LeetCode:Database 110.最近的三笔订单
Posted Xiao Miao
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode:Database 110.最近的三笔订单相关的知识,希望对你有一定的参考价值。
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;
以上是关于LeetCode:Database 110.最近的三笔订单的主要内容,如果未能解决你的问题,请参考以下文章