oracle练习题和答案 笔记2

 oracle练习题和答案 笔记2
使用scott/tiger用户下的emp表和dept表完成下列练习，表的结构说明如下

emp员工表(empno员工号/ename员工姓名/job工作/mgr上级编号/hiredate受雇日期/sal薪金/comm佣金/deptno部门编号)

dept部门表(deptno部门编号/dname部门名称/loc地点)

工资 ＝ 薪金 ＋ 佣金

------1．列出至少有一个员工的所有部门

select * from dept
where deptno in
(select deptno from emp group by deptno having count(*)>1);

------2．列出薪金比“SMITH”多的所有员工。
select * from emp
where sal>(select sal from emp where ename='SMITH');

------3．列出所有员工的姓名及其直接上级的姓名。

select ename,(select ename from emp where empno=a.mgr) from emp a;
或
 select ename, (select ename from emp where empno=a.mgr) as mgrname from emp a;

自连接
select a.ename,b.ename from emp a,emp b
where a.mgr=b.empno(+);外连接

 
------4．列出受雇日期晚于其直接上级的所有员工。

select ename from emp a
where hiredate>(select hiredate from emp where empno=a.mgr);

列出受雇日期早于其直接上级的所有员工。
select ename from emp a where
hiredate<(select hiredate from emp where empno=a.mgr);

------5．列出部门名称和这些部门的员工信息，同时列出那些没有员工的部门。

select dname,ename from dept left outer join emp
on dept.deptno=emp.deptno;

select dname,ename from dept a,emp b
where a.deptno = b.deptno(+);

------6．列出所有“CLERK”（办事员）的姓名及其部门名称。
select dname,ename from dept a,emp b
where a.deptno=b.deptno and job='CLERK';

select (select dname from dept where deptno=a.deptno) as dname ,ename
from emp a
where job='CLERK';

------7．列出最低薪金大于1500的各种工作。

select job,min(sal) msal from emp
group by job having min(sal)>1500;

------8．列出在部门“SALES”（销售部）工作的员工的姓名，假定不知道销售部的部门编号。

select ename from emp where deptno=(select deptno from dept where dname='SALES');

------9．列出薪金高于公司平均薪金的所有员工。

select ename from emp where sal>(select avg(sal) from emp);

------10．列出与“SCOTT”从事相同工作的所有员工。

select * from emp where job=(select job from emp where ename='SCOTT');

------11．列出薪金等于部门30中员工的薪金的所有员工的姓名和薪金。
select * from emp where sal in
(select sal from emp where deptno=30);
或
select * from emp where sal = any
(select sal from emp where deptno=30);

 

------12．列出薪金高于在部门30工作的所有员工的薪金的员工姓名和薪金。
--最大值>all
select * from emp where sal>all
(select sal from emp where deptno=30);

--最小值select * from emp where sal < all
(select sal from emp where deptno=30);

------13．列出在每个部门工作的员工数量、平均工资和平均服务期限。

select deptno,count(*),
trunc(avg(sal+nvl(comm,0))) avgsal,
to_char(to_date('0001-01-01','yyyy-mm-dd') + avg(sysdate-hiredate)
-366-31,'yy"年"mm"月"dd') avgday
from emp group by deptno;

------14．列出所有员工的姓名、部门名称和工资。

select ename,dname,sal+nvl(comm,0) from emp,dept where emp.deptno=dept.deptno;

------15．列出从事同一种工作但属于不同部门的员工的一种组合。

select emp.ename,emp.job,emp.deptno from emp ,
(select distinct a.job,a.deptno from emp a,emp b
where (a.job=b.job)and(a.deptno!=b.deptno) order by a.job) c
where empno = (select max(empno) from
emp where job = c.job and deptno = c.deptno)
and emp.job=c.job and emp.deptno= c.deptno

select emp.ename,emp.job,emp.deptno from emp ,
(select distinct a.job,a.deptno from emp a,emp b
where (a.job=b.job)and(a.deptno!=b.deptno) order by a.job) c
where empno = (select min(empno) from
emp where job = c.job and deptno = c.deptno)
and emp.job=c.job and emp.deptno= c.deptno

 

select * from (select distinct a.* from emp a,emp b
where a.deptno <> b.deptno and
a.job = b.job) c
where empno in (select min(empno)
from (select distinct a.* from emp a,emp b
where a.deptno <> b.deptno and
a.job = b.job) group by deptno,job );

------16．列出所有部门的详细信息和部门人数。
select a.*,(select count(*) from emp where deptno=a.deptno) tot
from dept a ;

------17．列出各种工作的最低工资。

select job,min(sal+nvl(comm,0)) from emp group by job;

这样写有问题
select job,min(nvl(sal+comm,0)) from emp group by job;

------18．列出MANAGER（经理）的最低薪金。

select min(sal) from emp where job='MANAGER' ;

------19．列出所有员工的年工资,按年薪从低到高排序。

select ename,(sal+nvl(comm,0))*12 tot from emp order by tot;

-----orcle的等连接
SELECT d.* FROM EMP E ,DEPT D WHERE E.DEPTNO=D.DEPTNO;
-----orcla的外连接
SELECT d.* FROM EMP E ,DEPT D WHERE E.DEPTNO(+)=D.DEPTNO;
+放在没有匹配行的表一侧，所以dept表的记录完全显示

--标准等联结
select dept.* from emp join dept
on emp.deptno = dept.deptno;

--标准的右外联结
select dept.* from emp right outer join dept
on emp.deptno = dept.deptno;

select dept.* from dept left outer join emp
on emp.deptno = dept.deptno;