Mysql 同时使用ORDER BY、GROUP BY和multiple SELECT

我有两张桌子

表用户

uid  | username 
----------------
1    | brandon   
2    | john   
3    | nicole

表2.1.1-1-1-1

id   | from  | to | message      | time
-------------------------------------------
1    | 1     | 2  | hi sister    | 100
2    | 1     | 3  | hi brother   | 300
3    | 1     | 2  | hi again!    | 400
4    | 2     | 3  | lolz message | 600

（column time是一个实际的时间戳，我只是简化了这个示例）

“from”是用户的uid

“to”是用户的uid

我想从发送给特定用户的所有用户处获取最新消息

本例中的特定用户是

$logged_id = 2;

结果应返回id为3的行

我到目前为止有什么疑问

SELECT private_msgs.*, users.username FROM `private_msgs`, `users` 
WHERE private_msgs.to=$logged_id AND users.uid = private_msgs.from 
INNER JOIN (
  SELECT 
    from, max(time) as latest 
  FROM private_msgs 
  GROUP BY private_msgs.from
) r ON private_msgs.time = r.latest AND private_msgs.from = r.from
ORDER BY time DESC");

然而，我得到了错误

Error Number: 1064

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN ( SELECT from, max(time) as latest FROM private_msgs ' at line 3

---

查询中有几个错误。除了尾随逗号之外，

from

和

to

都是保留字，因此它们应该被引用：

SELECT pm.*, u.username
FROM `private_msgs` pm join
     `users` u
     on pm.`to` = $logged_id AND u.uid = pm.`from` INNER JOIN
     (SELECT `from`, max(time) as latest 
      FROM private_msgs 
      GROUP BY private_msgs.`from`
     ) r
    ON pm.time = r.latest AND pm.`from` = r.`from`
ORDER BY time DESC;

---

在这里，您将收到来自所有用户的最新消息。这是对您的answesr的查询

SELECT * FROM 
    (SELECT * FROM `user` a,` private_msgs` b 
     WHERE a.uid=b.from order by time DESC ) 
      t group by `from`

---

这个问题经常被问到。