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1127. User Purchase Platform
Description
Table: Spending
+-------------+---------+ | Column Name | Type | +-------------+---------+ | user_id | int | | spend_date | date | | platform | enum | | amount | int | +-------------+---------+ The table logs the history of the spending of users that make purchases from an online shopping website that has a desktop and a mobile application. (user_id, spend_date, platform) is the primary key (combination of columns with unique values) of this table. The platform column is an ENUM (category) type of ('desktop', 'mobile').
Write a solution to find the total number of users and the total amount spent using the mobile only, the desktop only, and both mobile and desktop together for each date.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Spending table: +---------+------------+----------+--------+ | user_id | spend_date | platform | amount | +---------+------------+----------+--------+ | 1 | 2019-07-01 | mobile | 100 | | 1 | 2019-07-01 | desktop | 100 | | 2 | 2019-07-01 | mobile | 100 | | 2 | 2019-07-02 | mobile | 100 | | 3 | 2019-07-01 | desktop | 100 | | 3 | 2019-07-02 | desktop | 100 | +---------+------------+----------+--------+ Output: +------------+----------+--------------+-------------+ | spend_date | platform | total_amount | total_users | +------------+----------+--------------+-------------+ | 2019-07-01 | desktop | 100 | 1 | | 2019-07-01 | mobile | 100 | 1 | | 2019-07-01 | both | 200 | 1 | | 2019-07-02 | desktop | 100 | 1 | | 2019-07-02 | mobile | 100 | 1 | | 2019-07-02 | both | 0 | 0 | +------------+----------+--------------+-------------+ Explanation: On 2019-07-01, user 1 purchased using both desktop and mobile, user 2 purchased using mobile only and user 3 purchased using desktop only. On 2019-07-02, user 2 purchased using mobile only, user 3 purchased using desktop only and no one purchased using both platforms.
Solutions
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# Write your MySQL query statement below WITH P AS ( SELECT DISTINCT spend_date, 'desktop' AS platform FROM Spending UNION SELECT DISTINCT spend_date, 'mobile' FROM Spending UNION SELECT DISTINCT spend_date, 'both' FROM Spending ), T AS ( SELECT user_id, spend_date, SUM(amount) AS amount, IF(COUNT(platform) = 1, platform, 'both') AS platform FROM Spending GROUP BY 1, 2 ) SELECT p.*, IFNULL(SUM(amount), 0) AS total_amount, COUNT(t.user_id) AS total_users FROM P AS p LEFT JOIN T AS t USING (spend_date, platform) GROUP BY 1, 2;