Hive-SQL查询连续活跃登录用户

2022/1/19 2:08:32

本文主要是介绍Hive-SQL查询连续活跃登录用户,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

连续活跃登陆的用户指至少连续2天都活跃登录的用户

解决类似场景的问题

创建数据

1 2 3 4 5 6 7 8 9 10 11 12 CREATE TABLE test5active( dt string, user_id string, age int) ROW format delimited fields terminated BY ',';   INSERT INTO TABLE test5active VALUES ('2019-02-11','user_1',23),('2019-02-11','user_2',19), ('2019-02-11','user_3',39),('2019-02-11','user_1',23), ('2019-02-11','user_3',39),('2019-02-11','user_1',23), ('2019-02-12','user_2',19),('2019-02-13','user_1',23), ('2019-02-15','user_2',19),('2019-02-16','user_2',19);
 

思路一:

1、因为每天用户登录次数可能不止一次,所以需要先将用户每天的登录日期去重。

2、再用row_number() over(partition by _ order by _)函数将用户id分组,按照登陆时间进行排序。

3、计算登录日期减去第二步骤得到的结果值,用户连续登陆情况下,每次相减的结果都相同。

4、按照id和日期分组并求和,筛选大于等于2的即为连续活跃登陆的用户。

第一步:用户登录日期去重

1 select DISTINCT dt,user_id from test5active;

第二步:用row_number() over()函数计数

1 2 3 4 5 6 7 select t1.user_id,t1.dt, row_number() over(partition by t1.user_id order by t1.dt) day_rank from ( select DISTINCT dt,user_id from test5active )t1;

第三步:日期减去计数值得到结果

1 2 3 4 5 6 7 8 9 10 11 select t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis from ( select t1.user_id,t1.dt, row_number() over(partition by t1.user_id order by t1.dt) day_rank from ( select DISTINCT dt,user_id from test5active )t1)t2;

第四步:根据id和结果分组并计算总和,大于等于2的即为连续登陆的用户,得到 用户id,开始日期,结束日期,连续登录天数

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 select t3.user_id,min(t3.dt),max(t3.dt),count(1) from ( select t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis from ( select t1.user_id,t1.dt, row_number() over(partition by t1.user_id order by t1.dt) day_rank from ( select DISTINCT dt,user_id from test5active )t1 )t2 )t3 group by t3.user_id,t3.dis having count(1)>1;

用户id 开始日期 结束日期 连续登录天数

最后:连续登陆的用户

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 select distinct t4.user_id from ( select t3.user_id,min(t3.dt),max(t3.dt),count(1) from ( select t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis from ( select t1.user_id,t1.dt, row_number() over(partition by t1.user_id order by t1.dt) day_rank from ( select DISTINCT dt,user_id from test5active )t1 )t2 )t3 group by t3.user_id,t3.dis having count(1)>1 )t4;

思路二:使用lag(向后)或者lead(向前)

1 2 3 4 5 6 7 select user_id,t1.dt, lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id from ( select DISTINCT dt,user_id from test5active )t1;

1 2 3 4 5 6 7 8 9 10 11 12 select distinct t2.user_id from ( select user_id,t1.dt, lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id from ( select DISTINCT dt,user_id from test5active )t1 )t2 where datediff(last_date_id,t2.dt)=1;


这篇关于Hive-SQL查询连续活跃登录用户的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!


扫一扫关注最新编程教程