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hive-sql 连续登录五天的用户

article 2025/1/3 3:03:22

with tmp as (
          select '梁牧泽' as uid, '2023-03-03' as `dt` union all
          select '梁牧泽' as uid, '2023-03-04' as `dt` union all
          select '梁牧泽' as uid, '2023-03-05' as `dt` union all
          select '梁牧泽' as uid, '2023-03-07' as `dt` union all
          select '梁牧泽' as uid, '2023-03-08' as `dt` union all
          select '梁牧泽' as uid, '2023-03-09' as `dt` union all
          select '梁牧泽' as uid, '2023-03-10' as `dt` union all
          select '梁牧泽' as uid, '2023-03-11' as `dt` union all
          select '梁牧泽' as uid, '2023-03-12' as `dt` union all
          select '夏初' as uid, '2023-03-10' as `dt` union all
          select '夏初' as uid, '2023-03-10' as `dt` union all
          select '夏初' as uid, '2023-03-11' as `dt` union all
          select '夏初' as uid, '2023-03-12' as `dt` union all
          select '夏初' as uid, '2023-03-13' as `dt` union all
          select '夏初' as uid, '2023-03-14' as `dt` union all
          select '夏初' as uid, '2023-03-15' as `dt`
) 
select 
uid,
dt,
follow_5_dt
from 
(select uid,dt
,last_value(dt) over(partition by uid order by dt asc rows between current row and 4 following) as follow_5_dt
,datediff(last_value(dt) over(partition by uid order by dt asc rows between current row and 4 following),dt) as diff_day
from tmp ) t where diff_day=4

结果：