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中等
数据库

2292. 连续两年有 3 个及以上订单的产品 🔒

English Version

题目描述

表: Orders

+---------------+------+
| Column Name   | Type |
+---------------+------+
| order_id      | int  |
| product_id    | int  |
| quantity      | int  |
| purchase_date | date |
+---------------+------+
order_id 包含唯一值。
该表中的每一行都包含订单 ID、购买的产品 ID、数量和购买日期。

 

编写解决方案，获取连续两年订购三次或三次以上的所有产品的 id。

以 任意顺序 返回结果表。

结果格式示例如下。

 

示例 1:

输入: 
Orders 表:
+----------+------------+----------+---------------+
| order_id | product_id | quantity | purchase_date |
+----------+------------+----------+---------------+
| 1        | 1          | 7        | 2020-03-16    |
| 2        | 1          | 4        | 2020-12-02    |
| 3        | 1          | 7        | 2020-05-10    |
| 4        | 1          | 6        | 2021-12-23    |
| 5        | 1          | 5        | 2021-05-21    |
| 6        | 1          | 6        | 2021-10-11    |
| 7        | 2          | 6        | 2022-10-11    |
+----------+------------+----------+---------------+
输出: 
+------------+
| product_id |
+------------+
| 1          |
+------------+
解释: 
产品 1 在 2020 年和 2021 年都分别订购了三次。由于连续两年订购了三次，所以我们将其包含在答案中。
产品 2 在 2022 年订购了一次。我们不把它包括在答案中。

解法

方法一

MySQL

# Write your MySQL query statement below
WITH
    P AS (
        SELECT product_id, YEAR(purchase_date) AS y, COUNT(1) >= 3 AS mark
        FROM Orders
        GROUP BY 1, 2
    )
SELECT DISTINCT p1.product_id
FROM
    P AS p1
    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id
WHERE p1.mark AND p2.mark;

方法二

MySQL

# Write your MySQL query statement below
WITH
    P AS (
        SELECT product_id, YEAR(purchase_date) AS y
        FROM Orders
        GROUP BY 1, 2
        HAVING COUNT(1) >= 3
    )
SELECT DISTINCT p1.product_id
FROM
    P AS p1
    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id;