id sender receiver time_sent Message SendDel RecDel
1 1 3 2011-08-17 14:00:00 [text] 0 0
2 3 1 2011-08-17 15:00:00 [text] 0 0
3 2 4 2011-08-18 14:19:28 [text] 1 0
4 4 2 2011-08-18 15:19:28 [text] 0 0
目标是检索最高价值的消息(MAX)并根据发送者,接收者对消息进行分组.因此消息ID 1和2将组合在一起,id 3和4将组合在一起.
示例:登录userid = 2应该返回
id sender receiver time_sent Message SendDel RecDel
4 4 2 2011-08-18 15:19:28 [text] 0 0
我不知道为什么,但我的查询没有将所有消息组合在一起.这是我的查询:
SELECT id, sender, receiver, MAX(time_sent), MAX(message)
FROM Messages
WHERE sender='$userid' OR receiver = '$userid'
Group By sender,receiver
Order BY time_sent DESC
任何方案?
解决方法:
你要:
So message id’s 1 and 2 would group together and id’s 3 and 4 would group together.
你必须在group by子句中做一些魔术来实现这一点.
SELECT id, sender, receiver, MAX(time_sent), MAX(message)
FROM Messages
WHERE sender='$userid' OR receiver = '$userid'
Group By (if(sender > receiver, sender, receiver))
, (if(sender > receiver, receiver, sender))
Order BY time_sent DESC
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。