连续活跃的用户

题目1

现在有一张用户活跃全量表，列有id，活跃日期（全部），分区
如何求得最大连续活跃天数的ID

way1

with t1 as 
(select id,row_number() over(partition by id order by lastactivetime asc) as ranks,
lastactivetime from table)

select id,ate_sub(lastactivetime,ranks) as difftime,
min(lastactiveimte) as min,max(lastactiveimte) as max
from t1 
group by id,ate_sub(lastactivetime,ranks) as difftime

可以很好的将多次连续值分开

t2 as 
(select id,difftime,min,max from t1
group by id,difftime,min,max)

select distinct id, max(datediff(max,min)+1) over(partition by id) as max_continuous_days
from t2

way2

with t1 as 
(select id,row_number() over(partition by id order by lastactivetime asc) as ranks,
lastactivetime from table)

select t1.id as id, 
datediff(t2.lastactiveitime) as datediff,
(t2.rank-t1.rank) as rankdiff
from t1 cross join t1 as t2
on t1.id = t2.id
having datediff = rankdiff
and datediff != 0