# 关键是找出 连续登录的 聚合key select user_id,max(consec_days) as max_consec_days from ( select user_id,grp_key,count(*) as consec_days from ( select * ,sum(is_break) over(partition by user_id order by fdate ) as grp_key from ( select user_id ,fdate ,lag(fdate) over(partition by user_id order by fdate) as pre_...
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