思路:对登录记录按照天数去重,筛选时间计算日期排名差,后续依次进行分组分组后,最大值即为连续登录最多天的 with user_login as ( select distinct user_id, fdate from tb_dau where fdate between '2023-01-01' and '2023-01-31' ) , user_rank as ( select user_id , fdate , row_number() over(partition by user_id order by fdate) as rn , date_sub(fdate, interval r...