文章目录
- 1. 题目
- 2. 解题
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
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2. 解题
# Write your MySQL query statement below
select distinct buyer_id
from Product p, Sales s
where p.product_id = s.product_idand buyer_id not in(select buyer_idfrom Product p, Sales swhere p.product_id = s.product_idand 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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