+----+--------+-------+------------+----------+
| 1  | Jack   | 0     | 18         | css,php  |
| 3  | Ali    | 1     | 5          | html     |
| 2  | Peter  | 0     | 0          | NULL     |
+----+--------+-------+------------+----------+
-- Note: It's ordered by reputation, score columns

以下是我尝试过的内容：

SELECT u.*,
      sum(r.score) as score,
      sum(r.reputation) as reputation
      /* WS_CONCAT(',', t.name) as tags */
FROM users u
LEFT JOIN reputation r ON r.user_id = u.id
/* I need more joins to get tags, I don't know how exactly */
WHERE r.date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)
GROUP BY u.id
ORDER BY reputation, score

如您所见，我的查询缺少最后一列（标记）。老实说，我不知道如何管理这样的事情。你知道吗？

Answer 1

您是否尝试过使用子查询？类似的东西：

SELECT u.id,u.name,
      sum(r.score) as score,
      sum(r.reputation) as reputation,
      (SELECT group_concat(t.name) FROM post_tag pt INNER JOIN reputations r ON r.post_id = pt.post_id INNER JOIN tags t ON pt.tag_id = t.id WHERE r.user_id = u.id ORDER BY sum(r.reputation) DESC LIMIT 2)
FROM users u
LEFT JOIN reputation r ON r.user_id = u.id
GROUP BY u.id, u.name, score, reputation
ORDER BY reputation, score;

Answer 2

你可以尝试一下。

SELECT DISTINCT u.*,
  sum(r.score) as score,
  sum(r.reputation) as reputation
  GROUP_CONCAT(',', t.name) as tags 
FROM users u
LEFT JOIN reputation r ON r.user_id = u.id
LEFT JOIN post_tag pt ON(pt.post_id=r.post_id)
LEFT JOIN tags t ON(t.id=pt.tag_id)
WHERE r.date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)
GROUP BY u.id , 
ORDER BY reputation, score

如何计算前两个标签？

2 个答案: