LeetCode刷题-数据库（MySQL）-1082. Sales Analysis I

1082. Sales Analysis I

一、题目描述

Table: Product

Column Name	Type
product_id	int
product_name	varchar
unit_price	int

product_id is the primary key of this table.

Table: Sales

Column Name	Type
seller_id	int
product_id	int
buyer_id	int
sale_date	date
quantity	int
price	int

This table has no primary key, it can have repeated rows.
product_id is a foreign key to Product table.

Write an SQL query that reports the best seller by total sales price, If there is a tie, report them all.

The query result format is in the following example:

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	2	3	2019-06-02	1	800
3	3	4	2019-05-13	2	2800

Result table:

seller_id
1
3

Both sellers with id 1 and 3 sold products with the most total price of 2800.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sales-analysis-i
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二、思路分析

Write an SQL query that reports the best seller by total sales price, If there is a tie, report them all.
本题要求找出销售冠军（如果有平手，那就把他们都找出来）。需要注意的有以下几点：

1. Sales表中的price是unit_price（单价）与quantity（数量）的乘积，所以直接将表中对应销售员每行的price加和即为销售额。所以，这就需要对seller_id使用GROUP BY后再SUM(price)。
2. 得到每个销售员的总销售额以后，按总销售额降序排列，用LIMIT 1取第一个，即为销冠。但这带来的问题是，如果有平手，则无法全部取出。解决方法是，先使用一个子查询得到销冠的销售额，然后再查询销售额等于销冠销售额的销售员即可。

三、代码实现

SELECT
	seller_id
FROM
	Sales
GROUP BY
	seller_id
HAVING
	SUM(price) = (
		SELECT
			SUM(price) AS pr
		FROM
			Sales
		GROUP BY seller_id
		ORDER BY pr DESC
		LIMIT 1
	);