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Formatted question description: https://leetcode.ca/all/2228.html

# 2228. Users With Two Purchases Within Seven Days

## Description

Table: Purchases

+---------------+------+
| Column Name   | Type |
+---------------+------+
| purchase_id   | int  |
| user_id       | int  |
| purchase_date | date |
+---------------+------+
purchase_id is the primary key for this table.
This table contains logs of the dates that users purchased from a certain retailer.

Write an SQL query to report the IDs of the users that made any two purchases at most 7 days apart.

Return the result table ordered by user_id.

The query result format is in the following example.

Example 1:

Input:
Purchases table:
+-------------+---------+---------------+
| purchase_id | user_id | purchase_date |
+-------------+---------+---------------+
| 4           | 2       | 2022-03-13    |
| 1           | 5       | 2022-02-11    |
| 3           | 7       | 2022-06-19    |
| 6           | 2       | 2022-03-20    |
| 5           | 7       | 2022-06-19    |
| 2           | 2       | 2022-06-08    |
+-------------+---------+---------------+
Output:
+---------+
| user_id |
+---------+
| 2       |
| 7       |
+---------+
Explanation:
User 2 had two purchases on 2022-03-13 and 2022-03-20. Since the second purchase is within 7 days of the first purchase, we add their ID.
User 5 had only 1 purchase.
User 7 had two purchases on the same day so we add their ID.

## Solutions

• # Write your MySQL query statement below
WITH
t AS (
SELECT
user_id,
datediff(
purchase_date,
lag(purchase_date, 1) OVER (
PARTITION BY user_id
ORDER BY purchase_date
)
) AS d
FROM Purchases
)
SELECT DISTINCT user_id
FROM t
WHERE d <= 7
ORDER BY user_id;