Formatted question description: https://leetcode.ca/all/1511.html

1511. Customer Order Frequency

Level

Easy

Description

Table: Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| country       | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information of the customers in the company.

Table: Product

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| description   | varchar |
| price         | int     |
+---------------+---------+
product_id is the primary key for this table.
This table contains information of the products in the company.
price is the product cost.

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| product_id    | int     |
| order_date    | date    |
| quantity      | int     |
+---------------+---------+
order_id is the primary key for this table.
This table contains information on customer orders.
customer_id is the id of the customer who bought "quantity" products with id "product_id".
Order_date is the date in format ('YYYY-MM-DD') when the order was shipped.

Write an SQL query to report the customer_id and customer_name of customers who have spent at least $100 in each month of June and July 2020.

Return the result table in any order.

The query result format is in the following example.

Customers
+--------------+-----------+-------------+
| customer_id  | name      | country     |
+--------------+-----------+-------------+
| 1            | Winston   | USA         |
| 2            | Jonathan  | Peru        |
| 3            | Moustafa  | Egypt       |
+--------------+-----------+-------------+

Product
+--------------+-------------+-------------+
| product_id   | description | price       |
+--------------+-------------+-------------+
| 10           | LC Phone    | 300         |
| 20           | LC T-Shirt  | 10          |
| 30           | LC Book     | 45          |
| 40           | LC Keychain | 2           |
+--------------+-------------+-------------+

Orders
+--------------+-------------+-------------+-------------+-----------+
| order_id     | customer_id | product_id  | order_date  | quantity  |
+--------------+-------------+-------------+-------------+-----------+
| 1            | 1           | 10          | 2020-06-10  | 1         |
| 2            | 1           | 20          | 2020-07-01  | 1         |
| 3            | 1           | 30          | 2020-07-08  | 2         |
| 4            | 2           | 10          | 2020-06-15  | 2         |
| 5            | 2           | 40          | 2020-07-01  | 10        |
| 6            | 3           | 20          | 2020-06-24  | 2         |
| 7            | 3           | 30          | 2020-06-25  | 2         |
| 9            | 3           | 30          | 2020-05-08  | 3         |
+--------------+-------------+-------------+-------------+-----------+

Result table:
+--------------+------------+
| customer_id  | name       |  
+--------------+------------+
| 1            | Winston    |
+--------------+------------+ 
Winston spent $300 (300 * 1) in June and $100 ( 10 * 1 + 45 * 2) in July 2020.
Jonathan spent $600 (300 * 2) in June and $20 ( 2 * 10) in July 2020.
Moustafa spent $110 (10 * 2 + 45 * 2) in June and $0 in July 2020.

Solution

Join the tables Customers, Orders and Product using the common fields. Use case statements and left statements to calculate the total spent amount of each customer in each month of June and July 2020.

# Write your MySQL query statement below
select c.customer_id, c.name
    from Customers c join Orders o on c.customer_id = o.customer_id
    join Product p on o.product_id = p.product_id
    group by c.customer_id, c.name
    having
    sum(case when left(o.order_date, 7) = '2020-06' then p.price * o.quantity else 0 end) >= 100
    and
    sum(case when left(o.order_date, 7) = '2020-07' then p.price * o.quantity else 0 end) >= 100;

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