leetcode183 从不订购的客户 Customers Who Never Order

假设一个网站包含两个表，Customers 表和 Orders 表。编写一个SQL语句找出所有从不订购任何东西的客户。

创建表和数据：

Create table If Not Exists Customers (Idint, Name varchar(255));
Create table If Not Exists Orders (Id int,CustomerId int);
Truncate table Customers;
insert into Customers (Id, Name) values('1', 'Joe');
insert into Customers (Id, Name) values('2', 'Henry');
insert into Customers (Id, Name) values('3', 'Sam');
insert into Customers (Id, Name) values('4', 'Max');
Truncate table Orders;
insert into Orders (Id, CustomerId) values('1', '3');
insert into Orders (Id, CustomerId) values('2', '1');

解法：

1.顾客表的id和订单表的customerid关联，得出的是买了的东西的顾客。用left join，没买东西的顾客，其对应的订单为空。这是一种求集合差的方法。

select C.name as Customers
from Customers as C left join Orders as O on (C.id = O.customerid)
where O.id is NULL;

先用子查询将买过东西的顾客id选出来。在应用left join求集合差。

select C.name as `Customers`
from Customers as C left join (
    select distinct customerid
    from Orders
) as O on (C.id = O.customerid)
where O.customerid is NULL;

2.用not in也可以。先用子查询将买过东西的顾客id选出来。然后排除这些顾客的id即可。

select C.name as Customers
from Customers as C 
where C.id not in (
    select distinct customerid
    from Orders
)

集合差定义：C=A-B。C中的元素等于在A中但是不在B中。因此，对A中的每个元素a，如果元素a不在B中，则元素a就是集合C的元素。

EXISTS是布尔运算符，常用于测试子查询。

SELECT 
    select_list
FROM
    a_table
WHERE
    [NOT] EXISTS(subquery);

当subquery返回任何行时，EXISTS返回true，否则返回false。

select C.name as `Customers`
from Customers as C 
where not exists (
    select distinct customerid
    from Orders as O
    where O.customerid = C.id
) ;

来源：https://www.cnblogs.com/forever-fortunate/p/11723039.html

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