3118. Friday Purchase III 🔒

题目描述

Table: Purchases

+---------------+------+
| Column Name   | Type |
+---------------+------+
| user_id       | int  |
| purchase_date | date |
| amount_spend  | int  |
+---------------+------+
(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
Each row contains user_id, purchase_date, and amount_spend.

Table: Users

+-------------+------+
| Column Name | Type |
+-------------+------+
| user_id     | int  |
| membership  | enum |
+-------------+------+
user_id is the primary key for this table.
membership is an ENUM (category) type of ('Standard', 'Premium', 'VIP').
Each row of this table indicates the user_id, membership type.

Write a solution to calculate the total spending by Premium and VIP members on each Friday of every week in November 2023. If there are no purchases on a particular Friday by Premium or VIP members, it should be considered as 0.

Return the result table ordered by week of the month, and membership in ascending order.

The result format is in the following example.

Example:

Input:

Purchases table:

+---------+---------------+--------------+
| user_id | purchase_date | amount_spend |
+---------+---------------+--------------+
| 11      | 2023-11-03    | 1126         |
| 15      | 2023-11-10    | 7473         |
| 17      | 2023-11-17    | 2414         |
| 12      | 2023-11-24    | 9692         |
| 8       | 2023-11-24    | 5117         |
| 1       | 2023-11-24    | 5241         |
| 10      | 2023-11-22    | 8266         |
| 13      | 2023-11-21    | 12000        |
+---------+---------------+--------------+

Users table:

+---------+------------+
| user_id | membership |
+---------+------------+
| 11      | Premium    |
| 15      | VIP        |
| 17      | Standard   |
| 12      | VIP        |
| 8       | Premium    |
| 1       | VIP        |
| 10      | Standard   |
| 13      | Premium    |
+---------+------------+

Output:

+---------------+-------------+--------------+
| week_of_month | membership  | total_amount |
+---------------+-------------+--------------+
| 1             | Premium     | 1126         |
| 1             | VIP         | 0            |
| 2             | Premium     | 0            |
| 2             | VIP         | 7473         |
| 3             | Premium     | 0            |
| 3             | VIP         | 0            |
| 4             | Premium     | 5117         |
| 4             | VIP         | 14933        |
+---------------+-------------+--------------+

Explanation:

During the first week of November 2023, a transaction occurred on Friday, 2023-11-03, by a Premium member amounting to $1,126. No transactions were made by VIP members on this day, resulting in a value of 0.
For the second week of November 2023, there was a transaction on Friday, 2023-11-10, and it was made by a VIP member, amounting to $7,473. Since there were no purchases by Premium members that Friday, the output shows 0 for Premium members.
Similarly, during the third week of November 2023, no transactions by Premium or VIP members occurred on Friday, 2023-11-17, which shows 0 for both categories in this week.
In the fourth week of November 2023, transactions occurred on Friday, 2023-11-24, involving one Premium member purchase of $5,117 and VIP member purchases totaling $14,933 ($9,692 from one and $5,241 from another).

Note: The output table is ordered by week_of_month and membership in ascending order.

解法

方法一：递归 + 连接

我们首先创建一个递归表 T，其中包含 week_of_month 列，表示月份的第几周。然后创建一个表 M，包含 membership 列，表示会员类型，取值为 'Premium' 和 'VIP'。

接着创建一个表 P，包含 week_of_month、membership 和 amount_spend 列，筛选出每个会员在每个月的第几周的周五的消费金额。最后，我们将 T 和 M 表连接，再左连接 P 表，并且按照 week_of_month 和 membership 列进行分组，计算每周每种会员的总消费金额。

MySQL

# Write your MySQL query statement below
WITH RECURSIVE
    T AS (
        SELECT 1 AS week_of_month
        UNION
        SELECT week_of_month + 1
        FROM T
        WHERE week_of_month < 4
    ),
    M AS (
        SELECT 'Premium' AS membership
        UNION
        SELECT 'VIP'
    ),
    P AS (
        SELECT CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month, membership, amount_spend
        FROM
            Purchases
            JOIN Users USING (user_id)
        WHERE DAYOFWEEK(purchase_date) = 6
    )
SELECT week_of_month, membership, IFNULL(SUM(amount_spend), 0) AS total_amount
FROM
    T
    JOIN M
    LEFT JOIN P USING (week_of_month, membership)
GROUP BY 1, 2
ORDER BY 1, 2;

3118. Friday Purchase III 🔒

题目描述

解法

方法一：递归 + 连接

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