紧急求助数据相减的SQL!!!!

select (select sum(f1.TAKEUP_CREDIT_AMT) from A f1,D t  
where f1.data_dt=t.date_key )-
(select sum(f1.CREDIT_AMT) from B f1,D t  
where f1.data_dt=t.date_key) 差值
from dual

SELECT SUM(f1.TAKEUP_CREDIT_AMT)-SUM(f1.CREDIT_AMT)
FROM A f1,B f2,D t
WHERE t.date_key=f1.data_dt(+)
AND     t.date_key=f2.data_dt(+)

-- 两种方法：其结果应该一样!(因为你是用的内连接：where f1.data_dt=t.date_key)

-- 方法一：每条记录的两个字段相减，然后累加!(即：先相减，再累加)
select sum(f1.TAKEUP_CREDIT_AMT-f1.CREDIT_AMT) as sum1
from A f1,D t 
where f1.data_dt=t.date_key;

-- 方法二：先对两个字段分别求和，再相减!(即：先累加，再相减)
select sum(f1.TAKEUP_CREDIT_AMT)-sum(f1.CREDIT_AMT) as sum2
from A f1,D t 
where f1.data_dt=t.date_key;

--连接查询和之差  
where f1.data_dt=t.date_key 
slect sum(f1.TAKEUP_CREDIT_AMT)-sum(f2.CREDIT_AMT) from A f1
left join D t on f1.data_dt=t.data_key
left join B f2 on f2.data_dt=t.data-key

-- 那可以用中间表D 作为主表，A、B表分别 left join D

with c as (select d.date_key,
                  nvl(a.TAKEUP_CREDIT_AMT,0) as TAKEUP_CREDIT_AMT,
                  nvl(b.CREDIT_AMT,0)        as CREDIT_AMT
           from d left join a on a.date_dt=d.date_key
                  left join b on b.date_dt=d.date_key )
select sum(c.TAKEUP_CREDIT_AMT-c.CREDIT_AMT) as sum1
from c;

with c as (select d.date_key,
                  nvl(a.TAKEUP_CREDIT_AMT,0) as TAKEUP_CREDIT_AMT,
                  nvl(b.CREDIT_AMT,0)        as CREDIT_AMT
           from d left join a on a.date_dt=d.date_key
                  left join b on b.date_dt=d.date_key )
select sum(c.TAKEUP_CREDIT_AMT)-sum(c.CREDIT_AMT) as sum2
from c;