7天週期活動，按日統計連續完成N天任務

景：產品作了一個週期7天活動，用戶能夠每一天解鎖一個任務，完成後發放獎勵；

需求：按日統計連續完成N天用戶數

目的：用戶參與活動的興趣度；

活動海報圖：

分析思路：

這就是連續登錄問題，用至關時間差斷定；

用戶明細以下：

借用lag函數斷定上一次登錄日期；

連續登錄幾天，其datediff日期差等於連續登錄天數

附上代碼：

,with  tmp_da_user as (

select

to_date(ds,'yyyymmdd') as ds

,uid

,count(distinct ds) as 完整天數

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 當天完成任務數

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

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

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

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

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

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

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

from tmp_da_user_tack

group by ds

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