一、题目

Table: Product

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id 是这张表的主键

Table: Sales

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+
这个表没有主键，它可以有重复的行.
product_id 是 Product 表的外键.

编写一个 SQL 查询，查询购买了 S8 手机却没有购买 iPhone 的买家。注意这里 S8 和 iPhone 是 Product 表中的产品。

查询结果格式如下图表示：

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+

Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 1          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 3        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+

Result table:
+-------------+
| buyer_id    |
+-------------+
| 1           |
+-------------+
id 为 1 的买家购买了一部 S8，但是却没有购买 iPhone，而 id 为 3 的买家却同时购买了这 2 部手机。

二、解决

1、count() 和 if()

思路：

该题关键在于筛选 购买了 S8 手机却没有购买 iPhone 的买家 转化为 某个买家购买S8的次数至少为1，购买iPhone的次数为0，先用 GROUP BY+COUNT+IF 来统计一个各个买家购买各手机的次数。

代码：

SELECT buyer_id AS 'buyer_id'
FROM Product AS p1
INNER JOIN Sales AS s1
ON p1.product_id = s1.product_id
GROUP BY buyer_id
HAVING SUM(IF(p1.product_name = 'S8', 1, 0)) > 0 AND SUM(IF(p1.product_name = 'iPhone', 1, 0)) = 0
# HAVING COUNT(IF(p1.product_name = 'S8', 1, NULL)) > 0 AND COUNT(IF(p1.product_name = 'iPhone', 1, NULL)) = 0
# HAVING SUM( CASE WHEN p1.product_name = 'S8' THEN 1 ELSE 0 END ) > 0 AND SUM( CASE WHEN p1.product_name = 'iPhone' THEN 1 ELSE 0 END ) = 0
# HAVING COUNT(CASE WHEN p1.product_name = 'S8' THEN 1 ELSE NULL END) > 0 AND COUNT(CASE WHEN p1.product_name = 'iPhone' THEN 1 ELSE NULL END) = 0

2、子查询

思路： 略。
代码：

SELECT DISTINCT buyer_id AS "buyer_id"
FROM
    (
        SELECT buyer_id,
            SUM(IF(product_name = 'S8', 1, 0)) AS s8_cnt,
            SUM(IF(product_name = 'iPhone', 1, 0)) AS iphone_cnt
        FROM Product AS p1
        INNER JOIN Sales s1 ON p1.product_id = s1.product_id
        GROUP BY buyer_id
        HAVING s8_cnt > 0 AND iphone_cnt = 0
    ) AS tmp1
;