Question

2688. Find Active Users

中文文档

Description

Table: Users

+-------------+----------+ 
| Column Name | Type   | 
+-------------+----------+ 
| user_id   | int    | 
| item    | varchar  |
| created_at  | datetime |
| amount    | int    |
+-------------+----------+
This table may contain duplicate records. 
Each row includes the user ID, the purchased item, the date of purchase, and the purchase amount.

Write a solution to identify active users. An active user is a user that has made a second purchase within 7 days of any other of their purchases.

For example, if the ending date is May 31, 2023. So any date between May 31, 2023, and June 7, 2023 (inclusive) would be considered "within 7 days" of May 31, 2023.

Return a list of user_id which denotes the list of active users in any order.

The result format is in the following example.

 

Example 1:

Input:
Users table:
+---------+-------------------+------------+--------+ 
| user_id | item        | created_at | amount |  
+---------+-------------------+------------+--------+
| 5     | Smart Crock Pot   | 2021-09-18 | 698882 |
| 6     | Smart Lock    | 2021-09-14 | 11487  |
| 6     | Smart Thermostat  | 2021-09-10 | 674762 |
| 8     | Smart Light Strip | 2021-09-29 | 630773 |
| 4     | Smart Cat Feeder  | 2021-09-02 | 693545 |
| 4     | Smart Bed     | 2021-09-13 | 170249 |
+---------+-------------------+------------+--------+ 
Output:
+---------+
| user_id | 
+---------+
| 6     | 
+---------+
Explanation: 
- User with user_id 5 has only one transaction, so he is not an active user.
- User with user_id 6 has two transaction his first transaction was on 2021-09-10 and second transation was on 2021-09-14. The distance between the first and second transactions date is <= 7 days. So he is an active user. 
- User with user_id 8 has only one transaction, so he is not an active user.  
- User with user_id 4 has two transaction his first transaction was on 2021-09-02 and second transation was on 2021-09-13. The distance between the first and second transactions date is > 7 days. So he is not an active user. 

Solutions

Solution 1

# Write your MySQL query statement
SELECT DISTINCT
  user_id
FROM Users
WHERE
  user_id IN (
    SELECT
      user_id
    FROM
      (
        SELECT
          user_id,
          created_at,
          LAG(created_at, 1) OVER (
            PARTITION BY user_id
            ORDER BY created_at
          ) AS prev_created_at
        FROM Users
      ) AS t
    WHERE DATEDIFF(created_at, prev_created_at) <= 7
  );