我如何找到销售一种类型而不是另一种类型的制造商？

Question

我必须:

找到销售PC而不是笔记本电脑的制造商.

此查询未输出正确的结果:

SELECT maker, type FROM product
WHERE type = 'PC' AND type <> 'Laptop'

从此输出和正确的输出:

此查询的结果集:

SELECT maker, type FROM product

表模式:

即便是这个也行不通:

SELECT maker, type FROM product
WHERE type IN('PC') AND type NOT IN('Laptop')

注 - type仅为澄清而添加

Answer 1

有许多奇特的方法可以解决这个问题。在这里，第一个想到的:)

1. 简单的解决方案。

   SELECT DISTINCT maker
      FROM Product
        WHERE type = 'PC' AND maker NOT IN (
            SELECT maker
            FROM Product
            WHERE type = 'Laptop')
   

2. 使用JOIN

   SELECT DISTINCT Pr.maker
   FROM Product AS Pr LEFT JOIN Product AS Pr2
       ON Pr.maker = Pr2.maker AND Pr2.type = 'Laptop'
   WHERE Pr.type = 'PC' AND Pr2.maker IS NULL
   

3. 使用EXCEPT。

   SELECT DISTINCT maker
       FROM product AS Pr1
       WHERE type = 'PC'
   EXCEPT
   SELECT DISTINCT Pr2.maker
       FROM product AS Pr2
       WHERE type = 'laptop'
   

4. 使用DISTINCT和NOT EXISTS.

   SELECT DISTINCT maker
   FROM Product AS PcP
   WHERE type = 'PC' AND
         NOT EXISTS (SELECT maker
                     FROM Product
                     WHERE type = 'laptop' AND
                     maker = PcP.maker
                    )
   

5. 使用CASE增量 和HAVING.

   SELECT maker
       FROM product
   GROUP BY maker
   HAVING SUM(CASE WHEN type = 'PC' THEN 1 ELSE 0 END) > 0
      AND SUM(CASE WHEN type = 'Laptop' THEN 1 ELSE 0 END) = 0
   

6. 使用DISTINCT和 比较子查询

   SELECT DISTINCT maker
   FROM Product AS Pr
   WHERE (SELECT COUNT(1)
          FROM Product AS Pt
          WHERE Pt.type = 'PC' AND Pt.maker = Pr.maker
         ) > 0 
       AND
         (SELECT COUNT(1)
          FROM Product AS Pt2
          WHERE Pt2.type = 'Laptop' AND
          Pt2.maker = Pr.maker
         ) = 0
   

7. 使用HAVING和作弊（HAVING不能包含没有聚合的列，因此我们使用无意义的（在本例中）聚合MIN（或者MAX- 这并不重要）

   SELECT maker
       FROM (SELECT DISTINCT maker, type
             FROM Product
             WHERE type IN ('PC', 'Laptop')
            ) AS T
   GROUP BY maker
   HAVING COUNT(*) = 1 AND MIN(type) = 'PC'
   

Answer 2

尝试:

SELECT maker
FROM product
GROUP BY maker
HAVING SUM(CASE WHEN type = 'PC' THEN 1 ELSE 0 END) > 0
   AND SUM(CASE WHEN type = 'Laptop' THEN 1 ELSE 0 END) = 0