Question

这应该相当简单，但我坚持使用它。

当我进行查询时

SELECT * FROM `inbox` where toid=4 or fromid=4 order by time desc

，我明白了：

id  toid    fromid  message           time
23  48101   4   hello call me     12/23/2011 12:27
6   34584   4   hi there      12/22/2011 15:42
5   34584   4   how are you   12/22/2011 14:08
4   34584   4   say hello     12/22/2011 14:07
3   34584   4   whats up      12/22/2011 14:07
2   4   34584   nice picture      11/24/2010 0:00
1   4   2   this is very interesting!   12/23/2008 0:00

现在，我需要将用户4和其他用户之间的对话分组到最后一条消息（如facebook消息）。

有人知道最好的方法吗？

谢谢！

Answer 1

SQL Fiddle

select i.id, toid, fromid, message, `time`
from
    inbox i
    inner join (
        select max(id) as id
        from inbox
        where toid = 4 or fromid = 4
        group by greatest(toid, fromid), least(toid, fromid)
    ) s on i.id = s.id

选择用户或用户发送的消息

1 个答案: