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2362. 生成发票

English Version

题目描述

表: Products

+-------------+------+
| Column Name | Type |
+-------------+------+
| product_id  | int  |
| price       | int  |
+-------------+------+
product_id 包含唯一值。
该表中的每一行显示了一个产品的 ID 和一个单位的价格。

 

表: Purchases

+-------------+------+
| Column Name | Type |
+-------------+------+
| invoice_id  | int  |
| product_id  | int  |
| quantity    | int  |
+-------------+------+
(invoice_id, product_id) 是该表的主键（具有唯一值的列的组合）
该表中的每一行都显示了从发票中的一种产品订购的数量。

 

编写解决方案，展示价格最高的发票的详细信息。如果两个或多个发票具有相同的价格，则返回 invoice_id 最小的发票的详细信息。

以 任意顺序 返回结果表。

结果格式示例如下。

 

示例 1:

输入: 
Products 表:
+------------+-------+
| product_id | price |
+------------+-------+
| 1          | 100   |
| 2          | 200   |
+------------+-------+
Purchases 表:
+------------+------------+----------+
| invoice_id | product_id | quantity |
+------------+------------+----------+
| 1          | 1          | 2        |
| 3          | 2          | 1        |
| 2          | 2          | 3        |
| 2          | 1          | 4        |
| 4          | 1          | 10       |
+------------+------------+----------+
输出: 
+------------+----------+-------+
| product_id | quantity | price |
+------------+----------+-------+
| 2          | 3        | 600   |
| 1          | 4        | 400   |
+------------+----------+-------+
解释: 
发票 1: price = (2 * 100) = $200
发票 2: price = (4 * 100) + (3 * 200) = $1000
发票 3: price = (1 * 200) = $200
发票 4: price = (10 * 100) = $1000

最高价格是 1000 美元，最高价格的发票是 2 和 4。我们返回 ID 最小的发票 2 的详细信息。

解法

方法一

# Write your MySQL query statement below
WITH
    P AS (
        SELECT *
        FROM
            Purchases
            JOIN Products USING (product_id)
    ),
    T AS (
        SELECT invoice_id, SUM(price * quantity) AS amount
        FROM P
        GROUP BY invoice_id
        ORDER BY 2 DESC, 1
        LIMIT 1
    )
SELECT product_id, quantity, (quantity * price) AS price
FROM
    P
    JOIN T USING (invoice_id);