有一个messages包含数据的表,如下所示:
messages
Id Name Other_Columns ------------------------- 1 A A_data_1 2 A A_data_2 3 A A_data_3 4 B B_data_1 5 B B_data_2 6 C C_data_1
如果我运行查询select * from messages group by name,我将得到的结果为:
select * from messages group by name
1 A A_data_1 4 B B_data_1 6 C C_data_1
什么查询将返回以下结果?
3 A A_data_3 5 B B_data_2 6 C C_data_1
即,应返回每个组中的最后一条记录。
目前,这是我使用的查询:
SELECT * FROM (SELECT * FROM messages ORDER BY id DESC) AS x GROUP BY name
但这看起来效率很低。还有其他方法可以达到相同的结果吗?
MySQL 8.0现在支持窗口功能,就像几乎所有流行的SQL实现一样。使用这种标准语法,我们可以编写每组最多n个查询:
WITH ranked_messages AS ( SELECT m.*, ROW_NUMBER() OVER (PARTITION BY name ORDER BY id DESC) AS rn FROM messages AS m ) SELECT * FROM ranked_messages WHERE rn = 1;
以下是我在2009年为此问题写的原始答案:
我这样写解决方案:
SELECT m1.* FROM messages m1 LEFT JOIN messages m2 ON (m1.name = m2.name AND m1.id < m2.id) WHERE m2.id IS NULL;
关于性能,一种解决方案可能会更好,这取决于数据的性质。因此,您应该测试两个查询,并使用给定数据库性能最好的查询。
例如,我有一个StackOverflow August数据转储的副本。我将其用于基准测试。该Posts表中有1,114,357行。它在Macbook Pro 2.40GHz的MySQL 5.0.75上运行。
Posts
我将编写查询以查找给定用户ID(我的用户)的最新帖子。
首先 在子查询中使用@Eric 所示的技术GROUP BY:
GROUP BY
SELECT p1.postid FROM Posts p1 INNER JOIN (SELECT pi.owneruserid, MAX(pi.postid) AS maxpostid FROM Posts pi GROUP BY pi.owneruserid) p2 ON (p1.postid = p2.maxpostid) WHERE p1.owneruserid = 20860; 1 row in set (1 min 17.89 sec)
甚至EXPLAIN分析也要花费超过16秒的时间:
EXPLAIN
+----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+ | 1 | PRIMARY | <derived2> | ALL | NULL | NULL | NULL | NULL | 76756 | | | 1 | PRIMARY | p1 | eq_ref | PRIMARY,PostId,OwnerUserId | PRIMARY | 8 | p2.maxpostid | 1 | Using where | | 2 | DERIVED | pi | index | NULL | OwnerUserId | 8 | NULL | 1151268 | Using index | +----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+ 3 rows in set (16.09 sec)
现在用产生同样的查询结果 我的技术有LEFT JOIN:
LEFT JOIN
SELECT p1.postid FROM Posts p1 LEFT JOIN posts p2 ON (p1.owneruserid = p2.owneruserid AND p1.postid < p2.postid) WHERE p2.postid IS NULL AND p1.owneruserid = 20860; 1 row in set (0.28 sec)
该EXPLAIN分析表明,这两个表都能够使用他们的指标:
+----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+ | 1 | SIMPLE | p1 | ref | OwnerUserId | OwnerUserId | 8 | const | 1384 | Using index | | 1 | SIMPLE | p2 | ref | PRIMARY,PostId,OwnerUserId | OwnerUserId | 8 | const | 1384 | Using where; Using index; Not exists | +----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+ 2 rows in set (0.00 sec)
这是我的Posts桌子的DDL :
CREATE TABLE `posts` ( `PostId` bigint(20) unsigned NOT NULL auto_increment, `PostTypeId` bigint(20) unsigned NOT NULL, `AcceptedAnswerId` bigint(20) unsigned default NULL, `ParentId` bigint(20) unsigned default NULL, `CreationDate` datetime NOT NULL, `Score` int(11) NOT NULL default '0', `ViewCount` int(11) NOT NULL default '0', `Body` text NOT NULL, `OwnerUserId` bigint(20) unsigned NOT NULL, `OwnerDisplayName` varchar(40) default NULL, `LastEditorUserId` bigint(20) unsigned default NULL, `LastEditDate` datetime default NULL, `LastActivityDate` datetime default NULL, `Title` varchar(250) NOT NULL default '', `Tags` varchar(150) NOT NULL default '', `AnswerCount` int(11) NOT NULL default '0', `CommentCount` int(11) NOT NULL default '0', `FavoriteCount` int(11) NOT NULL default '0', `ClosedDate` datetime default NULL, PRIMARY KEY (`PostId`), UNIQUE KEY `PostId` (`PostId`), KEY `PostTypeId` (`PostTypeId`), KEY `AcceptedAnswerId` (`AcceptedAnswerId`), KEY `OwnerUserId` (`OwnerUserId`), KEY `LastEditorUserId` (`LastEditorUserId`), KEY `ParentId` (`ParentId`), CONSTRAINT `posts_ibfk_1` FOREIGN KEY (`PostTypeId`) REFERENCES `posttypes` (`PostTypeId`) ) ENGINE=InnoDB;