LeetCode刷题-数据库（MySQL）- 586.订单最多的客户

586.订单最多的客户

一、题目描述

在表 order 中找到订单数最多客户对应的 customer_number 。
数据能够保证订单数最多的客户恰好只有一位。

表 orders 定义如下：

Column	Type
order_number (PK)	int
customer_number	int
order_date	date
required_date	date
shipped_date	date

样例输入

order_number	customer_number	order_date	required_date	shipped_date	status	comment
1	1	2017-04-09	2017-04-13	2017-04-12	Closed
2	2	2017-04-15	2017-04-20	2017-04-18	Closed
3	3	2017-04-16	2017-04-25	2017-04-20	Closed
4	3	2017-04-18	2017-04-28	2017-04-25	Closed

样例输出

customer_number
3

解释
customer_number 为 ‘3’ 的顾客有两个订单，比顾客 ‘1’ 或者 ‘2’ 都要多，因为他们只有一个订单
所以结果是该顾客的 customer_number ，也就是 3 。

**进阶：**如果有多位顾客订单数并列最多，你能找到他们所有的 customer_number 吗？

二、思路分析

因为题目数据能够保证订单数最多的客户恰好只有一位，所以GROUP BY后以COUNT(*)降序排列取第一行数据即可。

进阶： 先算出最高的订单数是多少，然后只需要在之前的基础上找出与最高订单数相等的行即所求。

三、代码实现

SELECT
	customer_number
FROM
	orders
GROUP BY customer_number
ORDER BY COUNT(*) DESC
LIMIT 1

进阶：

SELECT
	customer_number
FROM
	orders o1
GROUP BY o1.customer_number
HAVING COUNT(*) = (
			SELECT
				COUNT(*)
			FROM
				orders o2
			GROUP BY o2. customer_number
			ORDER BY COUNT(*) DESC
			LIMIT 1
	);