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?微信公眾號：山青詠芝（shanqingyongzhi）
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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 | +-----------+

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某網(wǎng)站包含兩個表，Customers?表和?Orders?表。編寫一個 SQL 查詢，找出所有從不訂購任何東西的客戶。

Customers?表：

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

Orders?表：

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

例如給定上述表格，你的查詢應(yīng)返回：

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

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方法：使用子查詢和NOT IN子句

算法：如果我們有一個訂購過的客戶列表，很容易知道誰從未訂購過。

我們可以使用以下代碼來獲取此類列表。

select customerid from orders;

然后，我們可以NOT IN用來查詢不在此列表中的客戶。

1 select customers.name as 'Customers' 2 from customers 3 where customers.id not in 4 ( 5 select customerid from orders 6 );

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222ms

1 # Write your MySQL query statement below 2 SELECT Name as 'Customers' from Customers where Id not in (select CustomerId from Orders)

?

轉(zhuǎn)載于:https://www.cnblogs.com/strengthen/p/9720985.html

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