一套oracle的练习题

create table student(
sno varchar2(10) 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 student values ('s001','张三',23,'男');
insert into student values ('s002','李四',23,'男');
insert into student values ('s003','吴鹏',25,'男');
insert into student values ('s004','琴沁',20,'女');
insert into student values ('s005','王丽',20,'女');
insert into student values ('s006','李波',21,'男');
insert into student values ('s007','刘玉',21,'男');
insert into student values ('s008','萧蓉',21,'女');
insert into student values ('s009','陈萧晓',23,'女');
insert into student 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;
 
 
练习:
注意:以下练习中的数据是根据初始化到数据库中的数据来写的SQL 语句,请大家务必注意。
 
 
1、查询“c001”课程比“c002”课程成绩高的所有学生的学号;
2、查询平均成绩大于60 分的同学的学号和平均成绩;
3、查询所有同学的学号、姓名、选课数、总成绩;
4、查询姓“刘”的老师的个数;
5、查询没学过“谌燕”老师课的同学的学号、姓名;
6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名;
7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名;
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]
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
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、查询同名同性学生名单,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10 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
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
 
 
 
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
*********************************
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)
where cno in(
select cno from course a,teacher b
where a.tno=b.tno
and b.tname='谌燕')
*********************************
14.
*********************************
select* from sc where sno<>'s001'
minus
(
select* from sc
minus
select * 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) AS "[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, course c
where  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
*********************************
22.
*********************************
select cno,count(sno)from sc group by cno;
*********************************
23.
*********************************
select sc.sno,st.sname,count(cno) from student st
left join sc
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 =1988
*********************************
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)
*********************************
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';

posted @ 2017-04-23 10:31  酸奶加绿茶  阅读(298)  评论(0编辑  收藏  举报