我正在尝试获取最新消息列表。情况是这样的:当有完全相同时间发送的消息时，sequel 语句会选择所有消息。我只需要来自每个聊天室的一条消息，即最新消息。在@JuanCarlosOropeza(以及他出色的 fiddle http://sqlfiddle.com/#!9/d506e/10，我将立即将其包含在我就此主题提出的每个问题中)的帮助下，我得到了下面的续集声明。
SELECT *
FROM ( SELECT t.*,
@rn := if(@friend = t.friend_id,
@rn + 1,
if( @friend := t.friend_id, 1, 1)
) as rn
FROM (
SELECT m.message,
m.message_read,
m.message_date,
CASE WHEN m.sender = 4
THEN m.receiver
ELSE m.sender
END as friend_id,
CASE WHEN m.sender = 4
THEN p2.nickname
ELSE p1.nickname
END as name,
CASE WHEN m.sender = 4
THEN p2.image
ELSE p1.image
END as image
FROM message as m
JOIN profile as p1
ON m.sender = p1.user_id -- sender
JOIN profile as p2
ON m.receiver = p2.user_id -- receiver
WHERE 4 IN (m.sender, m.receiver)
) as t
CROSS JOIN ( SELECT @rn := 0, @friend := 0 ) as var
ORDER BY t.friend_id, t.message_date desc
) q
WHERE q.rn = 1
上面的续集声明做得很好；然而，它有一个缺陷。 ORDER BY t.friend_id, t.message_date desc 不考虑“完全相同的 message_date”。同时发送消息很可能是不可能的，但谁知道呢？
好友表:
+----+-------+-------+---------+--------------+
ø id ø user1 ø user2 ø pending ø request_date ø
+----+-------+-------+---------+--------------+
ø 1 ø 4 ø 2 ø 0 ø 2018-05-09 ø
ø 2 ø 5 ø 2 ø 0 ø 2018-05-09 ø
ø 3 ø 1 ø 4 ø 0 ø 2018-05-09 ø
+----+-------+-------+---------+--------------+
配置文件表:
+----+---------+----------------------+---------------+-------+
ø id ø user_id ø nickname ø email ø image ø
+----+---------+----------------------+---------------+-------+
ø 1 ø 1 ø Welcome to MathMatch ø test@test.com ø NULL ø
ø 2 ø 2 ø user3 ø NULL ø NULL ø
ø 3 ø 3 ø USER4 ø NULL ø NULL ø
ø 4 ø 4 ø elokiller ø NULL ø NULL ø
ø 5 ø 5 ø USER6 ø NULL ø NULL ø
+----+---------+----------------------+---------------+-------+
消息表:
+----+--------+----------+-----------+--------------+-----------------+---------------------+
ø id ø sender ø receiver ø message ø message_read ø message_visible ø message_date ø
+----+--------+----------+-----------+--------------+-----------------+---------------------+
ø 1 ø 4 ø 2 ø lulz ø 1 ø 2 ø 2018-05-15 10:24:36 ø
ø 2 ø 1 ø 4 ø Hey Buddy ø 1 ø NULL ø 2018-05-10 11:58:39 ø
ø 3 ø 2 ø 4 ø nooo ø 1 ø 2 ø 2018-05-15 10:24:36 ø
ø 4 ø 4 ø 2 ø shut up ø 1 ø 2 ø 2018-05-15 10:24:36 ø
+----+--------+----------+-----------+--------------+-----------------+---------------------+
如果您知道如何解决此问题，那就太好了。当我运行上面的sequel语句时，此时得到的结果如下。
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
ø message ø message_read ø message_date ø friend_id ø name ø image ø rn ø
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
ø Hey Buddy ø 1 ø 2018-05-10 11:58:39 ø 1 ø Welcome to MathMatch ø NULL ø 1 ø
ø lulz ø 1 ø 2018-05-15 10:24:36 ø 2 ø user3 ø NULL ø 1 ø
ø nooo ø 1 ø 2018-05-15 10:24:36 ø 2 ø user3 ø NULL ø 1 ø
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
我期望的结果:
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
ø message ø message_read ø message_date ø friend_id ø name ø image ø rn ø
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
