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1532. The Most Recent Three Orders
Description
Table: Customers
+---------------+---------+ | Column Name | Type | +---------------+---------+ | customer_id | int | | name | varchar | +---------------+---------+ customer_id is the column with unique values 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 column with unique values for this table. This table contains information about the orders made by customer_id. Each customer has one order per day.
Write a solution to find the most recent three orders of each user. If a user ordered less than three orders, return all of their orders.
Return the result table ordered by customer_name in ascending order and in case of a tie by the customer_id in ascending order. If there is still a tie, order them by order_date in descending order.
The result format is in the following example.
Example 1:
Input: Customers table: +-------------+-----------+ | customer_id | name | +-------------+-----------+ | 1 | Winston | | 2 | Jonathan | | 3 | Annabelle | | 4 | Marwan | | 5 | Khaled | +-------------+-----------+ Orders table: +----------+------------+-------------+------+ | 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 | +----------+------------+-------------+------+ Output: +---------------+-------------+----------+------------+ | 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 | +---------------+-------------+----------+------------+ Explanation: 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: Could you write a general solution for the most recent n orders?
Solutions
Solution 1: Equi-Join + Window Function
We can use an equi-join to join the Customers table and the Orders table based on customer_id, and then use the window function row_number() to sort the orders for each customer by order_date in descending order and assign a row number to each order. Finally, we can filter out the orders with a row number less than or equal to $3$.
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# Write your MySQL query statement below WITH T AS ( SELECT *, ROW_NUMBER() OVER ( PARTITION BY customer_id ORDER BY order_date DESC ) AS rk FROM Orders JOIN Customers USING (customer_id) ) SELECT name AS customer_name, customer_id, order_id, order_date FROM T WHERE rk <= 3 ORDER BY 1, 2, 4 DESC;