mysql - Mariadb - 从用户 ID = 变量的所有消息列表中获取最新消息

Question

我正在尝试获取特定用户的最新消息列表。我的表格如下:

+----------------+---------------------+------+-----+---------+----------------+
| Field          | Type                | Null | Key | Default | Extra          |
+----------------+---------------------+------+-----+---------+----------------+
| id             | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| sender_user_id | int(11)             | NO   |     | NULL    |                |
| remote_user_id | int(11)             | NO   |     | NULL    |                |
| message        | longtext            | NO   |     | NULL    |                |
| read           | tinyint(4)          | NO   |     | NULL    |                |
| created_at     | timestamp           | YES  |     | NULL    |                |
| updated_at     | timestamp           | YES  |     | NULL    |                |
+----------------+---------------------+------+-----+---------+----------------+

创建表:

CREATE TABLE `messages` (
  `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `sender_user_id` int(11) NOT NULL,
  `remote_user_id` int(11) NOT NULL,
  `message` longtext COLLATE utf8mb4_unicode_ci NOT NULL,
  `read` tinyint(4) NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

到目前为止，这是我尝试过的: SELECT sender_user_id, remote_user_id FROM messages m1 WHERE id = (SELECT MAX(m2.id) FROM messages m2 WHERE m1.sender_user_id = m2.sender_user_id AND m1.remote_user_id = m2.remote_user_id)

这看起来很接近，但显然没有考虑到您搜索的用户 ID，它有时会返回重复项，例如:

MariaDB [snowdon]> SELECT sender_user_id, remote_user_id FROM messages m1 WHERE id = (SELECT MAX(m2.id) FROM messages m2 WHERE m1.sender_user_id = m2.sender_user_id AND m1.remote_user_id = m2.remote_user_id)
    -> ;
+----------------+----------------+
| sender_user_id | remote_user_id |
+----------------+----------------+
|              1 |              2 |
|              2 |              1 |
|              3 |              1 |
+----------------+----------------+
3 rows in set (0.00 sec)

第一行和第二行是同一对话的一部分，所以我只希望最新的出现在结果中。

有人可以帮我写 SQL 吗？

萨克斯

Answer 1

我想你想要:

SELECT (CASE WHEN m.sender_user_id = ? THEN m.remote_user_id ELSE m.sender_user_id END) as other_user_id
FROM messages m
WHERE ? IN (m.sender_user_id, m.remote_user_id) AND
      m.id = (SELECT MAX(m2.id)
              FROM messages m2
              WHERE (m2.sender_user_id, m2.remote_user_id) IN 
                      ( (m.sender_user_id, m.remote_user_id),
                        (m.remote_user_id, m.sender_user_id)
                      )
             );