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 the customers.

Table: Orders

+---------------+---------+ | Column Name | Type | +---------------+---------+ | order_id | int | | order_date | date | | customer_id | int | | product_id | int | +---------------+---------+ order_id is the primary key 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 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:

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.

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/the-most-recent-orders-for-each-product
著作權歸領扣網絡所有。商業轉載請聯系官方授權，非商業轉載請注明出處。

2. 解題

• 先找產品的最大日期，再 join+where + in 篩選

# Write your MySQL query statement belowselect product_name, o.product_id, order_id, order_date from Orders o left join Products p using(product_id) where (product_id, order_date) in (select product_id, max(order_date) order_datefrom Ordersgroup by product_id ) order by product_name, product_id, order_id

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