Oracle_SQL_练习题及答案(32页).doc

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1、-本文由NiCoBee贡献 Oracle SQL 练习题注：这些查询题目涵盖了各种查询的技术，请大家认真做，做好了的查询语句之后，有兴趣的同学可以再创建相应的视图和存储过程create table wkj_student(sno varchar2(10) primary key,sname varchar2(20),sage number(2),ssex varchar2(5);create table wkj_teacher(tno varchar2(10) primary key,tname varchar2(20);create table wkj_course(cno varchar2

2、(10),cname varchar2(20),tno varchar2(20),constraint pk_course primary key (cno,tno);create table wkj_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 i

3、nto 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 s

4、tudent 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,

5、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 c

6、ourse 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 va

7、lues (s003,c002,81.9);insert into sc values (s001,c003,59);insert into sc values (s002,c003,58);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

8、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

9、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, b.score;*select * from sc awhere o=c001and exists(select * from sc b where o=c002 and a.scoreb.scoreand a.

10、sno = b.sno)*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;*3. 3、查询所有同学的学号、姓名、选课数、总成绩；*select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno错错误示范 select st.*,count(o),sum(sc.score)from student st,sc

11、where st.sno=sc.snogroup by st.sno select count(o) sum (sc.score) 不是单数分组，要先group，不然和 select stu.sno关联不起来*4. 4、查询姓“刘”的老师的个数；*select count(*) from teacher where tname like 刘%;*5. 5、查询没学过“谌燕”老师课的同学的学号、姓名；*select a.sno,a.sname from student awhere a.sno not in (select distinct s.sno from sc s,(select c.*

12、 from course c ,(select tno from teacher t where tname=谌燕)t where c.tno=t.tno) bwhere o = o ) *select * from student st where st.sno not in(select distinct sno from sc s join course c on o=ojoin teacher t on c.tno=t.tno where tname=谌燕)*6. 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 st.* from studen

13、t stjoin sc a on st.sno=a.snojoin sc b on a.sno=b.snowhere o=c001 and o=c002 ;*7. 7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名；*select st.* from student st join sc s on st.sno=s.snojoin course c on o=ojoin teacher t on c.tno=t.tnowhere t.tname=谌燕*8. 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；*select * from student s

14、tjoin sc a on st.sno=a.snojoin sc b on st.sno=b.snowhere o=c002 and o=c001 and a.score b.score*9. 9、查询所有课程成绩小于60 分的同学的学号、姓名*select st.*,s.score from student stjoin sc s on st.sno=s.snojoin course c on o=owhere s.score 60*10. 查询没有学全所有课的同学的学号、姓名*select stu.sno,stu.sname,count(o) from student stuleft j

15、oin sc on stu.sno=sc.snogroup by stu.sno,stu.snamehaving count(o)(select count(distinct cno)from course)=select * from student where sno in(select sno from (select stu.sno,o from student stucross join course cminusselect sno,cno from sc)=*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*select st.* from stude

16、nt st,(select distinct a.sno from (select * from sc) a,(select * from sc where sc.sno=s001) bwhere o=o) hwhere st.sno=h.sno and st.snos001*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；*select * from scleft join student ston st.sno=sc.sno where sc.snos001and o in(select cno from scwhere sno=s001)*13. 把“SC”表中

17、“谌燕”老师教的课的成绩都更改为此课程的平均成绩*update sc c set score=(select avg(c.score) from course a,teacher bwhere a.tno=b.tnoand b.tname=谌燕and o=ogroup by o)where cno in(select cno from course a,teacher bwhere a.tno=b.tnoand b.tname=谌燕)*不会做14.(原来是错的我改过来了) 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；*dele

18、te from scwhere o in(select cno from course cleft join teacher t on c.tno=t.tnowhere t.tname=谌燕 )*16. 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；*insert into sc (sno,cno,score)select distinct st.sno,o,(select avg(score)from sc where cno=c002) from student st,scwhere not exists(se

19、lect * from sc where cno=c002 and sc.sno=st.sno) and o=c002;*17. 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*select cno ,max(score),min(score) from sc group by cno;*18. 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*select cno,avg(score),round(sum(case when score=60 then 1 else 0 end)/count(*),2)*100|% as 及格率from sc group

20、by cnoorder by avg(score) , 及格率 desc*19. 19、查询不同老师所教不同课程平均分从高到低显示*select max(t.tno),max(t.tname),max(o),max(ame),o,avg(score) from sc , course c,teacher t where o=o and c.tno=t.tnogroup by oorder by avg(score) desc*20. 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*答案是错的select o,ame,sum(case wh

21、en score between 100 and 85 then 1 else 0 end) AS 100-85,sum(case when score between 85 and 70 then 1 else 0 end) AS 85-70,sum(case when score between 70 and 60 then 1 else 0 end) AS 70-60,sum(case when score 60 then 1 else 0 end) AS 85 and score70 and score60 andthen 1 else 0 end) 60到70,sum(case wh

22、en score 60 then 1 else 0 end) 60以下from sc, course cwhere o=o group by o ,ame;*21. 查询各科成绩前三名的记录:(不考虑成绩并列情况)*select * from (select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)where rn1;*27. 27、1981 年出生的学生名单(注：Student 表中Sage 列的类型是number)*select sno,sname,sage,ssex f

23、rom student t where to_char(sysdate,yyyy)-sage =1988*28. 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*select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;*29. 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*select st.sno,st.sname,avg(score) from student stleft join sc on sc.sno=st.snogroup by s

24、t.sno,st.sname having avg(score)85;*30. 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*select sname,score from student st,sc,course cwhere st.sno=sc.sno and o=o and ame=Oracle and sc.score70*33. 33、查询不及格的课程，并按课程号从大到小排列*select sc.sno,ame,sc.score from sc,course cwhere o=o and sc.score80;*35. 35、求选了课程的学生人数*select c

25、ount(distinct sno) from sc;*y有问题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*select st.sname,score from student st,sc ,course c,teacher twherest.sno=sc.sno and o=o and c.tno=t.tnoand t.tname=谌燕 and sc.score=(select max(score)from sc where o=o)*37. 37、查询各个课程及相应的选修人数*select cno,count(sno) from sc group by cno;

26、*38. 38、查询不同课程成绩相同的学生的学号、课程号、学生成绩*select a.* from sc a ,sc b where a.score=b.score and oo*39. 39、查询每门功课成绩最好的前两名*select * from (select sno,cno,score,row_number()over(partition by cno order by score desc) my_rn from sc t)where my_rn10order by count(sno) desc,cno asc;*41. 1、检索至少选修两门课程的学生学号*select sno f

27、rom sc group by sno having count(cno)1; *（错的）42、查询全部学生都选修的课程的课程号和课程名*select sno, count (cno) from scgroup by sno having count (cno) = (select count (o) from course c)*43. 43、查询没学过“谌燕”老师讲授的任一门课程的学生姓名*select st.sname from student stwhere st.sno not in(select distinct sc.sno from sc,course c,teacher tw

28、here o=o and c.tno=t.tno and t.tname=谌燕)*44. 44、查询两门以上不及格课程的同学的学号及其平均成绩*select sno,avg(score)from sc where sno in(select sno from sc where sc.score1) group by sno*45. 45、检索“c004”课程分数小于60，按分数降序排列的同学学号*select sno from sc where cno=c004 and score90 order by score desc;*46. 46、删除“s002”同学的“c001”课程的成绩*delete from sc where sno=s002 and cno=c001;*第 33 页-