7天周期活动，按日统计连续完成N天任务

景：产品作了一个周期7天活动，用户能够每一天解锁一个任务，完成后发放奖励；

需求：按日统计连续完成N天用户数

目的：用户参与活动的兴趣度；

活动海报图：

分析思路：

这就是连续登录问题，用至关时间差断定；

用户明细以下：

借用lag函数断定上一次登录日期；

连续登录几天，其datediff日期差等于连续登录天数

附上代码：

,with  tmp_da_user as (

select

to_date(ds,'yyyymmdd') as ds

,uid

,count(distinct ds) as 完整天数

from tmp_t_user_takse_detail

where status_name = '完成'

and uid in ('203111313364823804','1833903495100023')

group by

ds

,uid

)

,tmp_da_user_tack as (

select *

,lag(ds,1)over(partition by uid order by ds asc) as lag_1

,lag(ds,2)over(partition by uid order by ds asc) as lag_2

,lag(ds,3)over(partition by uid order by ds asc) as lag_3

,lag(ds,4)over(partition by uid order by ds asc) as lag_4

,lag(ds,5)over(partition by uid order by ds asc) as lag_5

,lag(ds,6)over(partition by uid order by ds asc) as lag_6

,lag(ds,7)over(partition by uid order by ds asc) as lag_7

from tmp_da_user

)

select

ds

,count(distinct uid) as 当天完成任务数

,count(case when DATEDIFF(ds,lag_1,'dd') = 1 then uid end ) as 连续2天完成

,count(case when DATEDIFF(ds,lag_2,'dd') = 2 then uid end ) as 连续3天完成

,count(case when DATEDIFF(ds,lag_3,'dd') = 3 then uid end ) as 连续4天完成

,count(case when DATEDIFF(ds,lag_4,'dd') = 4 then uid end ) as 连续5天完成

,count(case when DATEDIFF(ds,lag_5,'dd') = 5 then uid end ) as 连续6天完成

,count(case when DATEDIFF(ds,lag_6,'dd') = 6 then uid end ) as 连续7天完成

,count(case when DATEDIFF(ds,lag_7,'dd') = 7 then uid end ) as 连续8天完成

from tmp_da_user_tack

group by ds

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