题解 | 最长连续登录天数

select
    user_id,
    max(continuous_days) as max_consec_days
from
    (
        select
            user_id,
            count(*) as continuous_days
        from
            (
                select
                    user_id,
                    fdate,
                    date_sub(fdate, interval (row_number() over (partition by user_id order by fdate)) day) as login_group
                from 
					tb_dau
			) a
        group by user_id, login_group
	) b
group by user_id;