leetcode----mysql

1251. 平均售价 - 力扣（LeetCode）

表：Prices

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| start_date    | date    |
| end_date      | date    |
| price         | int     |
+---------------+---------+
(product_id，start_date，end_date) 是 prices 表的主键（具有唯一值的列的组合）。
prices 表的每一行表示的是某个产品在一段时期内的价格。
每个产品的对应时间段是不会重叠的，这也意味着同一个产品的价格时段不会出现交叉。

表：UnitsSold

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| purchase_date | date    |
| units         | int     |
+---------------+---------+
该表可能包含重复数据。
该表的每一行表示的是每种产品的出售日期，单位和产品 id。

编写解决方案以查找每种产品的平均售价。average_price 应该 四舍五入到小数点后两位。如果产品没有任何售出，则假设其平均售价为 0。

返回结果表 无顺序要求 。

结果格式如下例所示。

示例 1：

输入：
Prices table:
+------------+------------+------------+--------+
| product_id | start_date | end_date   | price  |
+------------+------------+------------+--------+
| 1          | 2019-02-17 | 2019-02-28 | 5      |
| 1          | 2019-03-01 | 2019-03-22 | 20     |
| 2          | 2019-02-01 | 2019-02-20 | 15     |
| 2          | 2019-02-21 | 2019-03-31 | 30     |
+------------+------------+------------+--------+
UnitsSold table:
+------------+---------------+-------+
| product_id | purchase_date | units |
+------------+---------------+-------+
| 1          | 2019-02-25    | 100   |
| 1          | 2019-03-01    | 15    |
| 2          | 2019-02-10    | 200   |
| 2          | 2019-03-22    | 30    |
+------------+---------------+-------+
输出：
+------------+---------------+
| product_id | average_price |
+------------+---------------+
| 1          | 6.96          |
| 2          | 16.96         |
+------------+---------------+
解释：
平均售价 = 产品总价 / 销售的产品数量。
产品 1 的平均售价 = ((100 * 5)+(15 * 20) )/ 115 = 6.96
产品 2 的平均售价 = ((200 * 15)+(30 * 30) )/ 230 = 16.96

这道题的创建表的语句为

Create table If Not Exists Prices (product_id int, start_date date, end_date date, price int)
Create table If Not Exists UnitsSold (product_id int, purchase_date date, units int)
Truncate table Prices
insert into Prices (product_id, start_date, end_date, price) values ('1', '2019-02-17', '2019-02-28', '5')
insert into Prices (product_id, start_date, end_date, price) values ('1', '2019-03-01', '2019-03-22', '20')
insert into Prices (product_id, start_date, end_date, price) values ('2', '2019-02-01', '2019-02-20', '15')
insert into Prices (product_id, start_date, end_date, price) values ('2', '2019-02-21', '2019-03-31', '30')
Truncate table UnitsSold
insert into UnitsSold (product_id, purchase_date, units) values ('1', '2019-02-25', '100')
insert into UnitsSold (product_id, purchase_date, units) values ('1', '2019-03-01', '15')
insert into UnitsSold (product_id, purchase_date, units) values ('2', '2019-02-10', '200')
insert into UnitsSold (product_id, purchase_date, units) values ('2', '2019-03-22', '30')

这道题不会

我看答案了

1. 计算总销售额：

   • SUM(P.price * U.units) = 20.00 + 60.00 + 30.00 = 110.00

2. 计算总单位数：

   • SUM(U.units) = 2 + 3 + 1 = 6

3. 计算加权平均价格：

   • 110.00 / 6 = 18.33

4. 四舍五入到两位小数：

   • ROUND(18.33, 2) = 18.33

SELECT

    product_id,

    IFNULL(Round(SUM(sales) / SUM(units), 2), 0) AS average_price

FROM (

    SELECT

        Prices.product_id AS product_id,

        Prices.price * UnitsSold.units AS sales,

        UnitsSold.units AS units

    FROM Prices

    LEFT JOIN UnitsSold ON Prices.product_id = UnitsSold.product_id

    AND (UnitsSold.purchase_date BETWEEN Prices.start_date AND Prices.end_date)

) T

GROUP BY product_id