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
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select distinct dt,user_id from test5active; |
第二步：用row_number() over()函数计数
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selectt1.user_id,t1.dt,row_number() over(partition by t1.user_id order by t1.dt) day_rankfrom(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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selectt2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as disfrom(selectt1.user_id,t1.dt,row_number() over(partition by t1.user_id order by t1.dt) day_rankfrom(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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selectt3.user_id,min(t3.dt),max(t3.dt),count(1)from(selectt2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as disfrom(selectt1.user_id,t1.dt,row_number() over(partition by t1.user_id order by t1.dt) day_rankfrom(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_idfrom(selectt3.user_id,min(t3.dt),max(t3.dt),count(1)from(selectt2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as disfrom(selectt1.user_id,t1.dt,row_number() over(partition by t1.user_id order by t1.dt) day_rankfrom(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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selectuser_id,t1.dt,lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_idfrom(select distinct dt,user_id from test5active)t1; |
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selectdistinct t2.user_idfrom(selectuser_id,t1.dt,lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_idfrom(select distinct dt,user_id from test5active)t1)t2 where datediff(last_date_id,t2.dt)=1; |
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原文链接：https://www.cnblogs.com/yangms/p/14179403.html
