mysql - SQL:如何按两列的唯一组合进行分组？

Question

上下文:

• 表 message 包含列 from_user_id 和 to_user_id
• 用户应该看到最近的对话以及显示的最后一条消息
• 一个 session 由多条消息组成，这些消息具有相同的用户 ID 组合(用户发送消息，用户接收消息)

表格内容:

+-------------------------------------------------+--------------+------------+
| text                                            | from_user_id | to_user_id |
+-------------------------------------------------+--------------+------------+
| Hi there!                                       |           13 |         14 | <- Liara to Penelope
| Oh hi, how are you?                             |           14 |         13 | <- Penelope to Liara
| Fine, thanks for asking. How are you?           |           13 |         14 | <- Liara to Penelope
| Could not be better! How are things over there? |           14 |         13 | <- Penelope to Liara
| Hi, I just spoke to Penelope!                   |           13 |         15 | <- Liara to Zara
| Oh you did? How is she?                         |           15 |         13 | <- Zara to Liara
| Liara told me you guys texted, how are things?  |           15 |         14 | <- Zara to Penelope
| Fine, she's good, too                           |           14 |         15 | <- Penelope to Zara
+-------------------------------------------------+--------------+------------+

我的尝试是按from_user_id 和to_user_id 进行分组，但显然我得到了一组用户收到的消息和另一组用户发送的消息。

SELECT text, from_user_id, to_user_id,created FROM message 
WHERE from_user_id=13 or to_user_id=13
GROUP BY from_user_id, to_user_id
ORDER BY created DESC

让我:

+-------------------------------+--------------+------------+---------------------+
| text                          | from_user_id | to_user_id | created             |
+-------------------------------+--------------+------------+---------------------+
| Oh you did? How is she?       |           15 |         13 | 2017-09-01 21:45:14 | <- received by Liara
| Hi, I just spoke to Penelope! |           13 |         15 | 2017-09-01 21:44:51 | <- send by Liara
| Oh hi, how are you?           |           14 |         13 | 2017-09-01 17:06:53 |
| Hi there!                     |           13 |         14 | 2017-09-01 17:06:29 |
+-------------------------------+--------------+------------+---------------------+

虽然我想要:

+-------------------------------+--------------+------------+---------------------+
| text                          | from_user_id | to_user_id | created             |
+-------------------------------+--------------+------------+---------------------+
| Oh you did? How is she?       |           15 |         13 | 2017-09-01 21:45:14 | <- Last message of conversation with Zara
| Oh hi, how are you?           |           14 |         13 | 2017-09-01 17:06:53 |
+-------------------------------+--------------+------------+---------------------+

我怎样才能做到这一点？

编辑:使用 least 或 greatest 也不会导致所需的结果。它确实对条目进行了正确分组，但正如您在结果中看到的那样，最后一条消息不正确。

+----+-------------------------------------------------+------+---------------------+--------------+------------+
| id | text                                            | read | created             | from_user_id | to_user_id |
+----+-------------------------------------------------+------+---------------------+--------------+------------+
|  8 | Oh you did? How is she?                         | No   | 2017-09-01 21:45:14 |           15 |         13 |
|  5 | Could not be better! How are things over there? | No   | 2017-09-01 17:07:47 |           14 |         13 |
+----+-------------------------------------------------+------+---------------------+--------------+------------+

Answer 1

做你想做的事情的一种方法是使用相关子查询，找到匹配对话的最小创建日期/时间:

SELECT m.*
FROM message m
WHERE 13 in (from_user_id, to_user_id) AND
      m.created = (SELECT MAX(m2.created)
                   FROM message m2
                   WHERE (m2.from_user_id = m.from_user_id AND m2.to_user_id = m.to_user_id) OR
                         (m2.from_user_id = m.to_user_id AND m2.to_user_id = m.from_user_id) 
                  )
ORDER BY m.created DESC