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以下的文章,主要为大家在实际工作中提供一种解决方法。
---用户名:scott
---密 码:tiger
---*********Oracle表连接与子查询示例************
---求部门中哪些人的薪水最高
select ename,sal from emp
join (select max(sal) max_sal, deptno from emp group by deptno) t
on (emp.sal = t.max_sal and emp.deptno = t.deptno);
---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);
---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;
---雇员中哪些人是经理人
select ename from emp where empno in (select distinct mgr from emp);
---不用组函数,求薪水的最高值
select sal from emp where sal not in
(select distinct e1.sal from emp e1 join emp e2 on (e1.sal < e2.sal));
---用组函数,求薪水的最高值
select max(sal) from emp;
---求平均薪水最高的部门的部门编号
select deptno , avg_sal from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal =
(select max(avg_sal) from (select avg(sal) avg_sal,deptno from emp group by deptno) t);
----组函数嵌套的写法
select deptno , avg_sal from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal =
(select max(avg(sal)) from emp group by deptno);
---求平均薪水最高的部门的名称
select dname from dept
where deptno =
(
select deptno from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal =
(select max(avg_sal) from (select avg(sal) avg_sal,deptno from emp group by deptno) t)
);
---求平均薪水的等级最低的部门的部门名称
- select avg(sal) avg_sal,deptno from emp group by deptno
--部门平均薪水
select min(avg_sal) from
(
select avg(sal) avg_sal,deptno from emp group by deptno
)
--平均工资的最小值
select avg_sal,deptno from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal =
(
select min(avg_sal) from
(
select avg(sal) avg_sal,deptno from emp group by deptno
)
)
--平均工资的最小值及部门编号
select t.avg_sal,t.deptno,s.grade from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
join salgrade s on (t.avg_sal between s.losal and s.hisal)
where avg_sal =
(
select min(avg_sal) from
(
select avg(sal) avg_sal,deptno from emp group by deptno
)
)
--平均工资的最小值及部门编号和工资等级
select d.dname,t.avg_sal,t.deptno,s.grade from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
join salgrade s on (t.avg_sal between s.losal and s.hisal)
join dept d on (t.deptno = d.deptno)
where avg_sal =
(
select min(avg_sal) from
(
select avg(sal) avg_sal,deptno from emp group by deptno
)
)
--平均工资的最小值及部门编号和工资等级及部门名称
----Another 按照题意的写法
select t1.deptno,t1.avg_sal,grade,d.dname from
(
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)
) t1
join dept d on (t1.deptno = d.deptno)
where grade =
(
select min(grade) from
(
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)
)
);
---创建视图或者表,如果没有权限
--已连接。
- grant create table, create view to scott;
--授权成功。
---创建视图
create view v$_dept_avg_sal_info as
select deptno,avg_sal,grade from
(select deptno,avg(sal) avg_sal from emp group by deptno) t
join salgrade s on (t.avg_sal between s.losal and s.hisal);
--视图已建立。
---创建这个v$_dept_avg_sal_info视图可以简化上面那个查询的重复代码
select t1.deptno,t1.avg_sal,grade,d.dname from
v$_dept_avg_sal_info t1
join dept d on (t1.deptno = d.deptno)
where grade =
(
select min(grade) from
v$_dept_avg_sal_info
);
---求比普通员工的最高薪水还要高的经理的名称
select max(sal) from emp where empno not in
(select distinct mgr from emp where mgr is not null);
--普通员工的最高薪水
select ename from emp
where empno in (select distinct mgr from emp where mgr is not null)
and sal >
(
select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null)
);
--普通员工的最高薪水还要高的经理的名称
--- Oracle 联机归档日志 备份方式
---求薪水最高的第6名到第10名雇员(rownum)
select ename,sal from
(select ename,sal,rownum r from
(
select ename, sal from emp order by sal desc
)
) where r>=6 and r<=10;
---五种约束条件
create table stu
(
id number(2),
name varchar2(20) constraint stu_name_nn not null,--非空约束
sex number(2),
age number(3),
sdate date,
grade number(3) default 1,
class number(3),
email varchar2(50),
constraint stu_name_email_uin unique(name,email)--唯一主键
) ;
insert into stu(name,email) values('','tianyuexing@163.com')
--ORA-01400: 无法将 NULL 插入 ("SCOTT"."STU"."NAME")
insert into stu(name,email) values('tianyuexing','tianyuexing@163.com');
insert into stu(name,email) values('tianyuexing','tianyuexing@163.com');
--ORA-00001: 违反唯一约束条件 (SCOTT.STU_NAME_EMAIL_UIN) |
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