Group by Clause 仅针对特定记录

Question

我正计划为客户拉平我的开始和结束日期，如图所示below.If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，因为它还会合并非连续值。有什么好的建议

Answer 1

试试

select customer, min(start), min(end)
from
(
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  
) 
group by customer, cont_flg, case when cont_flg=0 then start end

Answer 2

如果您捕获实际的 lead/lag 日期而不是 0/1，那么您会得到类似这样的结果：

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

为了效果，我已将您的第二条记录拆分为两条相邻的记录，因此此处有一行机器人调整日期为空。然后你可以删除那些都为空的，因为它们反映了完全在一个范围内的记录，你剩下每个时期的开始和结束日期：

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，因为相邻的行（lead/lag）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

SQL Fiddle demo.

Group by Clause 仅针对特定记录

Group by Clause for specific records alone

sql

oracle

oracle11g