我正面临一个非常奇怪的问题,希望你能向我解释.
我要做的是根据子查询中的合并列来命令结果集.让我解释一下.
我有两张桌子：
CREATE TABLE `user` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
CREATE TABLE `user_favorites_user` (
`source_user_id` int(11) NOT NULL,
`favorited_user_id` int(11) NOT NULL,
KEY `source_user_id` (`source_user_id`),
KEY `favorited_user_id` (`favorited_user_id`),
CONSTRAINT `user_favorites_user_ibfk_1` FOREIGN KEY (`source_user_id`) REFERENCES `user` (`id`),
CONSTRAINT `user_favorites_user_ibfk_2` FOREIGN KEY (`favorited_user_id`) REFERENCES `user` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
当一个用户(假设ID = 1)正在浏览网站时,我想向他展示其他用户在底部订购了他的收藏夹.
所以,我从这个查询开始：
select user.*, coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user
left join (
select 1 as is_favorited, favorited_user_id from user_favorites_user
where source_user_id = '1'
) favorites on favorites.favorited_user_id = user.id
到目前为止一切都那么好,这就是我得到的和我期望的：
+----+-------+------------------------+
| id | name | is_favorited_coalesced |
+----+-------+------------------------+
| 3 | user3 | 1 |
| 4 | user4 | 1 |
| 1 | user1 | 0 |
| 2 | user2 | 0 |
+----+-------+------------------------+
4 rows in set (0.00 sec)
现在,我想订购结果集.我认为ORDER BY子句可能就足够了：
select user.*, coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user
left join (
select 1 as is_favorited, favorited_user_id from user_favorites_user
where source_user_id = '1'
) favorites on favorites.favorited_user_id = user.id
order by is_favorited_coalesced asc
此时,我得到与上面相同的结果：
+----+-------+------------------------+
| id | name | is_favorited_coalesced |
+----+-------+------------------------+
| 3 | user3 | 1 |
| 4 | user4 | 1 |
| 1 | user1 | 0 |
| 2 | user2 | 0 |
+----+-------+------------------------+
4 rows in set (0.00 sec)
然后我认为合并对于即时排序不好,所以我添加了一个包装器查询,但结果仍然相同.
为什么ORDER BY is_favorited_coalesced不起作用？我在这里错过了什么？
编辑：
我试过用：
order by coalesce(favorites.is_favorited,0) asc
而不是别名,但我得到了相同的结果：
select user.*, coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user left join ( select 1 as is_favorited, favorited_user_id from user_favorites_user where source_user_id = '1' ) favorites on favorites.favorited_user_id = user.id order by coalesce(favorites.is_favorited,0)
--------------
+----+-------+------------------------+
| id | name | is_favorited_coalesced |
+----+-------+------------------------+
| 3 | user3 | 1 |
| 4 | user4 | 1 |
| 1 | user1 | 0 |
| 2 | user2 | 0 |
+----+-------+------------------------+
4 rows in set (0.00 sec)
编辑2
我发现了另一种奇怪的行为.如果我尝试按ID列排序,这就是我得到的：
--------------
select user.*, coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user left join ( select 1 as is_favorited, favorited_user_id from user_favorites_user where source_user_id = '1' ) favorites on favorites.favorited_user_id = user.id order by id asc
--------------
+----+-------+------------------------+
| id | name | is_favorited_coalesced |
+----+-------+------------------------+
| 1 | user1 | 1 |
| 2 | user2 | 1 |
| 3 | user3 | 1 |
| 4 | user4 | 1 |
+----+-------+------------------------+
4 rows in set (0.00 sec)
我不知道为什么会这样.
我正在使用VirtualBox在Windows下的虚拟化Fedora 25上使用MySQL 5.7.20.
