我面临一个非常奇怪的问题，希望您能向我解释。 我想做的是基于子查询中的合并列对结果集进行排序。让我更好地解释。
我有两个表：
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）正在浏览该网站时，我想向他显示其他订购的用户，其最爱在底部。 因此，我从以下查询开始：
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，这是我所得到的，而且我期望如此：
+----+-------+------------------------+
| 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，我要订购结果集。我认为ORDER BY子句就足够了：
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
这时，我得到的结果与上面相同：
+----+-------+------------------------+
| id | name | is_favorited_coalesced |
+----+-------+------------------------+
| 3 | user3 | 1 |
| 4 | user4 | 1 |
| 1 | user1 | 0 |
| 2 | user2 | 0 |
+----+-------+------------------------+
4 rows in set (0.00 sec)
然后我认为合并不利于即时排序，因此我添加了包装查询，但结果仍然相同。
为什么ORDER BY is_favorited_coalesced无法正常工作？我在这里想念什么？
编辑： 我尝试使用：
order by coalesce(favorites.is_favorited,0) asc
代替别名，但得到相同的结果：
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 我发现了另一个奇怪的行为。如果我尝试按ID列订购，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。 我在使用VirtualBox的Windows下在虚拟化的Fedora 25上使用MySQL 5.7.20。
编辑3
我在评论中建议：
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：
我已经跑步：
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)
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);
答案 0 :(得分：2)
这是错误Query returns wrong data if order by is present（或至少密切相关）。
它仍然（以非常相似的形式）仍然存在于MySQL 8.0.12中（请参见dbfiddle中的your example，尽管希望它一旦修复就不会显示不正确的行为）：尽管实际上是正确排序的现在（可能是因为您对其进行了计算），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 |
+----+-------+--------------+------------------------+
这似乎是与（非）实现有关的优化程序问题（MySQL 5.7有很多这样的问题）。您可以通过强制实现派生表的实现（例如，通过添加limit）来解决大多数错误：
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提到的那样，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'的派生表合并。您还可以使用它来全局禁用该功能，直到错误修复为止，因此您不必检查每个查询是否已损坏，而只需为已验证它们不受影响的查询启用它（这样他们就可以从中受益）再次优化）。