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 ;