编辑3
正如我在评论中所建议的那样：
mysql> explain select user.*, coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user left join ( select 1 as is_favorited, favorited_user_id from user_favorites_user where source_user_id = '1' ) favorites on favorites.favorited_user_id = user.id order by is_favorited_coalesced asc;show warnings;
+----+-------------+---------------------+------------+-------+----------------------------------+----------------+---------+------+------+----------+----------------------------------------------------+
| id | select_type | table | partitions | type | possible_keys | key | key_len | ref | rows | filtered | Extra |
+----+-------------+---------------------+------------+-------+----------------------------------+----------------+---------+------+------+----------+----------------------------------------------------+
| 1 | SIMPLE | user | NULL | ALL | NULL | NULL | NULL | NULL | 4 | 100.00 | NULL |
| 1 | SIMPLE | user_favorites_user | NULL | range | source_user_id,favorited_user_id | source_user_id | 4 | NULL | 2 | 100.00 | Using where; Using join buffer (Block Nested Loop) |
+----+-------------+---------------------+------------+-------+----------------------------------+----------------+---------+------+------+----------+----------------------------------------------------+
2 rows in set, 1 warning (0.00 sec)
+-------+------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Level | Code | Message |
+-------+------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Note | 1003 | /* select#1 */ select `so_test`.`user`.`id` AS `id`,`so_test`.`user`.`name` AS `name`,coalesce(1,0) AS `is_favorited_coalesced` from `so_test`.`user` left join (`so_test`.`user_favorites_user`) on(((`so_test`.`user_favorites_user`.`favorited_user_id` = `so_test`.`user`.`id`) and (`so_test`.`user_favorites_user`.`source_user_id` = '1'))) where 1 order by `is_favorited_coalesced` |
+-------+------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
也：
mysql> SELECT @@sql_mode;
+-------------------------------------------------------------------------------------------------------------------------------------------+
| @@sql_mode |
+-------------------------------------------------------------------------------------------------------------------------------------------+
| ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION |
+-------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
编辑4：
我跑了：
mysql> SELECT @@optimizer_switch;
+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| @@optimizer_switch |
+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| index_merge=on,index_merge_union=on,index_merge_sort_union=on,index_merge_intersection=on,engine_condition_pushdown=on,index_condition_pushdown=on,mrr=on,mrr_cost_based=on,block_nested_loop=on,batched_key_access=off,materialization=on,semijoin=on,loosescan=on,firstmatch=on,duplicateweedout=on,subquery_materialization_cost_based=on,use_index_extensions=on,condition_fanout_filter=on,derived_merge=on |
+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
正如评论中所述.
包括用于快速测试的数据集：
SET NAMES utf8;
SET time_zone = '+00:00';
SET foreign_key_checks = 0;
SET sql_mode = 'NO_AUTO_VALUE_ON_ZERO';
CREATE TABLE `user` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
INSERT INTO `user` (`id`, `name`) VALUES
(1, 'user1'),
(2, 'user2'),
(3, 'user3'),
(4, 'user4');
CREATE TABLE `user_favorites_user` (
`source_user_id` int(11) NOT NULL,
`favorited_user_id` int(11) NOT NULL,
KEY `source_user_id` (`source_user_id`),
KEY `favorited_user_id` (`favorited_user_id`),
CONSTRAINT `user_favorites_user_ibfk_1` FOREIGN KEY (`source_user_id`) REFERENCES `user` (`id`),
CONSTRAINT `user_favorites_user_ibfk_2` FOREIGN KEY (`favorited_user_id`) REFERENCES `user` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
INSERT INTO `user_favorites_user` (`source_user_id`, `favorited_user_id`) VALUES
(1, 3),
(1, 4);
解决方法:
这是错误Query returns wrong data if order by is present(或至少密切相关).
它是(以非常相似的形式)仍然存在于MySQL 8.0.12中(参见例如dbfiddle中的your example,虽然它一旦被修复就不会显示不正确的行为)：虽然它现在实际上正确排序(可能是因为你做了一个对它进行计算),它仍然返回is_favorited的错误值：
select user.*, favorites.is_favorited,
coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user
left join (
select 1 as is_favorited, favorited_user_id from user_favorites_user
where source_user_id = '1'
) favorites on favorites.favorited_user_id = user.id
order by is_favorited_coalesced desc
+----+-------+--------------+------------------------+
| id | name | is_favorited | is_favorited_coalesced |
+----+-------+--------------+------------------------+
| 1 | user1 | | 1 |
| 2 | user2 | | 1 |
| 3 | user3 | | 0 |
| 4 | user4 | | 0 |
+----+-------+--------------+------------------------+
它似乎是一个与(非)物化相关的优化问题(MySQL 5.7有很多这些).您可以通过强制派生表的实现来解决大多数错误(例如,通过添加限制)：
select user.*, favorites.is_favorited,
coalesce(favorites.is_favorited,0) as is_favorited_coalesced from user
left join (
select 1 as is_favorited, favorited_user_id from user_favorites_user
where source_user_id = '1' limit 1000000
) favorites on favorites.favorited_user_id = user.id
order by is_favorited_coalesced desc
+----+-------+--------------+------------------------+
| id | name | is_favorited | is_favorited_coalesced |
+----+-------+--------------+------------------------+
| 1 | user1 | 1 | 1 |
| 2 | user2 | 1 | 1 |
| 3 | user3 | | 0 |
| 4 | user4 | | 0 |
+----+-------+--------------+------------------------+
正如@RaymondNijland所提到的,还有其他解决方法,例如在运行该查询之前禁用与set [GLOBAL|SESSION] optimizer_switch='derived_merge=off'合并的派生表.您还可以使用此功能全局禁用该功能,直到错误得到修复,因此您不必检查每个查询是否已损坏,只需启用它以查询您已经确认它们不受影响(因此他们可以从中获利)再次优化).