MySQL练习六

Mandy的名字被占用了

发布于 2021-09-02 11:14:03

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发布于 2021-09-02 11:14:03

题目表述

在表 orders 中找到订单数最多客户对应的 customer_number 。数据保证订单数最多的顾客恰好只有一位。表 orders 定义如下：

| Column            | Type      |
|-------------------|-----------|
| order_number (PK) | int       |
| customer_number   | int       |
| order_date        | date      |
| required_date     | date      |
| shipped_date      | date      |
| status            | char(15)  |
| comment           | char(200) |

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 。

解题思路

此题考查两个知识点，一个是group，一个是limit。使用group针对customer_number做分组统计出订单数最多的用户。limit直接返回第一个，即为数量最多的用户。

解题答案

// limit可以写为limit 1;
select customer_number from orders  group by customer_number order by count(customer_number) desc limit 0,1;

进阶:如果存在多个用户订单数一样多，如何返回对应所有的customer_number.

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
	);

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原始发表：2021-08-04，如有侵权请联系 cloudcommunity@tencent.com 删除

limit