题解 | #最长连续登录天数#

select user_id, max(con_nums) as max_consec_days
from
(
select user_id, count(*) as con_nums
from
(
select
    user_id,
    date_sub(fdate, interval ranking day) as day_groups
from
    (
        select
            *,
            dense_rank() over (partition by user_id order by fdate asc) as ranking
        from
            tb_dau
    ) a
) b
group by user_id,day_groups
) c
group by user_id