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408数据库SQL练习题02

mafa1993 2022-01-07 阅读 66

586. 订单最多的客户

难度简单

SQL架构

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 。

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?

题解:

select customer_number from orders group by customer_number order by count(customer_number) desc limit 1

511. 游戏玩法分析 I

难度简单

SQL架构

活动表 Activity

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
表的主键是 (player_id, event_date)。
这张表展示了一些游戏玩家在游戏平台上的行为活动。
每行数据记录了一名玩家在退出平台之前,当天使用同一台设备登录平台后打开的游戏的数目(可能是 0 个)。

写一条 SQL 查询语句获取每位玩家 第一次登陆平台的日期

查询结果的格式如下所示:

Activity 表:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-05-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result 表:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1         | 2016-03-01  |
| 2         | 2017-06-25  |
| 3         | 2016-03-02  |
+-----------+-------------+

题解:

select player_id,min(event_date) as first_login from Activity group by player_id ;

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