mysql - SQL查询从一个表中选择，其中要么不在另一个表中，要么在该表中具有特定值

Question

为简单起见，假设您有两张桌子...

users:
  user_id
  name
  age

clothing:
  user_id
  type
  color

您将如何选择所有 21 岁且根本没有任何衬衫或有衬衫但没有任何红色衬衫的用户？到目前为止，我的尝试看起来像......

SELECT users.user_id
FROM users LEFT JOIN clothing
ON users.user_id = clothing.user_id
AND clothing.type = shirt
AND clothing.color != red
WHERE users.age = 21;

如果我只想选择表中没有条目的用户的条目clothing（WHERE clothing.user_id IS NULL

Answer 1

SELECT users.user_id FROM users
WHERE users.age = 21
AND users.user_id NOT IN
    (SELECT user_id FROM clothing WHERE user_id IS NOT NULL AND type = 'shirt' AND color = 'red')

实际上更简单的方法是找出所有穿红衬衫的人，然后说，“把所有不在那个组的人都给我。” 话虽这么说，我认为你实际上需要clothing分成两张表：一张只列出衣服，另一张链接user_id到clothing_id.

Answer 2

怎么样：

SELECT users.user_id
FROM users u1 LEFT OUTER JOIN clothing c  # outer join makes sure you get a user back.
ON users.user_id = c.user_id
where c.type = shirt AND
(count(select * from users u2 where u1.user_id =- u2.user_id LEFT INNER JOIN clothing c2 on u2.user_id = c2.user_id) > 0 OR 
c.color != red) AND
users.age = 21;