Question

我在数据库中使用了两个表：

“成员”表：

  id   |   name    
-----------------------
  1    |   John
  2    |   Mike
  3    |   Jane
  4    |   Marry

“消息”表：

  message_id  |   text   |   from   |   to
 ---------------------------------------------  
      1       |  text 1  |     1    |    2 
      2       |  text 2  |     1    |    3
      3       |  text 3  |     1    |    4  
      4       |  text 4  |     2    |    1  
      5       |  text 5  |     1    |    2

我需要得到特定用户发送的分组消息，按发送的最后一条消息排序，并计算发送给特定用户的消息总数。

我当前的mysql查询看起来：

SELECT max(messages.message_id) as max_mess, messages.from, messages.to, members.name, messages.text 
FROM messages INNER JOIN members on members.id = messages.to  
WHERE messages.from = '1' 
GROUP BY messages.from  
ORDER BY max_mess DESC

所需结果应如下所示：

  max_mess  |   from   |    to   |   name  |   text  | total_messages
 --------------------------------------------------------------------
     5      |    1     |    2    |   Mike  |  text 5 |      3
     3      |    1     |    4    |   Marry |  text 3 |      1
     2      |    1     |    3    |   Jane  |  text 2 |      1

是否可以仅使用一个mysql查询以及当前查询中应该更改哪些内容？

Answer 1

您可以使用以下解决方案之一：

解决方案＃1（使用额外的JOIN）：

SELECT 
    m1.max_mess, 
    m1.from, 
    m1.to, 
    m1.name, 
    m2.text, 
    m1.total_messages 
FROM (
    SELECT 
        MAX(message_id) AS max_mess, 
        messages.from, 
        messages.to,
        members.name, 
        COUNT(message_id) AS total_messages 
    FROM messages INNER JOIN members ON members.id = messages.to  
    WHERE messages.from = 1 
    GROUP BY messages.from, messages.to  
) m1 LEFT JOIN messages m2 ON m1.max_mess = m2.message_id
ORDER BY m1.max_mess DESC;

演示： http://sqlfiddle.com/#!9/7c73b6/1/0

注意：此解决方案应该比第二个解决方案更快。您可以在StackOverflow上找到有关以下答案的更多详细信息：JOIN vs. Sub-Query。

解决方案＃2（使用子选择）：

SELECT 
    MAX(message_id) AS max_mess, 
    m1.from, 
    m1.to,
    members.name, 
    (SELECT messages.text FROM messages WHERE message_id = MAX(m1.message_id)) as text,
    COUNT(message_id) AS total_messages
FROM messages m1 INNER JOIN members ON members.id = m1.to 
WHERE m1.from = 1 
GROUP BY m1.from, m1.to  
ORDER BY max_mess DESC;

演示： http://sqlfiddle.com/#!9/7c73b6/2/0

Answer 2

这是一个查询。如果要删除JOIN语句，则应使用

SELECT max(messages.message_id) as max_mess, messages.from, messages.to, members.name, messages.text 
FROM messages, members  
WHERE messages.from = '1' AND members.id = messages.to
GROUP BY messages.from  
ORDER BY max_mess DESC

Answer 3

我认为你正在寻找这个：

SELECT MAX(messages.message_id) AS max_mess, messages.from, messages.to, members.name, messages.text,, SUM(`from`) total_messages 
FROM messages INNER JOIN members ON members.id = messages.to  
WHERE messages.from = '1' 
GROUP BY messages.from,messages.to  
ORDER BY max_mess DESC;

Mysql查询内部连接两个表，其中包含最后一个ID

3 个答案: