1398.(Medium)购买了产品 A 和产品 B 却没有购买产品 C 的顾客

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 。

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/customers-who-bought-products-a-and-b-but-not-c

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Solution

with t as 
(SELECT customer_id, 
    max('C'=product_name) c, 
    max('B'=product_name) b,
    max('A'=product_name) a
from  Orders GROUP BY customer_id)
SELECT cu.*
from Customers cu INNER JOIN t on 
    cu.customer_id=t.customer_id and 
    t.c=0 and t.a=1 and t.b=1;

Table Schema

Create table If Not Exists Customers (customer_id int, customer_name varchar(30));
Create table If Not Exists Orders (order_id int, customer_id int, product_name varchar(30));
Truncate table Customers;
insert into Customers (customer_id, customer_name) values ('1', 'Daniel');
insert into Customers (customer_id, customer_name) values ('2', 'Diana');
insert into Customers (customer_id, customer_name) values ('3', 'Elizabeth');
insert into Customers (customer_id, customer_name) values ('4', 'Jhon');
Truncate table Orders;
insert into Orders (order_id, customer_id, product_name) values ('10', '1', 'A');
insert into Orders (order_id, customer_id, product_name) values ('20', '1', 'B');
insert into Orders (order_id, customer_id, product_name) values ('30', '1', 'D');
insert into Orders (order_id, customer_id, product_name) values ('40', '1', 'C');
insert into Orders (order_id, customer_id, product_name) values ('50', '2', 'A');
insert into Orders (order_id, customer_id, product_name) values ('60', '3', 'A');
insert into Orders (order_id, customer_id, product_name) values ('70', '3', 'B');
insert into Orders (order_id, customer_id, product_name) values ('80', '3', 'D');
insert into Orders (order_id, customer_id, product_name) values ('90', '4', 'C');