用户连续登录天数hive

create table users(
    uid int,
    dt string)；

select uid,pdt,count(uid) as cnt_day
from(select t1.uid,date_sub(t1.dt,t1.rm) as pdt
from
(select uid,dt,
row_number() over(distribute by uid sort by dt) rm
from (select uid,dt from users group by uid,dt)m) t1)t2
group by uid,pdt;

对打点表中的用户计算连续登录天数

date_sub(t1.dt,t1.rm) 在t1.dt的基础上减去 t1.rm的天数，rm是按照差值为1递增的，所以如果得到的pdt值相同的同一个uid值，说明为用户连续登录的天数，但是这里的pdt值仅作为这个统计，没有其他意义

连续登录天数大于7天的用户

select uid,pdt
from(select t1.uid,date_sub(t1.dt,t1.rm) as pdt
from
(select uid,dt,
row_number() over(distribute by uid sort by dt) rm
from (select uid,dt from users group by uid,dt)m) t1)t2
group by uid,pdt having count(uid)>6;

最后

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