Hive-SQL查询连续活跃登录用户思路详解
这篇文章主要介绍了Hive-SQL查询连续活跃登陆的用户,活跃用户这里是指连续2天都活跃登录的用户,本文给大家分享解决思路及sql语句,对大家的学习或工作具有一定的参考借鉴价值,需要的朋友可以参考下
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; |
参考:2020年大厂面试题-数据仓库篇
SQL 查询连续登陆7天以上的用户
到此这篇关于Hive-SQL查询连续活跃登陆的用户的文章就介绍到这了
原文链接:https://www.cnblogs.com/yangms/p/14179403.html