ORCAL练习题46道（附答案）

程序员文章站 2022-07-23 12:41:36

create table student1(
sno varchar2(20) primary key,
sname varchar2(20),
sage number(2),
ssex varchar2(5)
);
create table teacher(
tno varchar2(10) primary key,
tname varchar2(20)
);
create table course(
cno varchar2(10),
cname varchar2(20),
tno varchar2(20),
constraint pk_course primary key (cno,tno)
);
create table sc(
sno varchar2(10),
cno varchar2(10),
score number(4,2),
constraint pk_sc primary key (sno,cno)
);
/*******初始化学生表的数据******/
insert into student1 values ('s001','张三',23,'男');
insert into student1 values ('s002','李四',23,'男');
insert into student1 values ('s003','吴鹏',25,'男');
insert into student1 values ('s004','琴沁',20,'女');
insert into student1 values ('s005','王丽',20,'女');
insert into student1 values ('s006','李波',21,'男');
insert into student1 values ('s007','刘玉',21,'男');
insert into student1 values ('s008','萧蓉',21,'女');
insert into student1 values ('s009','陈萧晓',23,'女');
insert into student1 values ('s010','陈美',22,'女');
commit;
/******************初始化教师表***********************/
insert into teacher values ('t001', '刘阳');
insert into teacher values ('t002', '谌燕');
insert into teacher values ('t003', '胡明星');
commit;
/***************初始化课程表****************************/
insert into course values ('c001','J2SE','t002');
insert into course values ('c002','Java Web','t002');
insert into course values ('c003','SSH','t001');
insert into course values ('c004','Oracle','t001');
insert into course values ('c005','SQL SERVER 2005','t003');
insert into course values ('c006','C#','t003');
insert into course values ('c007','JavaScript','t002');
insert into course values ('c008','DIV+CSS','t001');
insert into course values ('c009','PHP','t003');
insert into course values ('c010','EJB3.0','t002');
commit;
/***************初始化成绩表***********************/
insert into sc values ('s001','c001',78.9);
insert into sc values ('s002','c001',80.9);
insert into sc values ('s003','c001',81.9);
insert into sc values ('s004','c001',60.9);
insert into sc values ('s001','c002',82.9);
insert into sc values ('s002','c002',72.9);
insert into sc values ('s003','c002',81.9);
insert into sc values ('s001','c003','59');
commit;

select * from student1;
1、查询“c001”课程比“c002”课程成绩高的所有学生的学号；(每个学生和自己比)
select * from sc s1 join sc s2 on s1.sno=s2.sno
where s1.cno='c001' and s2.cno='c002' and s1.score>s2.score;

select a.sno from (select * from sc where cno='c001')a,
(select * from sc where cno='c002')b
where a.sno=b.sno and a.score>b.score;
2、查询平均成绩大于60 分的同学的学号和平均成绩；
select sno,avg(score) from sc group by sno
having avg(score)>60;

select * from(select AVG(score) as 平均成绩,
sno from sc group by sno) a
where a.平均成绩>60

3、查询所有同学的学号、姓名、选课数、总成绩；
--sno分组每一个学生选多少门课
--cno分组每门课被多少学生选
select st.sname,a.*from student1 st join
(select sno,count(cno),sum(score) from sc group by sno)a
on st.sno = a.sno;

4、查询姓“刘”的老师的个数；
like 模糊查询
_ 匹配一个字符
% 匹配0个或任意个字符
第二个字母为A 倒数第二个字母为E的老师 JAMES
select * from teacher where ename like '_A%E';
--escape 使用指定转义字符

select count(1) from teacher where tname like '刘%'

5、查询没学过“谌燕”老师课的同学的学号、姓名；
select sno,sname from student1 where sno not in
(select distinct(sno) from sc where cno in --所有学过谌燕老师的课的学生的学号（1门或2门）
(select cno from course where tno=
(select tno from teacher where tname='谌燕')))

select * from course;
select * from sc;
select * from student1;
select * from teacher;
6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名；
select sname,sno from student where sno in
(select sno from sc a join sc b on a.sno=b.sno and a.cno='c001' and b.cno='c002');

7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名；
select sno,sname from student1 where cno in
(select count(distinct(cno)) from sc where cno in
(select cno from course where tno=
（select tno from teacher where tname ='谌燕')))--老师上课的数量

