LeetCode：Database 110.最近的三笔订单

最新推荐文章于 2025-05-19 15:57:35 发布

Xiao Miao

最新推荐文章于 2025-05-19 15:57:35 发布

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分类专栏： LeetCode 文章标签： mysql

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本文介绍如何使用SQL语句从Customers和Orders表中查找每个用户最近的三笔订单，若订单不足三笔则返回全部，并按customer_name升序、customer_id升序和order_date降序排列。

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要求：写一个 SQL 语句，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。如果排名有相同，则继续按照 customer_id 升序排列，如果排名还有相同，则继续按照 order_date 降序排列。

表：Customers的结构

+---------------+---------+
| 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的结构

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| cost          | int     |
+---------------+---------+
order_id 是该表主键
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

Customers 表：

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

Orders 表：

+----------+------------+-------------+------+
| 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    |
+----------+------------+-------------+------+

Result Table：

+---------------+-------------+----------+------------+
| 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 |
+---------------+-------------+----------+------------+
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。

SQL语句：

with c as(select b.name as n1,a.customer_id as ci1,a.order_id as oi1,a.order_date as d1,row_number() over(partition by a.customer_id order by order_date desc) as r1
from orders a 
join customers b 
on a.customer_id=b.customer_id) 

select n1 as customer_name,ci1 as customer_id,oi1 as order_id,d1 as order_date
from c 
where r1<=3
order by n1 asc,ci1 asc,d1 desc;