题解 | #最长连续登录天数#

select user_id,max(seq) as max_consec_days#--计算最大值
from
(
select user_id,gap,count(*) as seq#--分别计算连续登陆天数，groupby user_id,头天#日期了
from
    (
        select user_id,SUBDATE(fdate,INTERVAL num day) as gap#--计算连续登录的头#天日期
        from
            (
                select user_id,fdate,
				        row_number() over (partition by user_id order by fdate) as num#--计算登录次序
                from tb_dau
                where year (fdate) = 2023
            )  t0
    ) t1
group by user_id,gap
)t2
group by user_id