根据第二列中具有最大日期值的列选择不同的值 [英] Select distinct values based on a column with maximum date value from second column

问题描述

标题有点复杂.我很抱歉，但对于非SQL用户来说，查询也很复杂.

The title is a bit complex. I apologize but the query is also complex for a non-SQL person.

我有一个表messages，该表具有以下结构:

I have a table messages that has the following structure:

消息(id，发送方ID，接收方ID，消息日期时间)

Message(id, sender_id, receiver_id, message_datetime)

我的目标是选择receiver_id从不同发件人处收到的最后一封邮件.

My goal is to select the last messages received by a receiver_id from distinct senders.

例如，当我这样做时:

SELECT * 
FROM  `message` 
WHERE  `receiver_id` =1

我得到类似的东西:

1005  |   2    |    1    |  2015-11-08
1004  |   3    |    1    |  2015-11-07
1003  |   3    |    1    |  2015-11-06
1002  |   2    |    1    |  2015-11-05 
1001  |   2    |    1    |  2015-11-04

当我需要类似的东西时:

While I need something like:

1005  |   2    |    1    |  2015-11-08
1004  |   3    |    1    |  2015-11-07

高度赞赏您通常的专家指导.我真的被这样的查询所困扰.再次感谢，对不起，格式化不正确.

Your usual expert guidance is highly appreciated. I am really stuck with such a query. Thanks again and sorry for the bad formatting.

推荐答案

您需要创建一个子查询，该子查询通过sender_id返回给定接收者的最大消息ID，并将其加入到message表中以获取所有其他字段:

You need to create a subquery that returns the maximum message id by sender_id for a given receiver and join it to the messages table to get all other fields:

SELECT m.* 
FROM  `message` AS m
INNER JOIN (SELECT sender_id, MAX(message_date) as md
                FROM message WHERE  `receiver_id` =1 GROUP BY sender_id) AS t
ON m.message_date=t.md and m.sender_id=t.sender_id
WHERE  `receiver_id` =1

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