题解 | 最长连续登录天数

select user_id,max(cnt) as max_consec_days
from(
    select user_id,date_gap,count(date_gap) as cnt
    from(
        select user_id,date_sub(fdate,interval ranked day) as date_gap
        from(
            select *,row_number()over(partition by user_id order by fdate asc) ranked
            from tb_dau
            where fdate between '2023-01-01' and '2023-01-31'
        ) as t1
    ) as t2
    group by user_id,date_gap
) as t3
group by user_id