选择具有相同值的上次更新的行

Question

我们有这个表，不断插入 Inrunning 字段的值，这是一个数字。有多个错误地插入。我想获取第一个插入发生重复值的行。在这种情况下，5000 值在几分钟内插入了 3 次。所以我想获取第一个插入的值，即第 3 行。

Answer 1

你可以在这里使用analytical函数如下：

Select * from
(Select t.*, 
       Sum(case when inrunningdelay_1 = leadval then 1 end)
           over (partition by writer order by wrriten_1) as sm
 from
(Select t.*,
        lead(inrunningdelay_1) over (partition by writer order by wrriten_1) as leadval
from your_table t) t)
Where sm = 1;

干杯！！

Answer 2

如果您使用Oracle DB 12c+，那么您可以通过FETCH 子句与1 ROW ONLY 选项以及ORDER BY 子句一起使用而无需使用额外的子查询，包括 LEAD()解析函数为

LEAD(inrunningdelay_1) OVER ( PARTITION BY writer ORDER BY written_1 )

从值zero 开始

ABS( inrunningdelay_1 - LEAD(inrunningdelay_1) OVER ( PARTITION BY writer ORDER BY written_1 ) )

通过使用差值的绝对值

SELECT t.*
  FROM t   
 ORDER BY ABS( inrunningdelay_1 - LEAD(inrunningdelay_1) 
                                  OVER ( ORDER BY written_1 ) )
 FETCH FIRST 1 ROW ONLY 

Demo

更新：PARTITION BY writer 在上述情况下是多余的，因为这仅适用于 Tejash 提到的一位作家。对于多个作者，请考虑以下 Select 语句：

WITH t2 AS
(
  SELECT t.*, 
         ABS( inrunningdelay_1 - LEAD(inrunningdelay_1) 
             OVER ( PARTITION BY writer ORDER BY written_1 ) ) as ld            
    FROM t      
   ORDER BY writer, ld, written_1   
)
SELECT written_1, writer, inrunningdelay_1  
  FROM t2  
 ORDER BY ROW_NUMBER() OVER  ( PARTITION BY writer ORDER BY ld )
 FETCH FIRST 1 ROW WITH TIES

您将只过滤掉 zero 返回的值，然后选择仅第一个返回的每个作者通过ROW_NUMBER() 解析函数在最深的ORDER BY 子句中的贡献：

Demo 2