1398. 购买了产品 A 和产品 B 却没有购买产品 C 的顾客

SQL架构

 Customers 表：

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| customer_id         | int     |
| customer_name       | varchar |
+---------------------+---------+
customer_id 是这张表的主键。
customer_name 是顾客的名称。

Orders 表：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| product_name  | varchar |
+---------------+---------+
order_id 是这张表的主键。
customer_id 是购买了名为 "product_name" 产品顾客的id。

请你设计 SQL 查询来报告购买了产品 A 和产品 B 却没有购买产品 C 的顾客的 ID 和姓名（ customer_id 和 customer_name ），我们将基于此结果为他们推荐产品 C 。
您返回的查询结果需要按照 customer_id 排序。

查询结果如下例所示。

Customers table:
+-------------+---------------+
| customer_id | customer_name |
+-------------+---------------+
| 1           | Daniel        |
| 2           | Diana         |
| 3           | Elizabeth     |
| 4           | Jhon          |
+-------------+---------------+

Orders table:
+------------+--------------+---------------+
| order_id   | customer_id  | product_name  |
+------------+--------------+---------------+
| 10         |     1        |     A         |
| 20         |     1        |     B         |
| 30         |     1        |     D         |
| 40         |     1        |     C         |
| 50         |     2        |     A         |
| 60         |     3        |     A         |
| 70         |     3        |     B         |
| 80         |     3        |     D         |
| 90         |     4        |     C         |
+------------+--------------+---------------+

Result table:
+-------------+---------------+
| customer_id | customer_name |
+-------------+---------------+
| 3           | Elizabeth     |
+-------------+---------------+
只有 customer_id 为 3 的顾客购买了产品 A 和产品 B ，却没有购买产品 C 。

select
c.customer_id,c.customer_name
from
(
select
customer_id,sum(if(product_name = 'A',1,0)) sumA,sum(if(product_name = 'B',1,0)) sumB,sum(if(product_name = 'C',1,0)) sumC
from
Orders
group by
customer_id
having 
sumA >=1 and sumB >=1 and sumC =0
) s1 left join Customers c
on s1.customer_id = c.customer_id
order by
customer_id 

sum(直接写判断语句):

select
c.customer_id,c.customer_name
from
(
select
customer_id,sum(product_name = 'A') sumA,sum(product_name = 'B') sumB,sum(product_name = 'C') sumC
from
Orders
group by
customer_id
having 
sumA >=1 and sumB >=1 and sumC =0
) s1 left join Customers c
on s1.customer_id = c.customer_id
order by
customer_id