我有接收部门名称和聚合操作(平均值、最大值、最小值)的函数,并将该操作应用于在给定部门工作的员工的薪水并返回结果
create or replace
function salaryData (p_depname varchar2, p_aggr_op varchar2)
return number
is
v_aggrr_op varchar2(20);
v_sal_max number;
v_sal_min number;
v_sal_avg number;
begin
select max(e.salary), min(e.salary), avg(e.salary)
into v_sal_max, v_sal_min, v_sal_avg
from employee e join department d
on e.deptno=d.deptno
where d.deptname=p_depname;
if p_aggr_op in ('max','MAX') then
return v_sal_max;
end if;
if p_aggr_op in ('min','MIN') then
return v_sal_min;
end if;
if p_aggr_op in ('avg','AVG') then
return v_sal_avg;
end if;
end salaryData;
/
当我使用函数调用时
select salaryData('FINANCE','max') as max_sal from employee;
我得到的输出如下:
max_sal|
-------|
20000 |
20000 |
20000 |
20000 |
20000 |
-------
我怎样才能证明这是employeeid,employeename在departmentname工作的最高薪水