问题：要知道两个表或视图中是否有相同的数据（基数和值）。考虑这个视图：
create view V
as
select * from emp where deptno !=10
union all
select * from emp where ename = ‘WARD‘
现要检测这个视图与表emp中的数据是否完全相同。员工“WARD”行重复，说明解决方案不仅要显示不同行，还要显示重复行。
解决方案：
1.首先，查找出表emp中存在而试图v中没有的行
2.然后合并（union all）在视图V中存在，而在表emp中没有的行。
select * from
(
select e.empno,e.ename,e.job,e.mgr,e.hiredate,
e.sal,e.comm,e.deptno,count(*) as cnt
from emp e
group by empno,ename,job,mgr,hiredate,sal,comm,deptno
) e
where not exists
(
select null from
(
select v.empno,v.ename,v.job,v.mgr,v.hiredate,v.sal,v.comm,v.deptno
count (*) as cnt
from v
group by empno,ename,job,mgr,hiredate,sal,comm,deptno
)v
where v.empno = e.empno
and v.ename = e.ename
and v.job = e.job
and v.mgr = e.mgr
and v.hiredate = e.hiredate
and v.sal = e.sal
and v.deptno = e.deptno
and v.cnt = e.cnt
and coalesce(v.comm.0) = coalesce(e.comm,0)
)
union all
select * from
(
select e.empno ,e.ename,e.ename,e.job,e.mgr,e.hiredate,e.sal ,e.comm,e.deptno,count(*) as cnt
from v
group by empno,ename,job,mgr,hiredate,sal,comm,deptno
)v
where not exists (
select null from (
select e.empno,e.ename,e.job,e.mgr,e.hiredate,e.sal,e.comm,e.deptno,count(*) as cnt
from emp e
group by empno,ename,job,mgr,hiredate,sal,comm,deptno
)e
where v.empno = e.empno
and v.ename = e.ename
and v.job = e.job
and v.mgr = e.mgr
and v.hiredate = e.hiredate
and v.sal = e.sal
and v.deptno = e.deptno
and v.cnt = e.cnt
and coalesce(v.comm.0) = coalesce(e.comm,0)
);