【sql学习】LeetCode之183. Customers Who Never Order

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       |
+-----------+

自己的代码：

Runtime: 276 ms, faster than 27.78% of MySQL online submissions for Customers Who Never Order.

SELECT
	Name Customers
FROM
	Customers
WHERE
	id NOT IN (
		SELECT
			t1.Id
		FROM
			Customers t1,
			Orders t2
		WHERE
			t1.Id = t2.CustomerId
	)

小结：

1）not in 的用法

2）这种嵌套查询的方式

别人的代码：

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Runtime: 264 ms, faster than 48.44% of MySQL online submissions for Customers Who Never Order.

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

小结：

1）左连接的用法与特点：右表无法匹配左表的那些记录行，会被置为NULL

Runtime: 249 ms, faster than 74.55% of MySQL online submissions for Customers Who Never Order.

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

小结：

使用这种方式，根本没有必要进行表连接，就可获得答案；