关系减法错误结果

Question

我有两个疑问：

查询 1：

SELECT DISTINCT maker FROM Product WHERE type='laptop'

给出结果 {A, B, E, F, G}

和

查询 2：

SELECT DISTINCT maker FROM Product WHERE type='pc'

给出结果 {A, B, C, D, E}

问题是：我怎样才能得到query1 - query2 = {A, B, E, F, G} - {A, B, C, D, E} = {F, G}

我试过：

SELECT DISTINCT maker FROM Product WHERE type='pc' AND NOT EXISTS (
    SELECT DISTINCT maker FROM Product WHERE type='b'
);

但它给了我空的记录。对不起，我是 MySQL 的新手

Answer 1

您可以NOT IN改用：

SELECT DISTINCT maker 
FROM Product 
WHERE type='pc' 
  AND maker  NOT IN
 (
    SELECT maker 
    FROM Product
    WHERE type='b'
      AND maker IS NOT NULL -- to prevent empty results
 );

注意：这对NULL制造商来说是不正确的。

推荐的解决方案是NOT EXISTS，但您需要一个相关的子查询：

SELECT DISTINCT maker 
FROM Product AS p1
WHERE type='pc'
  AND NOT EXISTS
 (
    SELECT maker
    FROM Product AS p2
    WHERE type='b'
      AND p1.maker = p2.maker -- correlation to the outer SELECT
 );

如果 MySQL 支持，EXCEPT这将是一个简单的

SELECT maker FROM Product WHERE type='pc'
EXCEPT
SELECT maker FROM Product WHERE type='b';