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SQL，力扣题目1549，每件商品的最新订单【窗口函数】

article 2024/11/5 12:15:03

一、力扣链接

LeetCode_1549

二、题目描述

表: Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id 是该表主键.
该表包含消费者的信息.

表: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| product_id    | int     |
+---------------+---------+
order_id 是该表主键.
该表包含消费者customer_id产生的订单.
不会有商品被相同的用户在一天内下单超过一次.

表: Products

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
| price         | int     |
+---------------+---------+
product_id 是该表主键.
该表包含所有商品的信息.

写一个解决方案, 找到每件商品的最新订单(可能有多个).

返回的结果以 product_name 升序排列, 如果有排序相同, 再以 product_id 升序排列. 如果还有排序相同, 再以 order_id 升序排列.

三、目标拆解

四、建表语句

Create table If Not Exists Customers (customer_id int, name varchar(10))
Create table If Not Exists Orders (order_id int, order_date date, customer_id int, product_id int)
Create table If Not Exists Products (product_id int, product_name varchar(20), price int)
Truncate table Customers
insert into Customers (customer_id, name) values ('1', 'Winston')
insert into Customers (customer_id, name) values ('2', 'Jonathan')
insert into Customers (customer_id, name) values ('3', 'Annabelle')
insert into Customers (customer_id, name) values ('4', 'Marwan')
insert into Customers (customer_id, name) values ('5', 'Khaled')
Truncate table Orders
insert into Orders (order_id, order_date, customer_id, product_id) values ('1', '2020-07-31', '1', '1')
insert into Orders (order_id, order_date, customer_id, product_id) values ('2', '2020-7-30', '2', '2')
insert into Orders (order_id, order_date, customer_id, product_id) values ('3', '2020-08-29', '3', '3')
insert into Orders (order_id, order_date, customer_id, product_id) values ('4', '2020-07-29', '4', '1')
insert into Orders (order_id, order_date, customer_id, product_id) values ('5', '2020-06-10', '1', '2')
insert into Orders (order_id, order_date, customer_id, product_id) values ('6', '2020-08-01', '2', '1')
insert into Orders (order_id, order_date, customer_id, product_id) values ('7', '2020-08-01', '3', '1')
insert into Orders (order_id, order_date, customer_id, product_id) values ('8', '2020-08-03', '1', '2')
insert into Orders (order_id, order_date, customer_id, product_id) values ('9', '2020-08-07', '2', '3')
insert into Orders (order_id, order_date, customer_id, product_id) values ('10', '2020-07-15', '1', '2')
Truncate table Products
insert into Products (product_id, product_name, price) values ('1', 'keyboard', '120')
insert into Products (product_id, product_name, price) values ('2', 'mouse', '80')
insert into Products (product_id, product_name, price) values ('3', 'screen', '600')
insert into Products (product_id, product_name, price) values ('4', 'hard disk', '450')

五、过程分析

1、最新订单为日期最晚的订单，需要降序

2、选择排序为1的即为最新

六、代码实现

with t1 as(
select order_id, order_date, product_id,
       rank() over(partition by product_id order by order_date desc) rn
from orders
)
select (select product_name from products p where p.product_id = t1.product_id) product_name, 
       product_id, order_id, order_date
from t1 where rn = 1
order by product_name, product_id, order_id;