我今天手头没有甲骨文，但这应该能让你接近。

select ( select sum(Records) from my_table where column1 ='1' and column2 = 'abc') -  
(select sum(Records) from my_table where column1 ='2' and column2 = 'adf') 
as Total from dual;

您只需将现有查询的值用作列，并取其差值。

with my_table (column1, column2, column3,column4,  records) as 
    ( select 1, 'aa', 'something', 'abc',  12   from dual union all       
      select 2, 'bb', 'something', 'acd',  25   from dual union all 
      select 2, 'aa', 'something', 'adf',  04   from dual union all 
      select 1, 'bb', 'something', 'abc',  21   from dual
    )   
select r1, r2, r1-r2 diff
  from  (select sum(Records) r1 from my_table where column1 ='1' and column4 = 'abc')
     ,  (select sum(Records) r2 from my_table where column1 ='2' and column4 = 'adf');

注意：为了匹配提供的示例日期，我将查询改为引用column4而不是column2。

使用条件聚合：

select sum(case when column1 = '1' and column2 = 'abc' then Records
                when column1 = '2' and column2 = 'adf' then - Records
                else 0
           end) as diff
from my_table ;

或者，如果要保留已编写的查询，请将它们用作CTE：

with 
a (suma) as 
  (select sum(Records) from my_table where column1 ='1' and column2 = 'abc'),
b (sumb) as
  (select sum(Records) from my_table where column1 ='2' and column2 = 'adf')
select a.suma - b.sumb as difference
from a cross join b