Question

我正在尝试通过最新消息在2个用户之间进行独特的对话。我迷路了。

我的表设置为

iD |  From  |  to  |  Message | Read |   Created_at
1  |   2    |   1  |  blah1   |   0  |   2019-01-11 02:15:24
2  |   1    |   2  |  blah2   |   0  |   2019-01-11 02:16:24
3  |   2    |   1  |  blah3   |   0  |   2019-01-11 02:17:24
4  |   2    |   3  |  blah4   |   0  |   2019-01-11 02:18:24
5  |   1    |   5  |  blah5   |   0  |   2019-01-11 02:19:24
6  |   1    |   3  |  blah6   |   0  |   2019-01-11 02:20:24

我希望每次对话的结果都没有重复。

例如。如果选择所有具有用户ID 1的对话，我需要这样的结果...

iD |  From  |  to  |  Message | Read |   Created_at
3  |   2    |   1  |  blah3   |   0  |   2019-01-11 02:17:24
5  |   1    |   5  |  blah5   |   0  |   2019-01-11 02:19:24
6  |   1    |   3  |  blah6   |   0  |   2019-01-11 02:20:24

我尝试过的所有操作都导致1到2以及2到1的对话重复

，我在这里搜索的所有内容都不适合我。

Answer 1

您可以使用该查询获取给定用户（在我的示例中，我使用的是用户1）与其他任何用户之间的最新对话时间：

SELECT CASE WHEN `From` = 1 THEN `To` ELSE `From` END AS other, MAX(Created_at) AS latest
FROM conversations
WHERE `From` = 1 OR `To` = 1
GROUP BY other

这只是将问题JOIN添加到整个表格中以获取两个用户之间最新消息的详细信息的问题：

SELECT c.*
FROM conversations c
JOIN (SELECT CASE WHEN `From` = 1 THEN `To` ELSE `From` END AS other, MAX(Created_at) AS latest
      FROM conversations
      WHERE `From` = 1 OR `To` = 1
      GROUP BY other) m
  ON (c.From = 1 AND c.To = m.other OR c.To = 1 AND c.From = m.other) AND c.Created_at = m.latest

输出：

iD  From    to  Message     Read    Created_at
3   2       1   blah3       0       2019-01-11 02:17:24
5   1       5   blah5       0       2019-01-11 02:19:24
6   1       3   blah6       0       2019-01-11 02:20:24

Demo on dbfiddle

MYSQL选择DISTINCT聊天对话

1 个答案: