我计划为客户展平我的开始结束日期,如下所示。如果日期范围是连续的,我将合并这些记录。否则我会保持原样 例如:
Input
Customer START END
A 2000 2001
A 2001 2007
A 2009 2010
A 2011 2015
Expected Output
A 2000 2007
A 2009 2010
A 2011 2015
使用分析函数,我能够用连续日期标记记录:
--TAG = 1 means continuous
select A *,
CASE WHEN LEAD (START) OVER (PARTITION BY CUSTOMER ORDER BY START,END) = END
OR LAG (END_DT) OVER (PARTITION BY CUSTOMER ORDER BY START,END) = START
THEN 1 ELSE 0 END AS CONT_FLG
From TABLE CUSTOMER
Customer START END CONT_FLG
A 2000 2001 1
A 2001 2007 1
A 2009 2010 0
A 2011 2015 0
但我无法继续按客户对 min (START) 和 mAx (END) 进行分组,因为它也会合并非连续值。有什么好的建议
最佳答案
如果您捕获实际的提前/滞后日期而不是 0/1,那么您会得到如下所示的结果:
select t.*,
case when lag(end_dt) over (partition by customer order by start_dt)
= start_dt then null else start_dt end as adj_start_dt,
case when lead(start_dt) over (partition by customer order by start_dt)
= end_dt then null else end_dt end as adj_end_dt
from t42 t
order by customer, start_dt;
CUSTOMER START_DT END_DT ADJ_START_DT ADJ_END_DT
-------- ---------- ---------- ------------ ----------
A 2000 2001 2000
A 2001 2003
A 2003 2007 2007
A 2009 2010 2009 2010
A 2011 2015 2011 2015
为了达到效果,我已将您的第二条记录拆分为两个相邻的记录,因此此处有一行机器人调整的日期为空。然后,您可以删除那些都为空的记录,因为它们反射(reflect)完全在一个范围内的记录,并且您留下每个期间的开始和结束日期:
select *
from (
select t.*,
case when lag(end_dt) over (partition by customer order by start_dt)
= start_dt then null else start_dt end as adj_start_dt,
case when lead(start_dt) over (partition by customer order by start_dt)
= end_dt then null else end_dt end as adj_end_dt
from t42 t
)
where adj_start_dt is not null or adj_end_dt is not null
order by customer, start_dt;
CUSTOMER START_DT END_DT ADJ_START_DT ADJ_END_DT
-------- ---------- ---------- ------------ ----------
A 2000 2001 2000
A 2003 2007 2007
A 2009 2010 2009 2010
A 2011 2015 2011 2015
然后您可以折叠那些带有空值的行,因为相邻行(带有超前/滞后)现在是相关的:
select distinct customer,
case when adj_start_dt is null then
lag(adj_start_dt) over (partition by customer order by start_dt)
else adj_start_dt end as grp_start_dt,
case when adj_end_dt is null then
lead(adj_end_dt) over (partition by customer order by start_dt)
else adj_end_dt end as grp_end_dt
from (
select t.*,
case when lag(end_dt) over (partition by customer order by start_dt)
= start_dt then null else start_dt end as adj_start_dt,
case when lead(start_dt) over (partition by customer order by start_dt)
= end_dt then null else end_dt end as adj_end_dt
from t42 t
)
where adj_start_dt is not null or adj_end_dt is not null
order by customer, grp_start_dt;
CUSTOMER GRP_START_DT GRP_END_DT
-------- ------------ ----------
A 2000 2007
A 2009 2010
A 2011 2015
关于sql - 仅针对特定记录按子句分组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27960180/