Question

如果我有以下数据

A | 01/01/2008 00:00:00

B | 01/01/2008 01:00:00

A | 01/01/2008 11:59:00

C | 02/01/2008 00:00:00

D | 02/01/2008 01:00:00

D | 02/01/2008 20:00:00

我想只选择标识符（A，B，C或D）在12小时内发生两次的记录。在上面的这个例子中，这只是'A'

任何人都可以帮忙（这是针对Oracle数据库的）

由于

中号

Answer 1

 Select Distinct A.Identifer  
  From Table A
    Join Table B --  EDIT to eliminate self Joins (to same row)
        On A.PrimKey <> B.PrimaryKey
           And A.Identifer = B.Identifer               
           -- EDIT to fix case where 2 at same time
           And A.OccurTime >=  B.OccurTime 
           And A.OccurTime < B.OccurTime + .5

并实施评论中提出的问题，（忽略不同日期的记录）

- 对于SQL Server，

   Select Distinct A.Identifer  
    From Table A
      Join Table B 
        On A.PrimKey <> B.PrimaryKey
           And A.Identifer = B.Identifer
           -- EDIT to fix case where 2 at same time
           And A.OccurTime >= B.OccurTime  
           And A.OccurTime < B.OccurTime + .5
    Where DateDiff(day, A.OccurTime, B.OccurTime) = 0

- 或oracle ...

 Select Distinct A.Identifer  
    From Table A
      Join Table B 
        On A.PrimKey <> B.PrimaryKey
           And A.Identifer = B.Identifer
           -- EDIT to fix case where 2 at same time
           And A.OccurTime >= B.OccurTime  
           And A.OccurTime < B.OccurTime + .5
 Where Trunc(A.OccurTime) = Trunc(B.OccurTime)

Answer 2

Select    
  A.Id
From    
  YourTable A 
Where    
  A.YourDateTime Between :StartDateTime and :EndDateTime 
Group By    
  A.Id
Having    
  COUNT(A.Id) = 2

Answer 3

我没有检查威廉的查询，但我会认真考虑使用他所拥有的东西。分析是炸弹。每当你发现自己加入一个表回到自身时，几乎可以肯定是一个使用分析的机会，并且会在每次引用两次的表中执行查询。

您将惊讶分析解决方案的速度会快多少。

Answer 4

SELECT identifier
  FROM table_name outer
 WHERE EXISTS( SELECT 1
                 FROM table_name inner
                WHERE inner.identifier  = outer.identifier
                  AND inner.date_column BETWEEN outer.date_column AND outer.date_column + interval '12' hour
                  AND inner.rowid != outer.rowid )

Answer 5

我不是100％确定您的要求，但这可能会为您提供有关如何做您需要的一些想法。例如，你说的确是2;如果有3次出现怎么办？等

create table t (ident varchar2(16), occurance timestamp);

insert into t (ident, occurance) values ('a', to_date('20080101000000', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('b', to_date('20080101010000', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('a', to_date('20080101115900', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('c', to_date('20080102000000', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('d', to_date('20080102010000', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('d', to_date('20080102200000', 'yyyymmddhh24miss'));

insert into t (ident, occurance) values ('d', to_date('20080103020000', 'yyyymmddhh24miss'));

select ident, occurance
from
(
select ident, occurance,
    lag(occurance) over (partition by ident order by occurance) previous, 
    lead(occurance) over (partition by ident order by occurance) next 
from t
)
where 
    ((occurance-previous<interval'12:00' hour to minute and extract(day from occurance) = extract(day from previous))
    or (next-occurance<interval'12:00' hour to minute and extract(day from occurance) = extract(day from next)))
/

Answer 6

SELECT namecol FROM tbl A
WHERE EXISTS (
  SELECT 1 from tbl B
  WHERE b.namecol = a.namecol
  AND b.timestamp > a.timestamp
  AND b.timestamp - 0.5 <= a.timestamp )

SQL查询 - 在12小时内选择重复项

6 个答案: