题解 | #查询连续登陆的用户#

select user_id from(
select user_id,
count(in_day-in_rank)
from(
select a.user_id,day(log_time) in_day,
row_number() over(partition by user_id order by day(log_time)) in_rank
from login_tb a join register_tb b
on a.user_id=b.user_id
)c
group by user_id
having count(in_day-in_rank)>=3
order by user_id
)d