Question

我有一个消息传递系统，该系统的表“ message”仅包含“主题”，然后是“ message_user”，其中包含消息正文，谁发送，谁发送消息以及是否删除/未读。

#Message Table

CREATE TABLE `message` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `subject` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

INSERT INTO `message` (`id`, `subject`)
VALUES
    (1, 'Test'),
    (2, 'Test Again');

#Message User Table

CREATE TABLE `message_user` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `message_id` int(11) NOT NULL,
  `user_id` int(11) NOT NULL,
  `interlocutor` int(11) DEFAULT NULL,
  `body` text,
  `folder` enum('inbox','sent') NOT NULL,
  `starmark` tinyint(1) NOT NULL DEFAULT '0',
  `unread` tinyint(1) NOT NULL DEFAULT '1',
  `deleted` enum('none','trash','deleted') NOT NULL DEFAULT 'none',
  `date` datetime DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=7 DEFAULT CHARSET=latin1;

INSERT INTO `message_user` (`id`, `message_id`, `user_id`, `interlocutor`, `body`, `folder`, `starmark`, `unread`, `deleted`, `date`)
VALUES
    (1, 1, 1, 2, 'Hi, how are you?', 'sent', 0, 1, 'none', '2018-10-23 09:36:02'),
    (2, 1, 2, 1, 'Hi, how are you?', 'inbox', 0, 1, 'none', '2018-10-23 09:36:02'),
    (3, 1, 2, 1, 'I am good thanks, you?', 'sent', 0, 1, 'none', '2018-10-23 09:46:02'),
    (4, 1, 1, 2, 'I am good thanks, you?', 'inbox', 0, 1, 'none', '2018-10-23 09:46:02'),
    (5, 2, 1, 3, 'Hi!', 'sent', 0, 1, 'none', '2018-10-23 09:50:22'),
    (6, 2, 3, 1, 'Hi!', 'inbox', 0, 1, 'none', '2018-10-23 09:50:22');

我写了以下查询：

SELECT
    *
FROM message m
JOIN message_user mu ON m.id = mu.message_id
WHERE mu.deleted = 'none'
    AND mu.user_id = 1 #user_id of person checking messages
ORDER BY mu.id DESC;

但是，尽管只有两次对话，但当前返回3行。我尝试使用GROUP BY，但它仍然显示3行。

我希望上面示例中的前两行不是最后一行。

我希望查询返回包含我（user_id）参与的最新消息的对话列表。

Answer 1

由于your MySQL version是8.0+，因此我们可以使用Window functions，例如Row_number()；否则解决方案将非常冗长，using Session variables。
对于m.id的分区（组），我们将确定行号值。行号值将按date的降序排列。
现在，我们只需要将此结果集用作Derived Table，并考虑那些行号值为1的行。
Date是keyword in MySQL。您应该避免使用它来命名列/表。仍然需要这样做，仍然需要使用反引号。

尝试以下操作（ DB Fiddle DEMO ）：

SELECT  dt.*
FROM (
      SELECT  m.id,
              m.subject,
              mu.id AS message_user_id,
              mu.message_id,
              mu.user_id,
              mu.interlocutor,
              mu.body,
              mu.folder,
              mu.starmark,
              mu.unread,
              mu.deleted,
              mu.`date`,
              Row_number()
                OVER (PARTITION BY m.id 
                      ORDER BY mu.`date` DESC) AS row_no
        FROM  message m
        JOIN  message_user mu
          ON  m.id = mu.message_id
        WHERE mu.deleted = 'none'
          AND mu.user_id = 1 ) AS dt
WHERE    dt.row_no = 1
ORDER BY dt.id DESC

Answer 2

尝试一下：

select 
    m.id as id_message, m.subject as subject_message,
    mu.id as id_message_user, mu.interlocutor, mu.body, mu.folder, mu.starmark, mu.deleted, mu.date 
from message as m
inner join message_user as mu on mu.message_id = m.id and mu.deleted = 'none' and mu.user_id = 1
group by id_message
order by id_message_user desc

我删除了

mu.user_id ：处于内部联接条件，因此始终为“无”
mu.unread ：相同，始终为1
mu.message_id ：id_message的副本

http://sqlfiddle.com/#!9/91a5e4/15

仅显示每次对话的最新消息

2 个答案: