- 一套Oracle SQL练习题及答案
- 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'; *********************************
作者 solookin
|