题解 | 最长连续登录天数

select user_id,
 max(consec_days) as max_consec_days
  from 
(
    select user_id,
    diff ,
    count(*) as consec_days 
    from (
    select user_id,
    subdate(fdate,row_number()over(partition by user_id order by fdate)) as diff
    from tb_dau
    where fdate between '2023-01-01' and '2023-12-31'
    ) b 
    group by user_id ,diff 
) a
group by user_id