按日期订购 SQL 与 LEFT JOIN 进行消息传递

Question

I have a problem that seems to be simple to solve but is not.

I have a MySQl database that storeS user, and chat sessions and chat lines of users.

Here's a part of UML schema of my database:

In chat_sessions we store only all session id (Auto increment)
In chat_sessions_members we store all of the user session
In chat_lines we store all line of the chat_sessions

Example:

The problem is how to get session of specific user and at the same time get the last lines if exist.

My last try gives me this code:

SELECT `chat_sessions_members`.`session_id`, ANY_VALUE(`send_at`) AS `send_at`, ANY_VALUE(`line`) AS `line`
FROM `chat_sessions_members`
LEFT JOIN `chat_lines` ON `chat_sessions_members`.`session_id` = `chat_lines`.`session_id`
WHERE `user_id` = '1'
GROUP BY `session_id`
ORDER BY DATE(`send_at`) DESC

But it's not an ordered reply.

I know a lot of post talk about that but I don't understand all of them.

Answer 1

If I understand correctly, you want the last (most recent) line for each session that has a given user. If so, you can use ROW_NUMBER() to identify the last line and LEFT JOIN to identify the user:

SELECT cl.*
FROM chat_sessions_members csm LEFT JOIN
     (SELECT cl.*, 
             ROW_NUMBER() OVER (PARTITION BY cl.session_id ORDER BY cl.send_at DESC) as seqnum
      FROM chat_lines cl
     ) cl
     ON cl.session_id = csm.session_id AND seqnum = 1
WHERE csm.user_id = 1
ORDER BY DATE(cl.send_at) DESC;