8、查询课程编号“c002”的成绩比课程编号“c001”课程低的所有同学的学号、姓名；

9、查询所有课程成绩小于60 分的同学的学号、姓名；

10、查询没有学全所有课的同学的学号、姓名；

11、查询至少有一门课与学号为“s001”的同学所学相同的同学的学号和姓名；

12、查询至少学过学号为“s001”同学所有一门课的其他同学学号和姓名；

13、把“SC”表中“谌燕”老师教的课的成绩都更改为此课程的平均成绩；

14、查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名；

15、删除学习“谌燕”老师课的SC 表记录；

16、向SC 表中插入一些记录，这些记录要求符合以下条件：
没有上过编号“c002”课程的同学学号、“c002”号课的平均成绩；

17、查询各科成绩最高和最低的分：
以如下形式显示：课程ID，最高分，最低分

18、按各科平均成绩从低到高和及格率的百分数从高到低顺序

19、查询不同老师所教不同课程平均分从高到低显示

20、统计列印各科成绩,各分数段人数:课程ID,课程名称,
[100-85],[85-70],[70-60],[ <60]
-- case when

21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
分区函数

22、查询每门课程被选修的学生数

23、查询出只选修了一门课程的全部学生的学号和姓名

24、查询男生、女生人数

25、查询姓“张”的学生名单

26、查询同名学生名单，并统计同名人数

27、1981 年出生的学生名单(注：Student 表中Sage 列的类型是number)

28、查询每门课程的平均成绩，结果按平均成绩升序排列，
平均成绩相同时，按课程号降序排列

29、查询平均成绩大于85 的所有学生的学号、
姓名和平均成绩

30、查询课程名称为“数据库”，
且分数低于60 的学生姓名和分数

31、查询所有学生的选课情况；

32、查询任何一门课程成绩在70 分以上的姓名、课程名称和分数；

33、查询不及格的课程，并按课程号从大到小排列

34、查询课程编号为c001 且课程成绩在80 分以上的学生的学号和姓名；

35、求选了课程的学生人数

36、查询选修“谌燕”老师所授课程的学生中，
成绩最高的学生姓名及其成绩

37、查询各个课程及相应的选修人数

38、查询不同课程成绩相同的学生的学号、课程号、学生成绩

39、查询每门功课成绩最好的前两名

40、统计每门课程的学生选修人数（超过2 人的课程才统计）。
要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，
按课程号升序排列

41、检索至少选修两门课程的学生学号

42、查询全部学生都选修的课程的课程号和课程名
查询被学生选了4次的课程的课程号和课程名

43、查询没学过“谌燕”老师讲授的任一门课程的学生姓名

44、查询两门以上不及格课程的同学的学号及其平均成绩

45、检索“c004”课程分数小于60，
按分数降序排列的同学学号

46、删除“s002”同学的“c001”课程的成绩

-- dual 虚表
select 9999*9999 from emp;
select 500*500 from dual
select sysdate from dual
select ascii('a') from dual;
select upper('abc') from dual;
select lower('ABC') from dual;
select trim(' AB CDEF ') from dual;
select ltrim(' aaaaa') from dual;
select rtrim('aaa ') from dual;
select concat(ename,sal) from emp;
||
select length('朱') from dual
select last_day(sysdate) from dual;

-- 查询当前距离月底还有多少天？
select last_day(sysdate) - sysdate from dual;

-- 常用数据库对象同义词视图序列索引表空间

create user zhu identified by a123;
grant connect,resource to zhu

grant select on emp to zhu
select * from scott.emp
delete from scott.emp
grant all on emp to zhu;

同义词
创建用户上官宇在上官宇用户下创建一个
scott用户的emp表的同义词

CREATE USER sgy IDENTIFIED BY sgy;
GRANT CONNECT,RESOURCE TO sgy;
-- 除了管理员用户之外所有用户想创建
-- 数据库对象必须由管理员进行权限赋予
GRANT CREATE SYNONYM TO sgy;
-- 创建同义词
CREATE SYNONYM MyEmp FOR SCOTT.EMP;

SELECT * FROM MYEMP;
-- 在 emp用户下进行操作
GRANT ALL ON EMP TO sgy;
SELECT ENAME,JOB,SAL FROM MyEmp WHERE SAL>2000;
select * from MyEmp
update MYEMP set ename = 'smith' where ename='SMITH'
SELECT * FROM SCOTT.EMP

DROP SYNONYM MYEMP;

一般来说所有的关键字都是大写
SelECt * fROM EMp

序列
drop sequence sgy
CREATE SEQUENCE sgy_seq
START WITH 1
INCREMENT BY 1
MAXVALUE 20
MINVALUE 1
noCYCLE
NOCACHE

CREATE SEQUENCE sgy_seq
start with 1
increment by 1
maxvalue 20
minvalue 1
nocycle
nocache

currval 序列的当前值
nextval 序列的下一个值
select sgy_seq.currval from dual;
select sgy_seq.nextval from dual;

--创建一个序列
初始值为1 增长间隔为1 没有最大值
没有最小值没有循环
create table shangguanyu(
sid number primary key,
sname varchar2(20)
)
向表中插入五条数据,使用序列生成的值代替
主键进行插入

视图 view
存放的是一段sql语句
-- 在sys用户下
grant create view to scott

create or replace view emp_view
as
select ename,sal,deptno from emp;

select * from emp_view

-- 简化复杂的查询语句
-- 屏蔽部分字段,保护数据安全性
-- 保护基表信息(只给用户查看其想查看和能看的内容)

create or replace view scott_view

select score,cname from sc join course c
on sc.cno = c.cno

CREATE UNIQUE INDEX UQ_ENAME_IDX ON EMP(ENAME);

select * from emp where ename='SMITH'

grant create view to scott

create or replace view zhu
as
select ename,e.deptno,dname from emp e join dept d
on e.deptno=d.deptno
with read only
rollback
update emp set ename='zhu' where job='SALESMAN'
commit;
select * from zhu
1.屏蔽部分基表字段
update zhu set ename='smith' where ename='SMITH'
rollback
select * from emp

select * from all_tables
select * from all_users
update all_tables set owner='zhu' where owner='SCOTT'

-- 什么是索引？
-- 在查询数据的时候数据库真的做了全表遍历吗？
-- 索引是基于字段的
-- emp表为例对ename字段建立了索引
-- 提前按照一定的规则对该字段的数据进行预排序
-- 当用户使用建立了索引的字段进行查询的时候
-- 通过读取数据的rowid 直接定位到指定行
-- 从而提高查询效率

-- ORACLE数据库在建表的时候
-- 会自动为所有的主键和唯一键建立索引
-- (提高查询效率)
-- 在我们建立索引时
-- 重复值越多索引的效率越低

-- 表数据的CRUD对索引是否有影响？有
-- 每当你对表进行增删改查的时候索引
-- 都需要重新创建
-- 数据量越大索引的效率越高

CREATE UNIQUE INDEX UQ_ENAME_IDX ON EMP(ENAME);

select * from emp where ename='SMITH'

-- 索引
create index index1 on emp1(ename)
-- 唯一索引
create unique index index2 on emp1(empno)

select * from emp where ename='sgy'
select * from emp where ename like '张%'
-- gender

-- 主键唯一约束

-- sql优化

select * from ename <> 'sgy'
select * from emp where sal in(2000,3000);
in 和 exists 的使用场景
select * from emp where sal = 2000
union
select * from emp where sal = 3000;
select * from emp where sal/2=3000;
select * from emp where sal = 6000;

答案:

1.
*********************************
select a.* from
(select * from sc a where a.cno='c001') a,
(select * from sc b where b.cno='c002') b
where a.sno=b.sno and a.score > b.score;
*********************************
select * from sc a
where a.cno='c001'
and exists(select * from sc b where b.cno='c002' and a.score>b.score
and a.sno = b.sno)
*********************************
2.
*********************************
select sno,avg(score) from sc group by sno having avg(score)>60;
*********************************
3.
*********************************
select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno
*********************************
4.
*********************************
select count(*) from teacher where tname like '刘%';
*********************************
5.
*********************************
select a.sno,a.sname from student a
where a.sno
not in
(select distinct s.sno
from sc s,
(select c.*
from course c ,
(select tno
from teacher t
where tname='谌燕')t
where c.tno=t.tno) b
where s.cno = b.cno )
*********************************
select * from student st where st.sno not in
(select distinct sno from sc s join course c on s.cno=c.cno
join teacher t on c.tno=t.tno where tname='谌燕')
*********************************
6.
*********************************
select st.* from sc a
join sc b on a.sno=b.sno
join student st
on st.sno=a.sno
where a.cno='c001' and b.cno='c002' and st.sno=a.sno;
*********************************
7.
*********************************
select st.* from student st join sc s on st.sno=s.sno
join course c on s.cno=c.cno
join teacher t on c.tno=t.tno
where t.tname='谌燕'
*********************************
8.
*********************************
select * from student st
join sc a on st.sno=a.sno
join sc b on st.sno=b.sno
where a.cno='c002' and b.cno='c001' and a.score < b.score
*********************************
9.
*********************************
select st.*,s.score from student st
join sc s on st.sno=s.sno
join course c on s.cno=c.cno
where s.score <60

select st.* s.score from student st, course c ,score sc
where st.sno=s.sno and s.cno=c.cno and s.score<60
*********************************
10.
*********************************
select stu.sno,stu.sname,count(sc.cno) from student stu
left join sc on stu.sno=sc.sno
group by stu.sno,stu.sname
having count(sc.cno)<(select count(distinct cno)from course)
===================================
select * from student where sno in
(select sno from
(select stu.sno,c.cno from student stu
cross join course c
minus
select sno,cno from sc)
)
===================================
*********************************
11.
*********************************
select st.* from student st,
(select distinct a.sno from
(select * from sc) a,
(select * from sc where sc.sno='s001') b
where a.cno=b.cno) h
where st.sno=h.sno and st.sno<>'s001'
*********************************
12.
*********************************
select * from sc
left join student st
on st.sno=sc.sno
where sc.sno<>'s001'
and sc.cno in
(select cno from sc
where sno='s001')
*********************************
13.
*********************************
update sc c set score =
(select avg(c.score) from course a,teacher b
where a.tno=b.tno
and b.tname='谌燕' and a.cno=c.cno
group by c.cno)--ZY老师教的平均成绩
where cno in(
select cno from course a,teacher b
where a.tno=b.tno
and b.tname='谌燕')--ZY老师教过的课的课程号
*********************************
14.

select email from Person group by email having
count(1)>1
*********************************
select sno,sname
from student
where sno = (select sc.sno -- 这些学生学过的课一定是s001学的
from sc
where sc.sno not in
(select sno -- 一节s001上过的课都没学过的人
from sc
where sc.cno not in
(select cno from sc where sno = 's001'))--s001学过的课
and sno != s001
group by sc.sno
having count(sc.cno) = (select count(cno) cou -- 学的cno数量相等
from sc
where sno = 's001'));

*********************************
15.
*********************************
delete from sc
where sc.cno in
(
select cno from course c
left join teacher t on c.tno=t.tno
where t.tname='谌燕'
)
*********************************
16.
*********************************
insert into sc (sno,cno,score)
select distinct st.sno,sc.cno,(select avg(score)from sc where cno='c002')
from student st,sc
where not exists
(select * from sc where cno='c002' and sc.sno=st.sno) and sc.cno='c002';
*********************************
17.
*********************************
select cno ,max(score),min(score) from sc group by cno;
*********************************
18.
*********************************

select cno,avg(score),sum(case when score>=60
then 1 else 0 end)/count(*)
as 及格率
from sc group by cno
order by avg(score) , 及格率 desc
*********************************
19.
*********************************
select max(t.tno),max(t.tname),max(c.cno),
max(c.cname),c.cno,avg(score) from sc , course c,teacher t
where sc.cno=c.cno and c.tno=t.tno
group by c.cno
order by avg(score) desc
*********************************
20.
*********************************
select sc.cno,c.cname,
sum(case when score between 85 and 100 then 1 else 0 end) [100-85],
sum(case when score between 70 and 85 then 1 else 0 end) AS "[85-70]",
sum(case when score between 60 and 70 then 1 else 0 end) AS "[70-60]",
sum(case when score <60 then 1 else 0 end) AS "[<60]"
from sc join course c
on sc.cno=c.cno
group by sc.cno ,c.cname;
*********************************
21.
*********************************
select * from
(select sno,cno,score,row_number()over
(partition by cno order by score desc)
rn from sc)
where rn<4
分区是和分组对应的
top-N

over partition by 决定了分区的字段
order by 决定了数据分区之后数据的排列顺序
rn 代表的是每一行数据在其所在分区内的编号
row_number 不考虑并列情况
rank 考虑并列并且后续排名
会受到并列人数的影响
dense_rank 考虑并列后续排名不受并列人数的影响

查询emp表中各个部门工资前二高的人员信息
(考虑并列情况)

*********************************
22.
*********************************
select cno,count(sno)from sc group by cno;
*********************************
23.
*********************************
select sc.sno,st.sname,count(cno) from student st
left join scselect
on sc.sno=st.sno
group by st.sname,sc.sno having count(cno)=1;
*********************************
24.
*********************************
select ssex,count(*)from student group by ssex;
*********************************
25.
*********************************
select * from student where sname like '张%';
*********************************
26.
*********************************
select sname,count(*)from student group by sname having count(*)>1;
*********************************
27.
*********************************
select sno,sname,sage,ssex from student t where to_char(sysdate,'yyyy')-sage =1981
*********************************
28.
*********************************
select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;
****
29.
*********************************
select st.sno,st.sname,avg(score) from student st
left join sc
on sc.sno=st.sno
group by st.sno,st.sname having avg(score)>85;
*********************************
30.
*********************************
select sname,score from student st,sc,course c
where st.sno=sc.sno and sc.cno=c.cno and c.cname='Oracle' and sc.score<60
*********************************
31.
*********************************
select st.sno,st.sname,c.cname from student st,sc,course c
where sc.sno=st.sno and sc.cno=c.cno;
*********************************
32.
*********************************
select st.sname,c.cname,sc.score from student st,sc,course c
where sc.sno=st.sno and sc.cno=c.cno and sc.score>70
*********************************
33.
*********************************
select sc.sno,c.cname,sc.score from sc,course c
where sc.cno=c.cno and sc.score<60 order by sc.cno desc;
*********************************
34.
*********************************
select st.sno,st.sname,sc.score from sc,student st
where sc.sno=st.sno and cno='c001' and score>80;
*********************************
35.
*********************************
select count(distinct sno) from sc;
*********************************
36.
*********************************
select st.sname,score from student st,sc ,course c,teacher t
where
st.sno=sc.sno and sc.cno=c.cno and c.tno=t.tno
and t.tname='谌燕' and sc.score=
(select max(score)from sc where sc.cno=c.cno)

create table a(
aid number,
aname char
);
insert into a values(1,'A');
insert into a values(2,'B');
create table b(
bid number,
bname char
);
insert into b values(1,'C');
insert into b values(2,'D');
insert into b values(3,'E');
查询表b中所有 bid和aid相同的数据信息
-- 不相关子查询
select * from b where bid in (select aid from a);
-- 相关子查询
select * from b where exists(select aid from a where a.aid=b.bid)
*********************************
37.
*********************************
select cno,count(sno) from sc group by cno;
*********************************
38.
*********************************
select a.* from sc a ,sc b where a.score=b.score and a.cno<>b.cno
*********************************
39.
*********************************
select * from (
select sno,cno,score,row_number()over(partition by cno order by score desc) my_rn from sc t
)
where my_rn<=2
*********************************
40.
*********************************
select cno,count(sno) from sc group by cno
having count(sno)>10
order by count(sno) desc,cno asc;
*********************************
41.
*********************************
select sno from sc group by sno having count(cno)>1;
||
select sno from sc group by sno having count(sno)>1;
*********************************
42.
*********************************
select distinct(c.cno),c.cname from course c ,sc
where sc.cno=c.cno
||
select cno,cname from course c
where c.cno in
(select cno from sc group by cno)
*********************************
43.
*********************************
select st.sname from student st
where st.sno not in
(select distinct sc.sno from sc,course c,teacher t
where sc.cno=c.cno and c.tno=t.tno and t.tname='谌燕')
*********************************
44.
*********************************
select sno,avg(score)from sc
where sno in
(select sno from sc where sc.score<60
group by sno having count(sno)>1
) group by sno
*********************************
45.
*********************************
select sno from sc where cno='c004' and score<90 order by score desc;
*********************************
46.
*********************************
delete from sc where sno='s002' and cno='c001';
*********************************

本文地址：https://blog.csdn.net/caotengnet/article/details/107630603

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