Question

我有一张这样的桌子。

| id-AI | from    | to    | text         |created_date
+----+---------+-------+-------------+-------------
| 1  | 7       | 1     | from 7 to 1  |2015-11-06 04:59:14 
| 2  | 1       | 2     | from 1 to 2  |2015-11-06 04:57:06
| 3  | 7       | 1     | from 7 to 1  |2015-11-06 04:25:45
| 4  | 1       | 6     | from 1 to 6  |2015-11-06 04:25:32 
| 5  | 1       | 5     | from 1 to 5  |2015-11-06 04:25:16

我想要一个这样的列表。一个用户列表敲我或我敲他/她或我们进行了长时间的对话。我想要的列表应该是最新的顺序，就像facebook消息一样。

| users |
+-------+
| 7    
| 2     
| 6
| 5

1是主user_id。获取列表，其中涉及1（在或从中）然后这称为对话。

Answer 1

您可以使用以下查询：

SELECT IF (`from` = 1, `to`, `from`)
FROM mytable
WHERE 1 IN (`from`, `to`)
ORDER BY created_date DESC

这将返回每个对话的创建者或收件人，其中涉及user_id = 1。

Demo here

获取有序的不同 user_id值集合使用此查询：

SELECT IF (`from` = 1, `to`, `from`)
FROM mytable
WHERE 1 IN (`from`, `to`)
GROUP BY IF (`from` = 1, `to`, `from`)
ORDER BY MAX(created_date) DESC

Demo here

Answer 2

使用UNION：

SELECT user
FROM (SELECT to AS user, created_date
      FROM mytable
      WHERE from = 1)
      UNION
      SELECT from AS user, created_date
      FROM mytable
      WHERE to = 1) x
GROUP BY user
ORDER BY MAX(created_date) DESC

获取每个会话的收件人的ID

2 个答案: