【LeetCode-SQL】1532. Last three orders

One 、 subject

surface ：Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id  Is the primary key of the table 
 This table contains consumer information 

surface ：Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| cost          | int     |
+---------------+---------+
order_id  Is the primary key of the table 
 The table contains id by customer_id Customer's order information 
 Every consumer   One order per day 

Write a SQL sentence , Find the last three orders of each user . If the user's order is less than 3 pen , Then return all his orders .

The returned results are as follows customer_name Ascending array . If the rankings are the same , Continue to follow customer_id Ascending array . If the ranking is the same , Continue to follow order_date Descending array .

The query result format is shown 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  Yes  4  Orders ,  Exclude  "2020-06-10"  The order of ,  Because it is the oldest order .
Annabelle  Only  2  Orders ,  All return .
Jonathan  There happens to be  3  Orders .
Marwan  Only  1  Orders .
 Result table we follow  customer_name  Ascending order ,customer_id  Ascending order ,order_date  Descending order .

Advanced ：

• You can write about the recent n A common solution for orders ?

Two 、 solve

1、rank() over()

Ideas ：
Code ：

select 	t2.name as customer_name 
        , t1.customer_id as customer_id
        , t1.order_id as order_id
        , t1.order_date as order_date
from (
	select
		customer_id
        , order_id
        , order_date
        , rank() over(partition by customer_id order by order_date desc) as rn
	from orders 
) t1
join Customers t2 on t1.customer_id  = t2.customer_id 
where rn <= 3
order by t2.name asc, t1.customer_id asc, t1.order_date desc;

3、 ... and 、 Reference resources

1、MySQL Window function rank() over() Grouping sorting