Oracle学习<五>

求部门平均薪水的等级

• A.求部门平均薪水的等级。

select deptno,avg_sal,grade from
(select deptno,avg(sal) avg_sal from emp group by deptno)t
join salgrade s on(t.avg_sal between s.losal and s.hisal)

• B.求部门平均的薪水等级

select deptno,avg(grade) from
(select deptno,ename, grade from emp join salgrade s on(emp.sal between s.losal and
s.hisal)) t
group by deptno

• C.那些人是经理

select ename from emp where empno in(select mgr from emp);
select ename from emp where empno in(select distinct mgr from emp);

• D.不准用组函数，求薪水的最高值（面试题）

select distinct sal from emp where sal not in(
select distinct e1.sal from emp e1 join emp e2 on (e1.sal<e2.sal));

• E.平均薪水最高的部门编号

select deptno,avg_sal from
(select avg(sal)avg_sal,deptno from emp group by deptno)
where avg_sal=
(select max(avg_sal)from
(select avg(sal) avg_sal,deptno from emp group by deptno)
)

• F.平均薪水最高的部门名称

select dname from dept where deptno=
(
select deptno from
(select avg(sal)avg_sal,deptno from emp group by deptno)
where avg_sal=
(select max(avg_sal)from
(select avg(sal) avg_sal,deptno from emp group by deptno)
)
)

• G.求平均薪水的等级最低的部门的部门名称

组函数嵌套
如：平均薪水最高的部门编号，可以E.更简单的方法如下：
select deptno,avg_sal from
(select avg(sal) avg_sal,deptno from emp group by deptno)
where avg_sal =
(select max(avg(sal)) from emp group by deptno)

 

组函数最多嵌套两层

分析：
首先求
1.平均薪水： select avg(sal) from group by deptno;

2.平均薪水等级： 把平均薪水当做一张表，需要和另外一张表连接salgrade
select deptno,grade avg_sal from
( select deptno,avg(sal) avg_sal from emp group by deptno) t
join salgrade s on(t.avg_sal between s.losal and s.hisal)

上面结果又可当成一张表。

DEPTNO GRADE AVG_SAL
-------- ------- ----------
30 3 1566.66667
20 4 2175
10 4 2916.66667

3.求上表平均等级最低值

select min(grade) from
(
select deptno,grade,avg_sal from
(select deptno,avg(sal) avg_sal from emp group by deptno)t
join salgrade s on(t.avg_sal between s.losal and s.hisa)
)

4.把最低值对应的2结果的那张表的对应那张表的deptno, 然后把2对应的表和另外一张表做连接。

select dname ,deptno,grade,avg_sal from
(
select deptno,grade,avg_sal from
(select deptno,avg(sal) avg_sal from emp group by deptno)t
join salgrade s on(t.avg_sal between s.losal and s.hisal)
) t1
join dept on (t1.deptno = dept.deptno)
where t1.grade =
(
select deptno,grade,avg_sal from
(select deptno,avg(sal) avg_sal from emp group by deptno) t
join salgrade s on(t.avg_sal between s.losal and s.hisal)
)
)
结果如下：

DNAME DEPTNO GRADE AVG_SAL
-------- ------- -------- --------
SALES 30 3 1566.6667

• H: 视图（视图就是一张表，一个字查询）

G中语句有重复，可以用视图来简化。
conn sys/bjsxt as sysdba;
grant create table,create view to scott;
conn scott/tiger
创建视图：
create view v$_dept_avg-sal_info as
select deptno,grade,avg_sal from
( select deptno,avg(sal) avg_sal from emp group by deptno)t
join salgrade s on 9t.avg_sal between s.losal and s.hisal)

然后
select * from v$_dept_avg-sal_info

结果如下：
DEPTNO GRADE AVG_SAL
-------- ------- ----------
30 3 1566.66667
20 4 2175
10 4 2916.66667

然后G中查询可以简化成：
select dname,t1.deptno,grade,avg_sal from
v$_dept_avg-sal_info t1
join dept on9t1.deptno =dept.deptno)
where t1.grade=
(
select min(grade) from v$_dept_avg-sal_info t1
)