Question

运动问题：

使用NOT EXISTS子句检查dogs表中不在users表中的所有用户。

我可以使用以下内部联接获得正确答案：

select d.user_guid as UserID, d.dog_guid as DogID 
from dogs d
left join users u
on d.user_guid = u.user_guid
where u.user_guid is null;

但是，我无法通过“不存在”子查询获得正确答案：

select d.user_guid as dUserID, d.dog_guid as dDogID
from dogs d
where not exists (select distinct u.user_guid from users u);

我上面错误查询的输出：

+---------+--------+
| dUserID | dDogID |
+---------+--------+

答案查询：

select d.user_guid as dUserID, d.dog_guid as dDogID
from dogs d
where not exists (select distinct u.user_guid 
                  from users u 
                  where u.user_guid = d.user_guid);

答案输出：

+---------+--------------------------------------+
| dUserID | dDogID                               |
+---------+--------------------------------------+
| None    | fd7c0a66-7144-11e5-ba71-058fbc01cf0b |
+---------+--------------------------------------+
| None    | fdbb6b7a-7144-11e5-ba71-058fbc01cf0b |
+---------+--------------------------------------+

我想知道什么可能的原因是为什么在Not Exists子句中没有u.user_guid = d.user_guid的情况下我的where子查询是错误的。

Answer 1

因为not exists (select distinct u.user_guid from users u)总是返回false

您可以像这样尝试NOT IN

SELECT
    d.user_guid AS dUserID,
    d.dog_guid AS dDogID
FROM
    dogs d
WHERE
    d.user_guid NOT IN (SELECT DISTINCT u.user_guid FROM user AS u);

如何使用不存在而不是内部联接？

1 个答案: