Oracle练习题习题答案张表题.docx

Oracle练习题+习题答案(张表+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);create table student(insert into student values (s003,吴鹏,25,男);sno varchar2(10) primary key,insert into student values (s004,琴沁,20,女);sname varchar2(20),insert into student values (s005,王丽,20,女);sage number(2),insert into student values (s006,李波,21,男);ssex varchar2(5)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);create table teacher(insert into teacher values (t002, 谌燕);tno varchar2(10) primary key,insert into teacher values (t003, 胡明星);tname varchar2(20)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);create table course(insert into course values (c004,Oracle,t001);cno varchar2(10),insert into course values (c005,SQL SERVER 2005,t003);cname varchar2(20),insert into course values (c006,C#,t003);tno varchar2(20),insert into course values (c007,JavaScript,t002);constraint pk_course primary key (cno,tno)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);create table sc(insert into sc values (s003,c001,81.9);sno varchar2(10),insert into sc values (s004,c001,60.9);cno varchar2(10),insert into sc values (s001,c002,82.9);score number(4,2),insert into sc values (s002,c002,72.9);constraint pk_sc primary key (sno,cno)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, <6021、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) bwhere a.sno=b.sno and a.score > b.score;*select * from sc awhere a.cno=c001and exists(select * from sc b where b.cno=c002 and a.score>b.scoreand 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 awhere a.snonot in(select distinct s.snofrom sc s,(select c.*from course c ,(select tnofrom teacher twhere tname=谌燕)twhere c.tno=t.tno) bwhere 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.cnojoin teacher t on c.tno=t.tno where tname=谌燕)*6.*select st.* from sc ajoin sc b on a.sno=b.snojoin student ston st.sno=a.snowhere 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.snojoin course c on s.cno=c.cnojoin teacher t on c.tno=t.tnowhere t.tname=谌燕*8.*select * from student stjoin sc a on st.sno=a.snojoin sc b on st.sno=b.snowhere a.cno=c002 and b.cno=c001 and a.score < b.score*9.*select st.*,s.score from student stjoin sc s on st.sno=s.snojoin course c on s.cno=c.cnowhere s.score <60*10.*select stu.sno,stu.sname,count(sc.cno) from student stuleft join sc on stu.sno=sc.snogroup by stu.sno,stu.sna mehaving 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 stucross join course cminusselect 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) bwhere a.cno=b.cno) hwhere st.sno=h.sno and st.sno<>s001*12.*select * from scleft join student ston st.sno=sc.snowhere sc.sno<>s001and sc.cno in(select cno from scwhere sno=s001)*13.*update sc c set score=(select avg(c.score) from course a,teacher bwhere a.tno=b.tnoand b.tname=谌燕and a.cno=c.cnogroup by c.cno)where cno in(select cno from course a,teacher bwhere a.tno=b.tnoand b.tname=谌燕)*14.*select* from sc where sno<>s001minus(select* from scminusselect * from sc where sno=s001)*15.*delete from scwhere sc.cno in(select cno from course cleft join teacher t on c.tno=t.tnowhere 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,scwhere 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 cnoorder 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 twhere sc.cno=c.cno and c.tno=t.tnogroup by c.cnoorder 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 cwhere sc.cno=c.cnogroup 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 stleft join scon sc.sno=st.snogroup 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 stleft join scon sc.sno=st.snogroup by st.sno,st.sname having avg(score)>85;*30.*select sname,score from student st,sc,course cwhere 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 cwhere sc.sno=st.sno and sc.cno=c.cno;*32.*select st.sname,c.cname,sc.score from student st,sc,course cwhere sc.sno=st.sno and sc.cno=c.cno and sc.score>70*33.*select sc.sno,c.cname,sc.score from sc,course cwhere sc.cno=c.cno and sc.score<60 order by sc.cno desc;*34.*select st.sno,st.sname,sc.score from sc,student stwhere 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 twherest.sno=sc.sno and sc.cno=c.cno and c.tno=t.tnoand 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 cnohaving count(sno)>10order 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 ,scwhere sc.cno=c.cno|select cno,cname from course cwhere c.cno in(select cno from sc group by cno)*43.*select st.sname from student stwhere st.sno not in(select distinct sc.sno from sc,course c,teacher twhere sc.cno=c.cno and c.tno=t.tno and t.tname=谌燕)*44.*select sno,avg(score)from scwhere sno in(select sno from sc where sc.score<60group 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;*