如果结果保持不变，如何获取最小日期和最大日期

Question

我有一个地址SCD类型2表，但有时输入的信息保持不变，我想编写一个查询，如果信息不变，则保留先前的开始数据并将结束日期设置为最大，例如

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OBJID          BEGDA      ENDDA     HASHROW_COL RK 
83022088    2012-03-30  2012-10-28  e1-ef-a9-36 1 
83022088    2012-10-29  2013-09-07  63-69-e5-25 2 
83022088    2013-09-08  2014-08-30  e1-ef-a9-36 3
83022088    2014-08-31  2016-11-26  e1-ef-a9-36 4
83022088    2016-11-27  9999-12-31  e1-ef-a9-36 5

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注意，在第3-5行中，HASHROW_COL保持不变。

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Desired result:
OBJID          BEGDA       ENDDA    HASHROW_COL RK 
83022088    2012-03-30  2012-10-28  e1-ef-a9-36 1 
83022088    2012-10-29  2013-09-07  63-69-e5-25 2
83022088    2013-09-08  9999-12-31  e1-ef-a9-36 3

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到目前为止查询

select a.objid, a.hashrow_col, 
case when a.objid <> b.objid then b.begda
    when a.hashrow_col = b.hashrow_col and (b.begda - interval '1' day <= a.endda) then 
    a.begda  end,
case when a.objid <> b.objid then b.endda
    when (a.hashrow_col = b.hashrow_col) and (b.begda - interval '1' day <= a.endda) and b.endda > a.endda  
    then b.endda 
    end,
from
(select objid, begda, endda, HASHROW_COL, 
from OTABLE )  a
inner join 
(select objid, begda, endda, HASHROW_COL, 
from OTABLE) b
on
a.objid = b.objid
where 
and a.objid = '83022088' 
order by a.OBJID, a.BEGDA,  a.HASHROW_COL;

Answer 1

这是一个孤岛问题。在这种情况下，我将使用：

select objid, min(begda), max(endda), HASHROW_COL,
       row_number() over (partition by objid order by min(begda)) as ranking
from (select t.*,
             sum(case when prev_endda = begda - interval '1' day then 0 else 1 end) over (partition by objid order by begda) as grouping
      from (select t.*,
                   lag(endda) over (partition by objid, HASHROW_COL order by begda) as prev_endda
            from t
           ) t
      ) t
group by grouping, objid, HASHROW_COL;

Answer 2

假设hashrow_col是保持不变的值，（即您需要分组的值），那么您所需要做的就是min（begda）和max（endda）

select objid, min(begda) as begda, max(endda) as enddat, HASHROW_COL,
       min(rnk) 
  from OTABLE
group by objid,hashrow_col