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发布于 2024-06-17 01:03:18 字数 4984 浏览 0 评论 0 收藏 0

1549. The Most Recent Orders for Each Product

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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 the customers.

 

Table: Orders

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| order_id    | int   |
| order_date  | date  |
| customer_id   | int   |
| product_id  | int   |
+---------------+---------+
order_id is the column with unique values for this table.
This table contains information about the orders made by customer_id.
There will be no product ordered by the same user more than once in one day.

 

Table: Products

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| product_id  | int   |
| product_name  | varchar |
| price     | int   |
+---------------+---------+
product_id is the column with unique values for this table.
This table contains information about the Products.

 

Write a solution to find the most recent order(s) of each product.

Return the result table ordered 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 order_id in ascending 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 | 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 table:
+------------+--------------+-------+
| product_id | product_name | price |
+------------+--------------+-------+
| 1      | keyboard   | 120   |
| 2      | mouse    | 80  |
| 3      | screen     | 600   |
| 4      | hard disk  | 450   |
+------------+--------------+-------+
Output: 
+--------------+------------+----------+------------+
| 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 |
+--------------+------------+----------+------------+
Explanation: 
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 do not include it in the result table.

Solutions

Solution 1: Equi-Join + Window Function

We can use an equi-join to join the Orders table and the Products table based on product_id, and then use the window function rank(), which assigns a rank to each product_id in the Orders table based on its order_date in descending order. Finally, we can select the rows with a rank of $1$ for each product_id.

# Write your MySQL query statement below
WITH
  T AS (
    SELECT
      *,
      RANK() OVER (
        PARTITION BY product_id
        ORDER BY order_date DESC
      ) AS rk
    FROM
      Orders
      JOIN Products USING (product_id)
  )
SELECT product_name, product_id, order_id, order_date
FROM T
WHERE rk = 1
ORDER BY 1, 2, 3;

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