题解 | #2021年11月每天新用户的次日留存率#

select min_time,round(count(t2.uid)/count(t1.uid),2) as uv_left_rate
from
    (select uid,min(date(in_time)) as min_time
    from tb_user_log
    group by uid
    )as t1
left join
    (select uid,date(in_time) as dt
     from tb_user_log
     union 
     select uid,date(out_time) as dt
     from tb_user_log
    )as t2
on t1.uid=t2.uid
and datediff(t2.dt,t1.min_time)=1
where date_format(min_time,'%Y-%m')='2021-11'
group by min_time
order by min_time