从多对多表中对 mysqli 进行排序

Question

我正在尝试使用 *amp 系统制作一个基于网络的内部消息系统，主要用于学习目的。 我不知道这是否是一个微不足道的话题，但我遇到了困难，所以请耐心等待。

目标是列出按发送/接收的最后一条消息排序的所有联系人。 目前没有对其进行排序，SQL 看起来像这样

$query = "SELECT username, user.id as user_id,  

(SELECT COUNT(message_read)  
FROM message_user  
WHERE message_read = 0 
AND sent_id = user_id  
AND receive_id = {$userId}) as unread  

FROM user  
WHERE user.id IN  
(SELECT contact_id FROM allowed_contact WHERE user_id = {$userId})   
;";

表的结构是：
user表有一个id ，
链接到具有sent_id和receive_id的message_user表，
message_user有一个message_id对应于message.id ，
message表有一个timestamp 。

我希望这可以在 SQL 中完成，但如果归结为 PHP，我就辞职求助于此。

Answer 1

这有效。

SELECT `u`.`id` AS user_id, username,
(SELECT COUNT(message_user.message_read)  
FROM message_user  
WHERE message_user.message_read = 0 
AND sent_id = user_id  
AND receive_id = {$userId}) as unread  

FROM `user` AS `u`
LEFT JOIN `message_user` AS `mu` 
ON 
    (CASE WHEN `u`.`id` != {$userId}
        THEN `u`.`id` = `mu`.`sent_id`
        WHEN `mu`.`sent_id` = {$userId} AND `mu`.`receive_id` = {$userId}
        THEN `u`.`id` = `mu`.`sent_id`
    END)
OR  
    (CASE WHEN `u`.`id` != {$userId}
        THEN `u`.`id` = `mu`.`receive_id`
    END)

LEFT JOIN `message` AS `m` ON `m`.`id` = `mu`.`message_id`

WHERE u.id IN  
(SELECT contact_id FROM allowed_contact WHERE user_id = {$userId})
GROUP BY u.id
ORDER BY MAX(`m`.`timestamp`) DESC;

这打破了我遇到的问题。

@Andreas 感谢您的时间和帮助。

Answer 2

使用 2 LEFT JOIN和DISTINCT （未经测试）：

SELECT DISTINCT `u`.`id`
FROM `user` AS `u`
LEFT JOIN `message_user` AS `mu` ON `u`.`id` = `mu`.`sent_id` OR `u`.`id` = `mu`.`receive_id`
LEFT JOIN `message` AS `m` ON `m`.`id` = `mu`.`message_id`
ORDER BY `m`.`timestamp` DESC;