题解 | 查询连续登陆的用户

select distinct user_id from(
select user_id,
    date_sub(time,interval ranks day) as startday,
    count(*) as consec_days
from(
    select
        distinct user_id,
        date(log_time) as time,
        row_number() over (partition by user_id order by log_time) as ranks
    from login_tb
    join register_tb using (user_id))t1
group by user_id, startday)t2
where consec_days>=3