网站设计及内容策划公司员工培训方案
1.背景
统计连续登录天数超过3天的用户，输出信息包括：用户id，登录天数，起始时间，结束时间；
2.准备数据
-- 建表
create table if not exists user_login_3days(user_id STRING,login_date date
);--插入数据
insert into user_login_3days values ('01','2023-08-02');
insert into user_login_3days values ('01','2023-08-03');
insert into user_login_3days values ('01','2023-08-04');
insert into user_login_3days values ('01','2023-11-02');
insert into user_login_3days values ('01','2023-12-09');
insert into user_login_3days values ('02','2023-01-01');
insert into user_login_3days values ('02','2023-04-23');
insert into user_login_3days values ('03','2023-09-10');
insert into user_login_3days values ('03','2023-09-11');
insert into user_login_3days values ('03','2023-09-12');
insert into user_login_3days values ('04','2023-04-23');
insert into user_login_3days values ('04','2023-04-24');
insert into user_login_3days values ('05','2023-09-11');
insert into user_login_3days values ('06','2023-09-12');-- 查询数据数据
select * from user_login_3days order by user_id;
3.解决思路以及实现
思路1:row_number()
- 1.通过对用户id进行开窗函数row_number,对登陆时间进行降序排列
- 2.使用date_sub(login_date,rn)函数进行日期求出差值日期
- 3.对user_id和diff_date分组求出时间的区间范围
- 4.对结果进行过滤操作
SELECTt2.user_id,count(1) as login_times,min(t2.login_date) as start_date,max(t2.login_date) as end_date
FROM
(SELECTt1.user_id,t1.login_date,date_sub(t1.login_date,rn) as diff_dateFROM(SELECTuser_id,login_date,row_number() over(partition by user_id order by login_date asc) as rnFROM user_login_3days) t1
) t2
group by t2.user_id, t2.diff_date
having login_times >= 3;
思路2:lag()/lead()
- 1.通过对用户id进行开窗函数lag/lead,求出前面第二个的日期与当前的日期差以及后面一个日期与当前日期的差值
- 2.对结果进行过滤操作
SELECTuser_id,lag_login_date,login_date
FROM(SELECTuser_id,login_date,lag(login_date,2,login_date) over(partition by user_id order by login_date) as lag_login_date,lead(login_date,1,login_date) over(partition by user_id order by login_date) as lead_login_dateFROM user_login_3days) t1
where datediff(login_date,lag_login_date) =2
4.总结
连续登陆问题解决的关键在于:如何判断连续?
通过对user_id分组排序后,使用登陆日期减去序号rn。如果连续，则得到的这个日期会相同。