LeetCode MySQL 1083. 销售分析 II

文章目录

1. 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。

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/sales-analysis-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

2. 解题

# Write your MySQL query statement below
select distinct buyer_id
from Product p, Sales s
where p.product_id = s.product_id
        and buyer_id not in
                    (
                        select buyer_id
                        from Product p, Sales s
                        where p.product_id = s.product_id
                                and product_name = 'iPhone'
                    )
        and product_name = 'S8'

or

# Write your MySQL query statement below
select buyer_id
from Product p, Sales s
where p.product_id = s.product_id
group by buyer_id
having sum(product_name = 'S8') > 0 and sum(product_name = 'iPhone')=0

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转载自blog.csdn.net/qq_21201267/article/details/107449727