Php Mysql正在通过获取组中的最后一个条目

嘿，这是我的MySQL查询

$userid="id of logged in user";

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它给出两个用户之间记录的第一条消息，而不是最后一条。我希望最后一封邮件显示在邮件收件箱中。我使用此查询在收件箱中显示与登录用户（或您可以说是特定用户）的不同用户的唯一对话 你可以检查demo@

试试这个

SELECT id, message, user_name, receiver_id, sender_id
FROM chat
WHERE post_time = MAX(post_time) AND (receiver_id =$userid
OR sender_id =$userid)
GROUP BY GREATEST( receiver_id, sender_id ) , LEAST( receiver_id, sender_id ) 
ORDER BY id DESC 
LIMIT 0 , 1000

这可能对你有帮助

SELECT id, message, user_name, receiver_id, sender_id
FROM chat
WHERE receiver_id =$userid
OR sender_id =$userid
HAVING post_time = MAX(post_time) 
GROUP BY GREATEST( receiver_id, sender_id ) ,         LEAST( receiver_id, sender_id ) 
ORDER BY id DESC 
LIMIT 0 , 1000

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我自己解决了这个密码。。。。。。。。。。。。。如果有人需要它，它将帮助其他人也谢谢你

SELECT m.* 
FROM (
SELECT sender_id + receiver_id -1 AS other_recipient_id, MAX( id ) AS id
FROM chat
WHERE (sender_id =*user's_id*
OR receiver_id =*user's_id*)
GROUP BY sender_id + receiver_id -1
)mm
INNER JOIN chat m ON mm.id = m.id
ORDER BY id ASC

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谢谢你的回复，但你的代码要么是错误的，要么不是我的工作。对不起…………我已经发布了新的答案。这可能会帮助您考虑提供适当的DDL（和/或SQLFIDLE）以及所需的结果集；查看与您的MySQL服务器版本对应的手册，了解使用near'或sender\u id=HAVING post\u time=MAX（post\u time）GROUP BY MASTER（receiver\u）的正确语法（第4行），只需在“GROUP BY”之后写入“HAVING”子句，然后尝试返回仅发送一条消息的对话抱歉……。）

SELECT m.* 
FROM (
SELECT sender_id + receiver_id -1 AS other_recipient_id, MAX( id ) AS id
FROM chat
WHERE (sender_id =*user's_id*
OR receiver_id =*user's_id*)
GROUP BY sender_id + receiver_id -1
)mm
INNER JOIN chat m ON mm.id = m.id
ORDER BY id ASC