ø Hey Buddy ø 1 ø 2018-05-10 11:58:39 ø 1 ø Welcome to MathMatch ø NULL ø 1 ø
ø lulz ø 1 ø 2018-05-15 10:24:36 ø 2 ø user3 ø NULL ø 1 ø
+-----------+--------------+---------------------+-----------+----------------------+-------+------+
如您所见，有两条消息来自同一个friend_id，即2。我只想检索该 friend 的一条消息。感谢您抽出宝贵时间。任何评论都会有帮助~
<小时/>另一个值得关注的场景。
当前消息表:
+----+--------+----------+-----------+--------------+-----------------+---------------------+
| id | sender | receiver | message | message_read | message_visible | message_date |
+----+--------+----------+-----------+--------------+-----------------+---------------------+
| 1 | 4 | 2 | lulz | 1 | 2 | 2018-05-15 10:24:36 |
| 2 | 1 | 4 | Hey Buddy | 1 | NULL | 2018-05-10 11:58:39 |
| 3 | 2 | 4 | nooo | 1 | 2 | 2018-05-15 10:24:36 |
| 4 | 4 | 2 | shut up | 1 | 2 | 2018-05-15 10:24:36 |
| 5 | 4 | 2 | heha | 1 | NULL | 2018-05-15 10:36:11 |
| 6 | 1 | 4 | SUP MATE | 1 | NULL | 2018-05-15 11:04:24 |
+----+--------+----------+-----------+--------------+-----------------+---------------------+
当前结果表:
+-----------+--------------+---------------------+-----------+-------+-------+------+
| message | message_read | message_date | friend_id | name | image | rn |
+-----------+--------------+---------------------+-----------+-------+-------+------+
| Hey Buddy | 1 | 2018-05-10 11:58:39 | 1 | JUAN | NULL | 1 |
| nooo | 1 | 2018-05-15 10:24:36 | 2 | user3 | NULL | 1 |
+-----------+--------------+---------------------+-----------+-------+-------+------+
预期结果
+-----------+--------------+---------------------+-----------+-------+-------+------+
| message | message_read | message_date | friend_id | name | image | rn |
+-----------+--------------+---------------------+-----------+-------+-------+------+
| SUP MATE | 1 | 2018-05-15 11:04:24 | 1 | JUAN | NULL | 1 |
| heha | 1 | 2018-05-15 10:36:11 | 2 | user3 | NULL | 1 |
+-----------+--------------+---------------------+-----------+-------+-------+------+
最佳答案
如果您将 ID 添加到结果表中，则可以通过friend_id 选择最新的 ID 和组。
假设您能够构建以下类型的 View
CREATE TABLE IF NOT EXISTS `MESSAGES` (
`id` int(6) unsigned NOT NULL,
`friend_id` int(3) unsigned NOT NULL,
`message_time` varchar(200) NOT NULL,
PRIMARY KEY (`id`)
) DEFAULT CHARSET=utf8;
INSERT INTO `MESSAGES` (`id`, `friend_id`, `message_time`) VALUES
('1', '1', '2018-05-10 11:58:39'),
('2', '2', '2018-05-15 10:24:36'),
('3', '2', '2018-05-15 10:24:36');
然后，您将首先选择按好友 ID 分组的最长时间消息，然后再选择其中 id 最长的消息。您可以使用以下 SQL 来实现。
SELECT MESSAGES.id, MESSAGES.friend_id, MESSAGES.message_time FROM
(
SELECT max(id) as id, friend_id FROM
(SELECT id,
m.friend_id,
m.message_time
FROM MESSAGES m
JOIN
(SELECT friend_id,
MAX(message_time) AS message_time
FROM MESSAGES
GROUP BY friend_id) mx ON mx.friend_id = m.friend_id
AND mx.message_time = m.message_time) b GROUP BY friend_id
) a JOIN
MESSAGES on a.id = MESSAGES.id ;
获取上述查询中 MESSAGES 表所需的结果应该相当简单，因为您已经按照您的问题做了类似的操作。
关于mysql - 从 MySQL 获取唯一数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50353371/