Mysql GROUP BY和ORDER BY DESC

Question

这是我的完整查询:

SELECT * FROM `clientgroupassign`
LEFT JOIN `clients` ON `clientgroupassign`.clientId = `clients`.clientId
LEFT JOIN `users` ON `carerId` = `userId`
LEFT JOIN 
    (SELECT * FROM 
        (SELECT * FROM `contacts` WHERE `contactGroup` = 4 ORDER BY `contactId` DESC) 
        as `contacts` 
    GROUP BY (`contactClientId`)
    )  AS `contacts` ON `contactClientId` = `clients`.clientId
WHERE groupId = 4
ORDER BY `clients`.clientId

第三次连接存在问题导致脚本执行约1分钟.当我在PMA中分开运行时:

SELECT * FROM (SELECT * FROM `contacts` WHERE `contactGroup` = 4 ORDER BY `contactId` DESC) AS `contacts` GROUP BY (`contactClientId`)

执行仍然很长.

我想要的是为contacts第4组中的每个客户端添加一个,最后添加的行(客户端可以在不同的组中).

谢谢.

Answer 1

要获取"组4中每个客户的联系人最后添加的行",请尝试以下操作:

SELECT
    c.*
FROM(
    SELECT
        contactClientId,
        MAX(contactId) as cid
    FROM
        contacts
    WHERE
        contactGroup = 4
    GROUP BY
        contactClientId
    ORDER BY
        NULL
) as tmp
INNER JOIN contacts as c
    ON c.contactId = tmp.cid
    AND c.contactClientId = tmp.contactClientId 

如果contactId是PK,contacts则不需要第二个连接子句.

默认情况下,MySQL会对所有GROUP BY col1,col2,...查询进行排序,就好像您在查询中指定了ORDER BY col1,col2,.... 如果显式包含一个包含相同列列表的ORDER BY子句,MySQL会在没有任何速度惩罚的情况下对其进行优化,尽管排序仍然存在.如果查询包含GROUP BY但您希望避免对结果进行排序的开销,则可以通过指定ORDER BY NULL来禁止排序.

完整的文档.

完整查询:

SELECT * FROM `clientgroupassign`
LEFT JOIN `clients` ON `clientgroupassign`.clientId = `clients`.clientId
LEFT JOIN `users` ON `carerId` = `userId`
LEFT JOIN 
    (
        SELECT
            c.*
        FROM(
            SELECT
                contactClientId,
                MAX(contactId) as cid
            FROM
                contacts
            WHERE
                contactGroup = 4
            GROUP BY
                contactClientId
            ORDER BY
                NULL
        ) as tmp
        INNER JOIN contacts as c
            ON c.contactId = tmp.cid
            AND c.contactClientId = tmp.contactClientId 
    )  AS `contacts` ON `contactClientId` = `clients`.clientId
WHERE groupId = 4
ORDER BY `clients`.clientId