题目8：MySQL----------Duplicate Emails

标签：database   mysql   

Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.

Table: Customers.

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

Table: Orders.

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

Using the above tables as example, return the following:

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

题目解答：

SELECT A.Name from Customers A
WHERE A.Id NOT IN (SELECT B.CustomerId from Orders B)

另外还可以写成下面两种方式：

SELECT A.Name from Customers A
WHERE NOT EXISTS (SELECT 1 FROM Orders B WHERE A.Id = B.CustomerId)

SELECT A.Name from Customers A
LEFT JOIN Orders B on  a.Id = B.CustomerId
WHERE b.CustomerId is NULL

题目8：MySQL----------Duplicate Emails

标签：database   mysql   

原文地址：http://blog.csdn.net/chenxun_2010/article/details/45939973