题解 | 最长连续登录天数

with tb_dau1 as(
select fdate,
       user_id,
       fdate-row_number() over(partition by user_id order by fdate) gap
from tb_dau
where year(fdate)=2023 and month(fdate)=1),
tb_dau2 as(
select user_id,
       count(*) over(partition by user_id,gap) consec_days
from tb_dau1)
select user_id,max(consec_days) max_consec_days
from tb_dau2
group by user_id