Skip to content

Latest commit

 

History

History
166 lines (135 loc) · 4.97 KB

File metadata and controls

166 lines (135 loc) · 4.97 KB
comments difficulty edit_url tags
true
中等
数据库

English Version

题目描述

表: Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id 是该表主键.
该表包含消费者的信息.

 

表: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| product_id    | int     |
+---------------+---------+
order_id 是该表主键.
该表包含消费者customer_id产生的订单.
不会有商品被相同的用户在一天内下单超过一次.

 

表: Products

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
| price         | int     |
+---------------+---------+
product_id 是该表主键.
该表包含所有商品的信息.

 

写一个解决方案, 找到每件商品的最新订单(可能有多个).

返回的结果以 product_name 升序排列, 如果有排序相同, 再以 product_id 升序排列. 如果还有排序相同, 再以 order_id 升序排列.

查询结果格式如下例所示。

 

示例 1:

输入:
Customers表:
+-------------+-----------+
| customer_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+
Orders表:
+----------+------------+-------------+------------+
| order_id | order_date | customer_id | product_id |
+----------+------------+-------------+------------+
| 1        | 2020-07-31 | 1           | 1          |
| 2        | 2020-07-30 | 2           | 2          |
| 3        | 2020-08-29 | 3           | 3          |
| 4        | 2020-07-29 | 4           | 1          |
| 5        | 2020-06-10 | 1           | 2          |
| 6        | 2020-08-01 | 2           | 1          |
| 7        | 2020-08-01 | 3           | 1          |
| 8        | 2020-08-03 | 1           | 2          |
| 9        | 2020-08-07 | 2           | 3          |
| 10       | 2020-07-15 | 1           | 2          |
+----------+------------+-------------+------------+
Products表:
+------------+--------------+-------+
| product_id | product_name | price |
+------------+--------------+-------+
| 1          | keyboard     | 120   |
| 2          | mouse        | 80    |
| 3          | screen       | 600   |
| 4          | hard disk    | 450   |
+------------+--------------+-------+
输出:
+--------------+------------+----------+------------+
| product_name | product_id | order_id | order_date |
+--------------+------------+----------+------------+
| keyboard     | 1          | 6        | 2020-08-01 |
| keyboard     | 1          | 7        | 2020-08-01 |
| mouse        | 2          | 8        | 2020-08-03 |
| screen       | 3          | 3        | 2020-08-29 |
+--------------+------------+----------+------------+
解释:
keyboard 的最新订单在2020-08-01, 在这天有两次下单.
mouse 的最新订单在2020-08-03, 在这天只有一次下单.
screen 的最新订单在2020-08-29, 在这天只有一次下单.
hard disk 没有被下单, 我们不把它包含在结果表中.

解法

方法一:等值连接 + 窗口函数

我们可以使用等值连接,将 Orders 表和 Products 表按照 product_id 连接起来,然后使用窗口函数 rank(),对 Orders 表中的每个 product_id 进行分组,按照 order_date 降序排列,然后取出每个分组中排名第一的记录。

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            *,
            RANK() OVER (
                PARTITION BY product_id
                ORDER BY order_date DESC
            ) AS rk
        FROM
            Orders
            JOIN Products USING (product_id)
    )
SELECT product_name, product_id, order_id, order_date
FROM T
WHERE rk = 1
ORDER BY 1, 2, 3;