实现Mysql中的n日留存率计算

select user_id,min(date) as first_day
from a2_userbehavior_csv
group by user_id

select user_id,date
from a2_userbehavior_csv
group by user_id,date

select t1.*,t2.date,datediff(t2.date,t1.first_day) as day_diff
from 
(select user_id,min(date) as first_day
from a2_userbehavior_csv
group by user_id) as t1
left join 
(select user_id,date
from a2_userbehavior_csv
group by user_id,date) as t2
on t1.user_id=t2.user_id

select first_day as dt,
	   concat(round(100*count(case when day_diff=1 then user_id end)/count(case when day_diff=0 then user_id end),2),"%") as '次日留存率',
	   concat(round(100*count(case when day_diff=3 then user_id end)/count(case when day_diff=0 then user_id end),2),"%") as '三日留存率',
	   concat(round(100*count(case when day_diff=7 then user_id end)/count(case when day_diff=0 then user_id end),2),"%") as '七日留存率'
from
(select t1.*,t2.date,datediff(t2.date,t1.first_day) as day_diff
from 
(select user_id,min(date) as first_day
from a2_userbehavior_csv
group by user_id) as t1
left join 
(select user_id,date
from a2_userbehavior_csv
group by user_id,date) as t2
on t1.user_id=t2.user_id) as t3
group by first_day 
order by first_day