题解 | #最长连续登录天数#

select
    user_id,
    max(consec_days) as max_consec_days
from
    (
        select
            user_id,
            count(*) as consec_days
        from
            (
                select
                    fdate,
                    user_id,
                    (
                        dense_rank() over (
                            partition by
                                user_id
                            order by
                                fdate
                        )
                    ) as rk
                from
                    tb_dau
            ) a
        group by
            user_id,
            (fdate - rk)
    ) b
group by user_id