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数据库每日一题---第22天:最后一次登录
2022-07-07 21:49:00 【InfoQ】
一、问题描述
Logins
+----------------+----------+
| 列名 | 类型 |
+----------------+----------+
| user_id | int |
| time_stamp | datetime |
+----------------+----------+
(user_id, time_stamp) 是这个表的主键。
每一行包含的信息是user_id 这个用户的登录时间。
SQL
2020
2020
二、题目要求
样例
输入:
Logins 表:
+---------+---------------------+
| user_id | time_stamp |
+---------+---------------------+
| 6 | 2020-06-30 15:06:07 |
| 6 | 2021-04-21 14:06:06 |
| 6 | 2019-03-07 00:18:15 |
| 8 | 2020-02-01 05:10:53 |
| 8 | 2020-12-30 00:46:50 |
| 2 | 2020-01-16 02:49:50 |
| 2 | 2019-08-25 07:59:08 |
| 14 | 2019-07-14 09:00:00 |
| 14 | 2021-01-06 11:59:59 |
+---------+---------------------+
输出:
+---------+---------------------+
| user_id | last_stamp |
+---------+---------------------+
| 6 | 2020-06-30 15:06:07 |
| 8 | 2020-12-30 00:46:50 |
| 2 | 2020-01-16 02:49:50 |
+---------+---------------------+
解释:
6号用户登录了3次,但是在2020年仅有一次,所以结果集应包含此次登录。
8号用户在2020年登录了2次,一次在2月,一次在12月,所以,结果集应该包含12月的这次登录。
2号用户登录了2次,但是在2020年仅有一次,所以结果集应包含此次登录。
14号用户在2020年没有登录,所以结果集不应包含。
考察
1.聚合函数
2.建议用时10~25min
三、问题分析
2020
2020
max
四、编码实现
select user_id, max(time_stamp) as 'last_stamp'
from Logins
where year(time_stamp)=2020
group by user_id
五、测试结果
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