oracle - 根据参数 -Oracle 重新启动 sum(over) 函数
问题描述
我试图做的是使用 sum over 函数,使其重置为 0 并在遇到参数值时重新开始。我附上的是一些带有开始和结束日期的基本会员数据。现在,如果会员期限之间的间隔超过 5 天,则它会再次重新启动计数器。我尝试使用延迟函数,但 oracle 不允许窗口函数。
数据:
ID BEG_DATE END_DATE MONTHS DAYS_IN_COVERAGE_BREAK tot_membership_months
123 01-JAN-15 31-DEC-15 30.3 30.3 first term
123 01-JAN-16 28-FEB-17 35.3 1 65.6 --Adds first & 2nd term
123 01-JUN-17 31-DEC-17 17.8 93 17.8 **--restarts- lapse in coverage**
123 01-JAN-18 30-MAR-19 37.8 1 55.6 --Adds from restart time
询问:
With TL as (select '123' as id, to_Date('01JAN2015') as Beg_date,to_date('31DEC2015') as end_Date from dual union
select '123' as id, to_Date('01JAN2016') as Beg_date,to_date('28FEB2017') as end_Date from dual union
select '123' as id, to_Date('01JUN2017') as Beg_date,to_date('31DEC2017') as end_Date from dual union
select '123' as id, to_Date('01JAN2018') as Beg_date,to_date('30MAR2019') as end_Date from dual)
select TL.*, round((END_DATE-BEG_DATE) /12,1) as MONTHS,
BEG_DATE-LAG(END_DATE) OVER (PARTITION BY ID ORDER BY BEG_DATE) as DAYS_IN_COVERAGE_BREAK
--case when BEG_DATE-LAG(END_DATE) OVER (PARTITION BY ID ORDER BY BEG_DATE)<5
-- then sum(round((END_DATE-BEG_DATE) /12,1) over (partition by mrn order by beg_date)) else round((END_DATE-BEG_DATE) /12,1) end as member_cum_months
from TL;
任何见解表示赞赏。谢谢
解决方案
这是一个间隙和岛屿问题,每行开始一个新岛屿,与前一行有 5 天以上的差距。
这是一个使用窗口函数的选项:lag()
给我们前一个end_date
,然后我们使用窗口总和,对于超过 5 天的每个间隔递增来定义组。
select id, beg_date, end_date,
end_date - min(beg_date) over(partition by id, grp order by beg_date) total_membership_days
from (
select t.*,
sum(case when beg_date > lag_end_date + interval '5' day then 1 else 0 end) over(partition by id order by beg_date) grp
from (
select t.*,
lag(end_date) over(partition by id order by beg_date) lag_end_date
from tl t
) t
) t
如果您希望结果为几个月而不是几天,那么:
select id, beg_date, end_date,
months_between(end_date, min(beg_date) over(partition by id, grp order by beg_date)) total_membership_months
from (
select t.*,
sum(case when beg_date > lag_end_date + interval '5' day then 1 else 0 end) over(partition by id order by beg_date) grp
from (
select t.*,
lag(end_date) over(partition by id order by beg_date) lag_end_date
from tl t
) t
) t
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