SQL面试题练习 —— 合并用户浏览行为
目录
- 1 题目
- 2 建表语句
- 3 题解
1 题目
有一份用户访问记录表，记录用户id和访问时间，如果用户访问时间间隔小于60s则认为时一次浏览，请合并用户的浏览行为。
样例数据
+----------+--------------+
| user_id | access_time |
+----------+--------------+
| 1 | 1736337600 |
| 1 | 1736337660 |
| 2 | 1736337670 |
| 1 | 1736337710 |
| 3 | 1736337715 |
| 2 | 1736337750 |
| 1 | 1736337760 |
| 3 | 1736337820 |
| 2 | 1736337850 |
| 1 | 1736337910 |
+----------+--------------+
2 建表语句
--建表语句
CREATE TABLE user_access_log (user_id INT,access_time BIGINT
) ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t';
--插入数据
insert into user_access_log (user_id,access_time)
values
(1,1736337600),
(1,1736337660),
(2,1736337670),
(1,1736337710),
(3,1736337715),
(2,1736337750),
(1,1736337760),
(3,1736337820),
(2,1736337850),
(1,1736337910);
3 题解
（1）分用户计算出每次点击时间差；
select user_id,access_time,last_access_time,access_time - last_access_time as time_diff
from (select user_id,access_time,lag(access_time) over (partition by user_id order by access_time) as last_access_timefrom user_access_log) t
执行结果
+----------+--------------+-------------------+------------+
| user_id | access_time | last_access_time | time_diff |
+----------+--------------+-------------------+------------+
| 1 | 1736337600 | NULL | NULL |
| 1 | 1736337660 | 1736337600 | 60 |
| 1 | 1736337710 | 1736337660 | 50 |
| 1 | 1736337760 | 1736337710 | 50 |
| 1 | 1736337910 | 1736337760 | 150 |
| 2 | 1736337670 | NULL | NULL |
| 2 | 1736337750 | 1736337670 | 80 |
| 2 | 1736337850 | 1736337750 | 100 |
| 3 | 1736337715 | NULL | NULL |
| 3 | 1736337820 | 1736337715 | 105 |
+----------+--------------+-------------------+------------+
（2）确认是否是新的访问
select user_id,access_time,last_access_time,if(access_time - last_access_time >= 60, 1, 0) as is_new_group
from (select user_id,access_time,lag(access_time) over (partition by user_id order by access_time) as last_access_timefrom user_access_log) t
执行结果
+----------+--------------+-------------------+---------------+
| user_id | access_time | last_access_time | is_new_group |
+----------+--------------+-------------------+---------------+
| 1 | 1736337600 | NULL | 0 |
| 1 | 1736337660 | 1736337600 | 1 |
| 1 | 1736337710 | 1736337660 | 0 |
| 1 | 1736337760 | 1736337710 | 0 |
| 1 | 1736337910 | 1736337760 | 1 |
| 2 | 1736337670 | NULL | 0 |
| 2 | 1736337750 | 1736337670 | 1 |
| 2 | 1736337850 | 1736337750 | 1 |
| 3 | 1736337715 | NULL | 0 |
| 3 | 1736337820 | 1736337715 | 1 |
+----------+--------------+-------------------+---------------+
（3）得出结果
使用sum()over(partition by …… order by ……)累加计算，给出组ID。聚合函数开窗使用order by 计算结果是从分组开始计算到当前行的结果。
这里的技巧：需要新建组的时候就给标签赋值1，否则0，然后累加计算结果在新建组的时候值就会变化，根据聚合值分组，得到合并结果。
with t_group as(select user_id,access_time,last_access_time,if(access_time - last_access_time >= 60, 1, 0) as is_new_groupfrom (select user_id,access_time,lag(access_time) over (partition by user_id order by access_time) as last_access_timefrom user_access_log) t)
select user_id,access_time,last_access_time,is_new_group,sum(is_new_group) over (partition by user_id order by access_time asc) as group_id
from t_group
执行结果
+----------+--------------+-------------------+---------------+-----------+
| user_id | access_time | last_access_time | is_new_group | group_id |
+----------+--------------+-------------------+---------------+-----------+
| 1 | 1736337600 | NULL | 0 | 0 |
| 1 | 1736337660 | 1736337600 | 1 | 1 |
| 1 | 1736337710 | 1736337660 | 0 | 1 |
| 1 | 1736337760 | 1736337710 | 0 | 1 |
| 1 | 1736337910 | 1736337760 | 1 | 2 |
| 2 | 1736337670 | NULL | 0 | 0 |
| 2 | 1736337750 | 1736337670 | 1 | 1 |
| 2 | 1736337850 | 1736337750 | 1 | 2 |
| 3 | 1736337715 | NULL | 0 | 0 |
| 3 | 1736337820 | 1736337715 | 1 | 1 |
+----------+--------------+-------------------+---------------+-----------+