文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

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?

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/the-most-recent-three-orders
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2. 解題

• 窗口函數(shù)，計(jì)算出排名，選出排名小于等于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) rnkfrom 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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