1. 题目

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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/customer-order-frequency
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• 返回6月和7月都花了至少100美元的人

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

762 ms

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