题解 | 查询连续登陆的用户
select user_id
from
(
select
rt.user_id,
day(lt.log_time) - row_number() over(partition by rt.user_id order by lt.log_time) as fake_time
from register_tb as rt
left join login_tb as lt
on rt.user_id = lt.user_id
) as a
group by user_id, fake_time
having count(fake_time) >= 3
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