SQL计算用户留存率问题

select 
    r.uid,
	date(r.register_date) as rt,
	date(l.login_time) as lt,
	datediff(l.login_time, r.register_date) as tdiff
from register_info r
left join login_info l 
on r.uid=l.uid
and date(l.login_time) between date(r.register_date)+interval 1 day and date(r.register_date)+interval 7 day		
order by uid asc

select 
	rt as 日期,
	count(distinct uid) as 新增用户数,
	count(distinct case when tdiff=1 then uid end)/count(distinct uid) as 次日留存率,
	count(distinct case when tdiff=3 then uid end)/count(distinct uid) as 三日留存率,
	count(distinct case when tdiff=7 then uid end)/count(distinct uid) as 七日留存率
from 
(
	select 
		r.uid,
		date(r.register_date) as rt,
		date(l.login_time) as lt,
		datediff(l.login_time, r.register_date) as tdiff
	from register_info r
	left join login_info l 
	on r.uid=l.uid
	and date(l.login_time) between date(r.register_date)+interval 1 day and date(r.register_date)+interval 7 day		
	order by uid asc
) t1
group by rt

select
	count(distinct uid) 新增用户数,
    count(distinct case when tdiff=1 then uid end)/count(distinct uid) as 次日留存率,
    count(distinct case when tdiff=3 then uid end)/count(distinct uid) as 三日留存率,
    count(distinct case when tdiff=7 then uid end)/count(distinct uid) as 七日留存率
from
(
	select t1.uid, t1.rt, t2.lt, datediff(t2.lt,t1.rt) as tdiff
	from
	(
		(select user_id as uid, min(date(login_time)) as rt from user_login_info group by user_id) t1
		LEFT JOIN
		(select user_id as uid, date(login_time) as lt from user_login_info) t2
		on t1.uid=t2.uid and t2.lt between t1.rt + interval 1 day and t1.rt + interval 7 day
	)
) t3
group by rt
order by rt asc