Question

我打算将客户的START END日期展平，如下所示。如果日期范围是连续的，我将合并这些记录。另外，我会保持这样 EG：

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

使用分析功能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

但我不能继续使用客户的min（START）和mAx（END）组，因为它也会合并非连续值。任何好的建议

Answer 1

尝试类似

的内容

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

如果您捕获实际的超前/滞后日期而不是0/1，那么您会得到类似的结果：

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

我已将您的第二条记录拆分为两条相邻的记录以实现效果，因此此处有一行，其中机器人调整日期为null。然后你可以删除那些同时为null的那些，因为它们完全反映了一个范围内的记录，并且你留下了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空值的那些，因为相邻的行（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

SQL Fiddle demo

仅按特定记录分组

2 个答案: