一． 常见问题

1) 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);

2) 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.lowsal and s.highsal);

3) 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. lowsal and s.highsal)) t

Group by deptno;

4) 求雇员中有那些人是经理人

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

5) 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));

6) 求平均薪水最高的部门的部门编号

Select deptno from (select avg(sal) avgsal, deptno from emp group by deptno)

Where avgsal = (select max(avgsal) from (select avg(sal) avgsal, deptno from emp group by deptno));

7) 求平均薪水最高的部门名称

Select dname from dept where deptno =

(

Select deptno from (select avg(sal) avgsal, deptno from emp group by deptno)

Where avgsal = (select max(avgsal) from

(select avg(sal) avgsal, deptno from emp group by deptno)

);

);

8) 求平均薪水的等级最低的部门的部门名称

select deptname

from (

select deptno, grade

from (

select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade s on

(t.avg_sal between s.lowsal and s.highsal)

) t1 join dept d on (t1.deptno = d.deptno);

where t1.grade =

(

select min(grade)

from (

select deptno, grade

from (select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade s on

(t.avg_sal between s.lowsal and s.highsal)

)

);

9) 部门经理人中平均薪水最低的部门名称

Create view v_mgr_avgsal_by_dept as

Select avg(salary) avg_sal, deptno

from emp

where empno in (select distinct mgr from emp)

group by deptno;

select dname

from v_mgr_avgsal_by_dept t join dept d on (t.deptno = d.deptno)

where t.avg_sal =

(

Select min(avg_sal) from v_mgr_avgsal_by_dept

);

10) 求比普通员工的最高薪水还要高的经理人名称

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)

);

11) 求薪水最高的前五名雇员

Select ename, sal

From (select ename, sal from emp order by sal desc) where rownum<=5

12) 求薪水最高的第 6 到第 10 名雇员

方法一：

select salary

from (

select salary, rownum r

from

(select salary from employees order by salary desc)

)

where r between 6 and 10

方法二：

select salary from

(

select salary, rownum r from

(select salary from employees order by salary desc)

where rownum <= 10

)

where r between 6 and 10

13) 比较效率

Select * from emp where deptno = 10 and ename like ‘%A%’; // 理论上效率稍高

Select * from emp where ename like ‘%A%’ and deptno = 10; // 因为判断数字比字符快

14) 有三个表 S ， C ， SC

S ( SNO ， SNAME )代表(学号，姓名)

C ( CNO ， CNAME ， CTEACHER )代表(课程号，课名，老师)

SC ( SNO ， CNO ， GRADE )代表(学号，课程号，成绩)

问题：

1， 找出没选过“黎明”老师的所有学生姓名

select sname

from s

where sno not in (

select s.sno

from sc join s on(sc.sno = s.sno) join c on(sc.cno = c.cno)

where c.cteacher = ‘liming’)

2， 列出 2 门以上(含 2 门)不及格学生姓名及平均成绩

select sname

from s

where sno in

(

select sno

from sc

where grade < 60

group by sno

having count(*) >= 2

)

3， 既学过一号课程又学过二号课程所有学生的姓名

select sname

from s

where sno in

(

select s.sno

from sc join s on (sc.sno = s.sno)

where cno = 1 or cno =2

group by s.sno

having count(*) = 2

)

15) 树状结构的存储与展示

Create table article

(

Id number primary key,

Context varchar2(4000), -- 内容

Pid number, -- 父 ID

Isleaf number(1), --0 代表非叶子节点， 1 代表叶子节点

aLevel number(2) -- 深度

);

Insert into article values (1, ‘ 蚂蚁大战大象 ’, 0, 0, 0);

Insert into article values (2, ‘ 大象被打趴了 ’, 1, 0, 1);

Insert into article values (3, ‘ 蚂蚁也不好过 ’, 2, 1, 2);

Insert into article values (4, ‘ 瞎说 ’, 2, 0, 2);

Insert into article values (5, ‘ 没有瞎说 ’, 4, 1, 3);

Insert into article values (6, ‘ 怎么可能 ’, 1, 0, 1);

Insert into article values (7, ‘ 怎么没有可能 ’, 6, 1, 2);

Insert into article values (8, ‘ 可能性是很大的 ’, 6, 1, 2);

Insert into article values (9, ‘ 大象进医院了 ’, 2, 0, 2);

Insert into article values (10, ‘ 护士是蚂蚁 ’, 9, 1, 3);

Commit;

-- 显示树状结构

create or replace procedure p

(v_pid article.pid%type, v_level binary_integer)

is

cursor c is select * from article where pid = v_pid;

v_split varchar2(1024) := '';

begin

for i in 1..v_level loop

v_split := v_split || '*****';

end loop;

for v_article in c loop

dbms_output.put_line(v_split || v_article.context);

if (v_article.isleaf = 0) then

p(v_article.id, v_level + 1);

end if;

end loop;

end;

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