Oracle - 从行组中获取最小和最大日期

Question

初步数据：

select t.AGENHI, t.TACTHI, t.DTEFHI, t.DTFIHI            
      from mytable t
      where agenhi = '81000040' ;

AGENHI      TACTHI   DTEFHI     DTFIHI            
81000040    1        24/02/92   08/03/92
81000040    1        09/03/92   28/02/93
81000040    1        01/03/93   31/05/97
81000040    0,8      01/06/97   31/12/97
81000040    1        01/01/98   31/12/98
81000040    1        01/01/99   

使用此查询：

SELECT AGENHI, 
DECODE(TACTHI, 0.05, '005', 0.07, '007', 0.1, '010', 0.137, '013', 0.15, 
'015', 0.2, '020', 0.21, '021', 0.23, '023', 0.25, '025', 0.3, '030', 
0.34, '034', 0.4, '040', 0.45, '045', 0.5, '050', 0.6, '060', 0.63, 
'063',0.7, '070', 0.75, '075', 0.8,'080', 0.84, '084',0.9, '090', 1, '100', 
TACTHI) as QUOTITE,
dtefhi as START_DATE,
DECODE(LEAD (DTFIHI, 1) OVER (ORDER BY DTFIHI NULLS LAST) ,null, 
to_date('31122099','ddmmyyyy'), LEAD (DTFIHI, 1) OVER (ORDER BY DTFIHI NULLS 
LAST)) AS END_DATE
FROM MYTABLE 
WHERE AGENHI = '81000040' AND DTFIHI IS NOT NULL;

我明白了：

AGENHI      QUOTITE     START_DATE    END_DATE
81000040    100         08/03/92      28/02/93
81000040    100         28/02/93      31/05/97
81000040    100         31/05/97      31/12/97
81000040    080         31/12/97      31/12/98
81000040    100         31/12/98      31/12/99

但是当下一行相同时我需要对“QUOTITE”进行分组，并显示第一行的开始日期和最后一行的日期。

预期结果 ：

AGENHI      QUOTITE     START_DATE    END_DATE
81000040    100         24/02/92      31/12/97
81000040    080         01/06/97      31/12/98
81000040    100         01/01/98      31/12/99

gordon提供的解决方案：

选择agenhi，tacthi，min（dtfihi）作为start_date，引导（max（dtfihi））over（由agenhi划分，seqnum - seqnum_2 order by max（dtfihi））作为end_date from（select t。*，row_number（）over（partition由dqfihi的agenhi命令）作为seqnum，row_number（）over（由agenhi分区，由dtfihi分类）作为seqnum_2来自HIA@CHRONOS_TO_S2.WORLD t其中agenhi ='81000040'和dtfihi IS NOT NULL）t group by agenhi，（ seqnum - seqnum_2），tacthi;

结果：

81000040    1   08/03/92   null 
81000040    1   31/12/98   null   
81000040    0,8 31/12/97   null 

我怎样才能获得end_date？我会看看我是否找到了解决方案，谢谢！

Answer 1

你有一个缺口和岛屿的问题。我会从原始数据开始，所以像这样：

select agenhi, tacthi, min(dtfihi) as start_date,
       lead(min(dtfihi)) over (partition by agenhi order by min(dtfihi)) as end_date
from (select t.*,
             row_number() over (partition by agenhi, order by dtfihi) as seqnum,
             row_number() over (partition by agenhi, tacthi order by dtfihi) as seqnum_2
      from mytable t
      where agenhi = '81000040' an dtfihi IS NOT NULL
     ) t
group by agenhi, (seqnum - seqnum_2), tacthi;

您可以添加decode()逻辑（我将其写为case表达式），但它并没有真正改变解决方案。

Answer 2

@Boneist，@ GordonLinoff再次感谢您的帮助。

select 
agenhi, 
DECODE(TACTHI, 0.05, '005', 0.07, '007', 0.1, '010', 0.137, '013', 0.15, '015', 0.2, '020', 0.21, '021', 0.23, '023', 0.25, '025', 0.3, '030', 
0.34, '034', 0.4, '040', 0.45, '045', 0.5, '050', 0.6, '060', 0.63, '063',0.7, '070', 0.75, '075', 0.8,'080', 0.84, '084',0.9, '090', 1, '100', TACTHI) as QUOTITECPAGE,  
min(dtefhi) as start_date,
nvl(lead(min(dtfihi)) over (partition by agenhi order by min(dtfihi)), 
to_date('31122099','ddmmyyyy')) as end_date
from (select t.*,
         row_number() over (partition by agenhi order by dtfihi) as seqnum,
         row_number() over (partition by agenhi, tacthi order by dtefhi) as seqnum_2
  from HIA@CHRONOS_TO_S2.WORLD t
  where agenhi = '81000040' order by DTEFHI
 ) t
 group by agenhi, (seqnum - seqnum_2), tacthi;