The Most Recent Orders For Each Product Problem

Description

LeetCode Problem 1549.

Table: Customers

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information about the customers.
``````

Table: Orders

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| product_id    | int     |
+---------------+---------+
order_id is the primary key for this table.
There will be no product ordered by the same user more than once in one day.
``````

Table: Products

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
| price         | int     |
+---------------+---------+
product_id is the primary key for this table.
This table contains information about the Products.
``````

Write an SQL query to find the most recent order(s) of each product.

Return the result table sorted by product_name in ascending order and in case of a tie by the product_id in ascending order. If there still a tie, order them by the order_id in ascending order.

The query result format is in the following example:

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Customers
+-------------+-----------+
| customer_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+

Orders
+----------+------------+-------------+------------+
| order_id | order_date | customer_id | product_id |
+----------+------------+-------------+------------+
| 1        | 2020-07-31 | 1           | 1          |
| 2        | 2020-07-30 | 2           | 2          |
| 3        | 2020-08-29 | 3           | 3          |
| 4        | 2020-07-29 | 4           | 1          |
| 5        | 2020-06-10 | 1           | 2          |
| 6        | 2020-08-01 | 2           | 1          |
| 7        | 2020-08-01 | 3           | 1          |
| 8        | 2020-08-03 | 1           | 2          |
| 9        | 2020-08-07 | 2           | 3          |
| 10       | 2020-07-15 | 1           | 2          |
+----------+------------+-------------+------------+

Products
+------------+--------------+-------+
| product_id | product_name | price |
+------------+--------------+-------+
| 1          | keyboard     | 120   |
| 2          | mouse        | 80    |
| 3          | screen       | 600   |
| 4          | hard disk    | 450   |
+------------+--------------+-------+

Result table:
+--------------+------------+----------+------------+
| product_name | product_id | order_id | order_date |
+--------------+------------+----------+------------+
| keyboard     | 1          | 6        | 2020-08-01 |
| keyboard     | 1          | 7        | 2020-08-01 |
| mouse        | 2          | 8        | 2020-08-03 |
| screen       | 3          | 3        | 2020-08-29 |
+--------------+------------+----------+------------+
keyboard's most recent order is in 2020-08-01, it was ordered two times this day.
mouse's most recent order is in 2020-08-03, it was ordered only once this day.
screen's most recent order is in 2020-08-29, it was ordered only once this day.
The hard disk was never ordered and we don't include it in the result table.
``````

MySQL Solution

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select p.product_name, o.product_id, o.order_id, o.order_date
from Orders o
left join Products p
on o.product_id = p.product_id
join (select product_id, max(order_date) as 'recent_date'
from Orders
group by product_id) m
on m.product_id = o.product_id and
m.recent_date = o.order_date
order by p.product_name, o.product_id, o.order_id
``````