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简单
数据库

English Version

题目描述

产品表:Product

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id 是这个表的主键(具有唯一值的列)。
该表的每一行显示每个产品的名称和价格。

销售表:Sales

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+
这个表它可以有重复的行。 
product_id 是 Product 表的外键(reference 列)。
该表的每一行包含关于一个销售的一些信息。

 

编写解决方案,找出总销售额最高的销售者,如果有并列的,就都展示出来。

任意顺序 返回结果表。

返回结果格式如下所示。

 

示例 1:

输入:
Product 表:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+
Sales 表:
+-----------+------------+----------+------------+----------+-------+
| 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  |
+-----------+------------+----------+------------+----------+-------+
输出:
+-------------+
| seller_id   |
+-------------+
| 1           |
| 3           |
+-------------+
解释:Id 为 1 和 3 的销售者,销售总金额都为最高的 2800。

解法

方法一

MySQL

# Write your MySQL query statement below
SELECT seller_id
FROM Sales
GROUP BY seller_id
HAVING
    SUM(price) >= ALL(
        SELECT SUM(price)
        FROM Sales
        GROUP BY seller_id
    );

方法二

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            seller_id,
            SUM(price) AS tot,
            RANK() OVER (ORDER BY SUM(price) DESC) AS rk
        FROM Sales
        GROUP BY seller_id
    )
SELECT seller_id
FROM T
WHERE rk = 1;