【发布时间】:2021-06-07 21:25:04
【问题描述】:
5.7 中的查询在 8.0.23 中不再有效。 针对 windows 8.0.23 x64 和 linux docker 容器 8.0.23 进行了测试。
当前的sql模式: ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_ENGINE_SUBSTITUTION
这不是最聪明的 sql,但这不是重点。
TL;DR: 关键是 'bob' 行应该在最后一个查询中显示(如 5.7 中),但在 8 中不再显示。
详情:
在 4 个表设置和插入后,查询将逐步构建,并使用早期的查询来证明错误。最里面的选择应该什么也找不到。中间选择只是重复了这一点。最后一个查询将 uu 连接到那个“无”,哦,令人惊讶的是,什么都没有变成什么! (最外面的 where 子句失败了。
问题:
发生了什么变化?我们做错了什么?我们在 5.7 中假设什么不再正确?还是bug?
数据:
drop table if exists uu, ww, kk, aa;
CREATE TABLE uu (
id int NOT NULL AUTO_INCREMENT,
name varchar(30),
PRIMARY KEY (`id`)
);
CREATE TABLE ww (
id int NOT NULL,
PRIMARY KEY (`id`)
);
CREATE TABLE kk (
id int NOT NULL AUTO_INCREMENT,
uid int,
wid int,
state int,
PRIMARY KEY (`id`),
CONSTRAINT `kk_uid` FOREIGN KEY (uid) REFERENCES uu (id),
CONSTRAINT `kk_wid` FOREIGN KEY (wid) REFERENCES ww (id)
);
CREATE TABLE aa (
id int NOT NULL AUTO_INCREMENT,
uid int,
wid int,
PRIMARY KEY (`id`),
CONSTRAINT `aa_uid` FOREIGN KEY (uid) REFERENCES uu (id),
CONSTRAINT `aa_wid` FOREIGN KEY (wid) REFERENCES ww (id)
);
insert into uu (name) values ('alice'), ('bob'), ('chuck');
insert into ww (id) values (10),(11);
insert into kk (uid,wid,state) values (2,10,99);
insert into kk (uid,wid,state) values (2,11,99);
insert into aa (uid,wid) values (2,10), (2,11);
-- plain joins:
select u.*, "][", k.*, "][", a.*
from uu u
left join kk k on k.uid = u.id
left join aa a on u.id = a.uid
order by u.id,k.id;
select a2.*, '][', k3.* -- distinct a2.wid
from aa a2
left join kk k3 on a2.uid = k3.uid and a2.wid = k3.wid
where k3.wid is null
;
SELECT a.*
FROM aa a
where a.wid in
(
select distinct a2.wid
from aa a2
left join kk k3 on a2.uid = k3.uid and a2.wid = k3.wid
where k3.wid is null
)
;
select u.*, '][k', k.*, '][k1', k1.*,'][tmp', tmp.*
FROM uu u
LEFT JOIN kk k ON k.uid = u.id
LEFT JOIN kk k1 ON k1.uid = u.id AND k1.state != 99
LEFT JOIN
(
SELECT a.*
FROM aa a
where a.wid in
(
select distinct a2.wid
from aa a2
left join kk k3 on a2.uid = k3.uid and a2.wid = k3.wid
where k3.wid is null
)
) tmp ON tmp.uid = u.id
-- ;
WHERE tmp.wid IS NULL
AND (
k.uid IS NULL
OR
(k.state = 99 AND k1.id IS NULL)
)
;
【问题讨论】:
-
确实看起来像一个错误; tmp 子查询本身什么也没找到,但是当加入时找到 bobs:dbfiddle.uk/… 仅在 mysql 8 中,而不是 5.7 或 mariadb
-
哇,这个小提琴网站真是太棒了。谢谢!
标签: mysql select subquery left-join mysql-8.0