Question

我有两张桌子：

user：id，name message：sender_id，receiver_id，message，read_at，created_at

我需要检索2个结果，并且我正在尝试找到最佳解决方案。我已经包含了我最后使用的查询。

我需要检索一个用户列表，并且每个用户都有可用的信息是否有来自每个用户的任何未读消息（他们作为发件人，我作为接收者）以及是否有任何读取消息在我们之间（他们发送我是接收者或我发送他们是接收者）
我需要与上面相同，但只包括我们之间有任何消息传递的成员，先按未读，然后按最后收到的消息排序。

在第一种情况下，我不需要计数，我只需要知道是否至少有一条未读消息。我正在发布代码和我当前的查询，请在有机会时查看：

我担心的是：在第二个查询中，我想通过messages.created_at订购，但我不认为我可以通过分组来做到这一点？而且我也不知道这种方法是否是最优化和最快速的。

CREATE TABLE `user` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
)

INSERT INTO `user` VALUES (1,'User 1'),(2,'User 2'),(3,'User 3'),(4,'User 4'),(5,'User 5');

CREATE TABLE `message` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `sender_id` bigint(20) DEFAULT NULL,
  `receiver_id` bigint(20) DEFAULT NULL,
  `message` text,
  `read_at` datetime DEFAULT NULL,
  `created_at` datetime NOT NULL,
  PRIMARY KEY (`id`)
)

INSERT INTO `message` VALUES (1,3,1,'Messge',NULL,'2010-10-10 10:10:10'),(2,1,4,'Hey','2010-10-10 10:10:12','2010-10-10 10:10:11'),(3,4,1,'Hello','2010-10-10 10:10:19','2010-10-10 10:10:15'),(4,1,4,'Again','2010-10-10 10:10:25','2010-10-10 10:10:21'),(5,3,1,'Hiii',NULL,'2010-10-10 10:10:21');

SELECT u.*, m_new.id as have_new, m.id as have_any
FROM user u 
LEFT JOIN message m_new ON (u.id = m_new.sender_id AND m_new.receiver_id = 1 AND m_new.read_at IS NULL)
LEFT JOIN message m ON ((u.id = m.sender_id AND m.receiver_id = 1) OR (u.id = m.receiver_id AND m.sender_id = 1))
GROUP BY u.id

SELECT u.*, m_new.id as have_new, m.id as have_any
FROM user u 
LEFT JOIN message m_new ON (u.id = m_new.sender_id AND m_new.receiver_id = 1 AND m_new.read_at IS NULL)
LEFT JOIN message m ON ((u.id = m.sender_id AND m.receiver_id = 1) OR (u.id = m.receiver_id AND m.sender_id = 1))
where m.id IS NOT NULL
GROUP BY u.id

Answer 1

第一个查询：

SELECT  u.*,
        EXISTS
        (
        SELECT  NULL
        FROM    message
        WHERE   sender_id = u.id
                AND receiver_id = 1
                AND read_at IS NULL
        ) have_new,
        EXISTS
        (
        SELECT  NULL
        FROM    message
        WHERE   (
                sender_id = u.id AND receiver_id = 1
                OR sender_id = 1 AND receiver_id = u.id
                )
                AND read_at IS NOT NULL
        ) have_any
FROM    user u

第二个问题：

SELECT  u.*,
        EXISTS
        (
        SELECT  NULL
        FROM    message
        WHERE   sender_id = u.id
                AND receiver_id = 1
                AND read_at IS NULL
        ) have_new,
        (
        SELECT  created_at
        FROM    message
        WHERE   (
                sender_id = u.id AND receiver_id = 1
                OR sender_id = 1 AND receiver_id = u.id
                )
        ORDER BY
                created_at DESC
        LIMIT 1
        ) last_message
FROM    (
        SELECT  DISTINCT sender_id AS party
        FROM    message
        WHERE   receiver_id = 1
        UNION
        SELECT  DISTINCT receiver_id
        FROM    message
        WHERE   sender_id = 1
        ) m
JOIN    user u
ON      u.id = m.party
ORDER BY
        have_new DESC, last_message DESC

创建以下索引：

messages (sender_id, receiver_id, last_message)
messages (receiver_id, sender_id, last_message)

Answer 2

我非常确定JOIN的性能比子查询更高。我看到的一个问题是表创建脚本中没有任何日期或外键索引。你会想要那些。如果我正确理解你的问题，这就是我如何完成查询＃2：

SELECT u.*,
    m_new.id AS have_new,
    MAX(m_new.created_at) AS new_created,
    m.id AS have_any,
    MAX(m.created_at) AS created
FROM USER u 
LEFT JOIN message AS m_new
    ON u.id = m_new.sender_id
    AND m_new.receiver_id = 1
    AND m_new.read_at IS NULL
LEFT JOIN message AS m
    ON (u.id = m.sender_id AND m.receiver_id = 1)
    OR (u.id = m.receiver_id AND m.sender_id = 1)
WHERE m.id IS NOT NULL
GROUP BY u.id
ORDER BY new_created DESC,
    created DESC
;

以下是JOIN s vs子查询中的SO中的几个好链接：

需要帮助2 MySql查询。加入vs子查询

2 个答案: