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中等
数据库

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题目描述

表:Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id 是该表具有唯一值的列
该表包含消费者的信息

 

表:Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| cost          | int     |
+---------------+---------+
order_id 是该表具有唯一值的列
该表包含 id 为 customer_id 的消费者的订单信息
每一个消费者 每天一笔订单

 

写一个解决方案,找到每个用户的最近三笔订单。如果用户的订单少于 3 笔,则返回他的全部订单。

返回的结果按照 customer_name 升序 排列。如果有相同的排名,则按照 customer_id 升序 排列。如果排名还有相同,则按照 order_date 降序 排列。

结果格式如下例所示:

 

示例 1:

输入:
Customers
+-------------+-----------+
| customer_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+

Orders
+----------+------------+-------------+------+
| order_id | order_date | customer_id | cost |
+----------+------------+-------------+------+
| 1        | 2020-07-31 | 1           | 30   |
| 2        | 2020-07-30 | 2           | 40   |
| 3        | 2020-07-31 | 3           | 70   |
| 4        | 2020-07-29 | 4           | 100  |
| 5        | 2020-06-10 | 1           | 1010 |
| 6        | 2020-08-01 | 2           | 102  |
| 7        | 2020-08-01 | 3           | 111  |
| 8        | 2020-08-03 | 1           | 99   |
| 9        | 2020-08-07 | 2           | 32   |
| 10       | 2020-07-15 | 1           | 2    |
+----------+------------+-------------+------+
输出:
+---------------+-------------+----------+------------+
| customer_name | customer_id | order_id | order_date |
+---------------+-------------+----------+------------+
| Annabelle     | 3           | 7        | 2020-08-01 |
| Annabelle     | 3           | 3        | 2020-07-31 |
| Jonathan      | 2           | 9        | 2020-08-07 |
| Jonathan      | 2           | 6        | 2020-08-01 |
| Jonathan      | 2           | 2        | 2020-07-30 |
| Marwan        | 4           | 4        | 2020-07-29 |
| Winston       | 1           | 8        | 2020-08-03 |
| Winston       | 1           | 1        | 2020-07-31 |
| Winston       | 1           | 10       | 2020-07-15 |
+---------------+-------------+----------+------------+
解释:
Winston 有 4 笔订单, 排除了 "2020-06-10" 的订单, 因为它是最老的订单。
Annabelle 只有 2 笔订单, 全部返回。
Jonathan 恰好有 3 笔订单。
Marwan 只有 1 笔订单。
结果表我们按照 customer_name 升序排列,customer_id 升序排列,order_date 降序排列。

 

进阶:

  • 你能写出最近 n 笔订单的通用解决方案吗?

解法

方法一:等值连接 + 窗口函数

我们可以使用等值连接,将 Customers 表和 Orders 表按照 customer_id 进行连接,然后使用 row_number() 窗口函数来为每个消费者的订单按照 order_date 降序排列,并为每个消费者的订单添加一个序号,最后筛选出序号小于等于 $3$ 的订单即可。

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            *,
            ROW_NUMBER() OVER (
                PARTITION BY customer_id
                ORDER BY order_date DESC
            ) AS rk
        FROM
            Orders
            JOIN Customers USING (customer_id)
    )
SELECT name AS customer_name, customer_id, order_id, order_date
FROM T
WHERE rk <= 3
ORDER BY 1, 2, 4 DESC;