Question

当我合并包含DENSE_RANK()的查询以生成IN子句的输入时，我遇到了一些奇怪的结果。

证明我把它们分开了;

此查询

select *
from ALL_QUOTE
where ID in ('G002WMLS')

返回1个结果：

ID        LongID  StartDate  EndDate
G002WMLS  67888   01/10/2011 30/11/2011

此查询

select ID
from (

select LongId, ID, DENSE_RANK() over (partition by LongId order by end_date desc, substr(ID, 2, 7) desc, start_date desc) d
from WithoutPD
)

where d = 1 and LongId = '67888'

也会返回1个结果：

ID
G002WMLS

然而，当我把它们组合在一起时：

select *
from ALL_QUOTE
where ID in (

select ID
from (

select LongId, ID, DENSE_RANK() over (partition by LongId order by end_date desc, substr(ID, 2, 7) desc, start_date desc) d
from WithoutPD
)

where d = 1

)
and LongId = '67888';

我最终得到了两个结果：

ID        LongID StartDate   EndDate 
G002MIMQ  67888  01/10/2010  30/09/2011 
G002WMLS  67888  01/10/2011  30/11/2011

我根本无法理解G002MIMQ如何包含在结果中。我正在使用Oracle 11.2.0.1.0，但我知道这可能是一个我误解的通用SQL功能。

希望你能解释一下这个奇怪的问题。

Answer 1

您已将and LongID='67888'移到子查询的where子句之外。

请尝试此查询...

select *
from ALL_QUOTE
where ID in (    
    select ID
    from (        
        select LongId, ID, DENSE_RANK() over (partition by LongId order by end_date desc, substr(ID, 2, 7) desc, start_date desc) d
        from WithoutPD        
    )        
    where d = 1
    and LongId = '67888'
);

修改

select AllQuote.* from AllQuote inner join ( select ID from ( select LongId, ID, DENSE_RANK() over (partition by LongId order by end_date desc, substring(ID, 2, 7) desc, start_date desc) d from WithoutPD ) t where d = 1 and LongId = '67888' ) v on AllQuote.ID = v.ID

DENSE_RANK和IN的奇怪行为

1 个答案: