LeetCode MySQL 1532. The Most Recent Three Orders（dense_rank + over窗口函数）

1. 题目

Table: 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: 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 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 customer_name in ascending order and in case of a tie by the customer_id in ascending order. If there still a tie, order them by the order_date in 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 n orders?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/the-most-recent-three-orders
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 窗口函数，计算出排名，选出排名小于等于3的

# Write your MySQL query statement below
select name customer_name, customer_id, order_id, order_date
from
(
    select customer_id, order_id, order_date, 
            dense_rank() over(partition by customer_id order by order_date desc) rnk
    from Orders
) t
left join Customers
using(customer_id)
where rnk <= 3
order by name, customer_id, order_date desc

694 ms

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