Oracle学习笔记8
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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