oracle一些常见题目

对于嵌套子查询的练习

–1.求部门中薪水最高的人

select * from emp e join (select max(e.sal) max from emp e group by e.deptno) m on m.max = e.sal;

–2.求部门平均薪水的等级

select m.*,sg.grade from (select avg(e.sal) sal,e.deptno from emp e group by e.deptno) m join salgrade sg on m.sal between sg.losal and sg.hisal;

–3.求部门平均的薪水等级

select m.*,sg.grade from (select avg(m.avg) sal from (select avg(e.sal) avg,e.deptno dp from emp e group by e.deptno) m) m join salgrade sg on m.sal between sg.losal and sg.hisal;

–4.雇员中有哪些人是经理人

select * from (select distinct e.mgr no from emp e where e.mgr is not null) m join emp e on e.empno = m.no;

–5.不准用组函数，求薪水的最高值

select e1.* from emp e1 where e1.sal >= all (select e.sal sal from emp e);

–6.求平均薪水最高的部门的部门编号

select e.dp from (select max(m.avg) max from (select avg(e.sal) avg from emp e group by e.deptno) m) m join (select avg(e.sal) avg,e.deptno dp from emp e group by e.deptno) e on e.avg = m.max;

–组函数嵌套写法(对多可以嵌套一次，group by 只对内层函数有效)

–7.求平均薪水最高的部门的部门名称

select d.dname from (select e.dp dp from (select max(m.avg) max from (select avg(e.sal) avg from emp e group by e.deptno) m) m join (select avg(e.sal) avg,e.deptno dp from emp e group by e.deptno) e on e.avg = m.max) m join dept d on d.deptno = m.dp ;

–8.求平均薪水的等级最低的部门的部门名称

select d.dname from (select n.dp dp from (select max(m.avg) max from (select avg(e.sal) avg from emp e group by e.deptno) m) m join (select avg(e.sal) avg,e.deptno dp from emp e group by e.deptno) n on m.max = n.avg) m join dept d on d.deptno = m.dp;

–9.求部门经理人中平均薪水最低的部门名称

select d.dname from (select n.dp dp from (select min(m.avg) min from (select avg(m.sal) avg from (select e.sal sal,e.deptno dp from (select distinct e.mgr no from emp e where e.mgr is not null) m join emp e on e.empno = m.no) m group by m.dp) m) m join (select avg(m.sal) avg,m.deptno dp from (select e.sal sal,e.deptno deptno from (select distinct e.mgr no from emp e where e.mgr is not null) m join emp e on e.empno = m.no) m group by m.deptno) n on m.min = n.avg) m join dept d on d.deptno = m.dp;

–10.求比普通员工的最高薪水还要高的经理人名称(not in)

select e.ename from (select m.no no from (select e.sal sal,e.empno no from (select distinct e.mgr no from emp e where e.mgr is not null) m join emp e on e.empno = m.no) m join (select max(m.sal) max from (select e.sal sal from emp e where e.empno not in (select distinct e.mgr no from emp e where e.mgr is not null)) m) n on m.sal > n.max) m join emp e on e.empno = m.no;

–11.求薪水最高的前5名雇员

select m.* from (select rownum r,m.* from (select * from emp e order by e.sal desc) m) m where m.r <= 5;

–12.求薪水最高的第6到第10名雇员(important)

select m.* from (select rownum r,m.* from (select * from emp e order by e.sal desc) m) m where m.r > 5 and m.r <= 10;

–13.求最后入职的5名员工

select m.* from (select rownum r,m.* from (select * from emp e order by e.hiredate desc) m) m where m.r <= 5;

对于以上习题，肯定还有更好和更简便的实现方式，本人只是练习使用嵌套子查询的使用，个人觉得这种查询很基础。

下面再多写个行转列吧

题目：

建表

create table STUDENT_SCORE ( name VARCHAR2(20),subject VARCHAR2(20),score NUMBER(4,1) )

– 添加数据

insert into student_score (NAME,SUBJECT,SCORE) values ('张三','语文',78.0);
insert into student_score (NAME,'数学',88.0);
insert into student_score (NAME,'英语',98.0);
insert into student_score (NAME,SCORE) values ('李四',89.0);
insert into student_score (NAME,76.0);
insert into student_score (NAME,90.0);
insert into student_score (NAME,SCORE) values ('王五',99.0);
insert into student_score (NAME,66.0);
insert into student_score (NAME,91.0);

– 希望得到下面的结果
– 姓名 语文 数学 英语
– 王五 89 56 89

使用case when then end

select ss.name,max(case 　　when ss.subject = '语文' then 　　ss.score end) 语文,max(case 　　when ss.subject = '数学' then 　　ss.score end) 数学,max(case 　　when ss.subject = '英语' then 　　ss.score end) 英语 from student_score ss group by ss.name;

使用decode

select ss.name,max(decode(ss.subject,ss.score)) 语文,ss.score)) 数学,ss.score)) 英语 from student_score ss group by ss.name;

使用多条子查询

select m1.name,m1.sc,m2.sc,m3.sc from (select ss.name,ss.score sc 　　　 from student_score ss 　　　　where ss.subject = '语文') m1 join (select ss.name,ss.score sc 　　　　from student_score ss 　　　　where ss.subject = '数学') m2 on m1.name = m2.name join (select ss.name,ss.score sc 　　　　from student_score ss 　　　　where ss.subject = '英语') m3 on m2.name = m3.name;

（编辑：李大同）

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