今日真题
题目介绍： The Most Recent Three Orders the-most-recent-three-orders
难度中等
SQL架构
Table:
代码语言：javascript复制Customers --------------- ---------
| Column Name | Type |
--------------- ---------
| customer_id | int |
| name | varchar |
--------------- ---------
customer_id is the primary key for this table.
This table contains information about customers.Table:
代码语言：javascript复制Orders --------------- ---------
| Column Name | Type |
--------------- ---------
| order_id | int |
| order_date | date |
| customer_id | int |
| cost | int |
--------------- ---------
order_id is the primary key for this table.
This table contains information about the orders made by customer_id.
Each customer has one order per day.Write an SQL query to find the most recent 3 orders of each user. If a user ordered less than 3 orders return all of their orders.
Return the result table sorted by
代码语言：javascript复制customer_namein ascending order and in case of a tie by the
代码语言：javascript复制customer_idin ascending order. If there still a tie, order them by the
代码语言：javascript复制order_datein descending order.
The query result format is in the following example:
``` 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 | ---------- ------------ ------------- ------
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 has 4 orders, we discard the order of "2020-06-10" because it is the oldest order. Annabelle has only 2 orders, we return them. Jonathan has exactly 3 orders. Marwan ordered only one time. We sort the result table by customer_name in ascending order, by customer_id in ascending order and by order_date in descending order in case of a tie. ```
Follow-up: Can you write a general solution for the most recent
代码语言：javascript复制norders?
代码语言：javascript复制sql
select name customer_name ,customer_id,order_id,order_date
from (
select name ,o.customer_id,order_id,order_date ,rank()over(partition by o.customer_id order by order_date desc) rk
from Orders o left join Customers c
on o.customer_id=c.customer_id
)t1
where rk <=3
order by customer_name ,customer_id,order_date desc- 已经有灵感了？在评论区写下你的思路吧！
