如何从所有用户获取最新消息

如果可以包含表结构，则更容易给出答案（messages表上是否有唯一 ID？但是从您问题中的信息来看，这应该有效：

SELECT 
  u.profile_pic,
  u.username,
  u.firstname,
  u.llastname,
  m.message_from,
  m2.message_body 
FROM  user u
JOIN (SELECT message_from,message_to,max(sent_time)
      FROM  messages
      GROUP BY message_from,message_to) m 
        ON user.user_no = messages.message_from 
        AND messages.message_to = '$userno'
JOIN messages m2 ON m.message_from = m2.message_from
                 AND m.message_to = m2.message_to
                 AND m.sent_time = m2.sent_time

基本上，子选择会将所有最新消息从一个用户拉出到另一个用户，然后您加入这些消息以过滤掉所有其他消息。

使用LEFT JOIN对同一用户和较新sent_time的消息表。每个用户的最新消息是没有此类消息的用户：

SELECT profile_pic,username,firstname,lastname,messages.message_from,messages.message_body 
FROM  user 
JOIN messages ON user.user_no = messages.message_from 
LEFT JOIN messages AS newermessages ON user.user_no = newermessages.message_from AND messages.sent_time < newermessages.sent_time
WHERE  messages.message_to = '$userno' 
AND newermessages.id IS NULL
ORDER BY messages.sent_time DESC

来自每个用户的最新消息：

SELECT a.*
  FROM messages a
  JOIN 
     ( SELECT message_from
            , MAX(sent_time) max_sent_time 
         FROM messages 
        GROUP 
           BY message_from
     ) b
    ON b.message_from = a.message_from
   AND b.max_sent_time = a.sent_time;

这个问题的其余部分留给读者练习。

尝试将 ORDER BY 子句的字段添加到您的选择中：

SELECT u.profile_pic
     , u.username
     , u.firstname
     , u.lastname
     , m.message_from
     , m.message_body 
     , m.sent_time
  FROM user u
  JOIN messages m
    ON m.message_from = u.user_no 
 WHERE m.message_to = $userno
 ORDER 
    BY m.sent_time DESC