Hive-SQL查询连续活跃登录用户

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

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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

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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

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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

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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

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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)

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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;

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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;