Question

select * from grouping;

以上查询给出：

WORKING_DAYS,DATE_TYPE,SEQ_START
10/11/2013  ,L        ,0
10/10/2013  ,L        ,0
10/8/2013   ,L        ,1
10/6/2013   ,H        ,0
10/5/2013   ,H        ,0
10/4/2013   ,L        ,0
10/3/2013   ,L        ,0
10/2/2013   ,L        ,0
10/1/2013   ,L        ,0

我想通过SEQ_START对这些行进行分组，这样当我运行它时应该给出以下结果。 rank列应该是由SEQ_START分组的WORKING_DAYS（或任何其他数字）。请注意，即使10/10/2013和10/11/2013的SEQ_START为0但仍然不应该有10/1/2013，而应该有10/10/2013，因为在10/8/2013，序列是破（SEQ_START为0）。

WORKING_DAYS,DATE_TYPE,SEQ_START,rank
10/1/2013   ,L        ,0        ,10/1/2013
10/2/2013   ,L        ,0        ,10/1/2013
10/3/2013   ,L        ,0        ,10/1/2013
10/4/2013   ,L        ,0        ,10/1/2013
10/8/2013   ,L        ,1        ,10/8/2013
10/10/2013  ,L        ,0        ,10/10/2013
10/11/2013  ,L        ,0        ,10/10/2013

我已经写了以下查询，其回报几乎是我的预期，但它给了10/10/2013和10/11/2013一个“0”值而不是他们应该有10/10/2013因为序列在10上被打破/ 8/2013，其中有1个用于SEQ_START

select working_days,  
       date_type, 
       seq_start, 
       FIRST_VALUE(working_days)  OVER (PARTITION BY seq_start 
                                        ORDER BY working_days) "rank"
  from grouping 
 where date_type = 'L'

Answer 1

select 
  working_days,  date_type, seq_start, 
  FIRST_VALUE(working_days) OVER (PARTITION BY grp ORDER BY working_days) "rank"
from (
  select 
    working_days, date_type, seq_start,
    sum(front) over(order by working_days) as grp
  from (
    select 
      working_days, date_type, seq_start,
      decode(seq_start, lag(seq_start)over(order by working_days), 0, 1) as front
    from t1
  )
)
where date_type = 'L'
order by 1

fiddle

Oracle first_value（PARTITION）中断序列

1 个答案: