php - 获取每个记录组的最后一条记录

Question

我不知道如何写获取最后一条记录的SQL语法(根据最近的帖子，未回复)。

我的 table

+-------------------+-----------------------+------+-----+---------+----------------+
| Field             | Type                  | Null | Key | Default | Extra          |
+-------------------+-----------------------+------+-----+---------+----------------+
| notification_id   | mediumint(8) unsigned | NO   | PRI | NULL    | auto_increment |
| user_id           | mediumint(8) unsigned | NO   |     | NULL    |                |
| notification_msg  | text                  | NO   |     | NULL    |                |
| notification_date | int(11) unsigned      | NO   |     | NULL    |                |
| private_message   | tinyint(1) unsigned   | NO   |     | 0       |                |
| has_replied       | tinyint(1) unsigned   | NO   |     | 0       |                |
| reply_id          | mediumint(8) unsigned | NO   |     | 0       |                |
+-------------------+-----------------------+------+-----+---------+----------------+

基本上对于每个线程通知，它应该获取每个通知记录的最后一条记录并检查 has_replied 是否为 0，如果为 0 那么它应该返回它所以PHP 可以读取是否有未回复的通知。所以像，它应该像这样返回(伪):

+--------------+-----+-----+
| username     | 1   | 4   |
| username2    | 0   | 2   |
+--------------+-----+-----+

第二列表示最后一个帖子是否已回复。

我当前的 SQL 语法(有效但没有得到最后一条记录，如果它被回复):

SELECT n.*,
       m.user_id,
       m.username
FROM notifications n
INNER JOIN members m ON n.user_id = m.user_id
WHERE private_message = 1
AND reply_id = 0
ORDER BY has_replied ASC,
         notification_date DESC

Answer 1

Select m.user_id, m.username
    , N...
From members As M
    Join    (
            Select user_id, Max( notification_id ) As notification_id
            From notifications 
            Group By user_id
            ) As UserLastNotification
        On UserLastNotification.user_id = m.user_id
    Join notifications As N
        On N.notification_id = UserLastNotification.notification_id
Where N.private_message = 1
    And N.reply_id = 0
Order By N.has_replied, N.notification_date Desc

请注意，这将过滤每个用户的最后一条通知是一条私有(private)消息，并且 reply_id 为零。