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1532. The Most Recent Three Orders
Medium
LeetCode
Database
Problem
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 by customer_id.
Each customer has one order per day.
Write an SQL query 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 query 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.
Code
SELECT a.name as customer_name,a.customer_id,b.order_id,b.order_date
FROM Customers as a
JOIN Orders as b
ON a.customer_id=b.customer_id
JOIN Orders as c
ON b.customer_id=c.customer_id AND b.order_date<=c.order_date
GROUP BY a.customer_id,b.order_id
HAVING COUNT(c.order_date)<=3
ORDER BY customer_name,customer_id,order_date DESC;
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