Oracle-SQL-練習題及答案

1、本文由NiCoBee奉獻 Oracle SQL 練習題注：這些查詢題目涵蓋了各種查詢的技術，請大家認真做，做好了的查詢語句之后，有興趣的同學可以再創(chuàng)立相應的視圖和存儲過程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);/*初始化學生表的數(shù)據(jù)*/insert into student values (s001,張三,23,男);insert into student values (s002,李四,23,男);insert in

3、to student values (s003,吳鵬,25,男);insert into student values (s004,琴沁,20,女);insert into student values (s005,王麗,20,女);insert into student values (s006,李波,21,男);insert into student values (s007,劉玉,21,男);insert into student values (s008,蕭蓉,21,女);insert into student values (s009,陳蕭曉,23,女);insert into st

4、udent values (s010,陳美,22,女);commit;/*初始化教師表*/insert into teacher values (t001, 劉陽);insert into teacher values (t002, 諶燕);insert into teacher values (t003, 胡明星);commit;/*初始化課程表*/insert into course values (c001,J2SE,t002);insert into course values (c002,Java Web,t002);insert into course values (c003,S

5、SH,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 co

6、urse values (c010,EJB3.0,t002);commit;/*初始化成績表*/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 val

7、ues (s003,c002,81.9);insert into sc values (s001,c003,59);insert into sc values (s002,c003,58);commit;練習：注意：以下練習中的數(shù)據(jù)是根據(jù)初始化到數(shù)據(jù)庫中的數(shù)據(jù)來寫的SQL 語句，請大家務必注意。1、查詢“c001”課程比“c002”課程成績高的所有學生的學號；2、查詢平均成績大于60 分的同學的學號和平均成績；3、查詢所有同學的學號、姓名、選課數(shù)、總成績；4、查詢姓“劉的老師的個數(shù)；5、查詢沒學過“諶燕老師課的同學的學號、姓名；6、查詢學過“c001”并且也學過編號“c002”課程的同學的學號

8、、姓名；7、查詢學過“諶燕老師所教的所有課的同學的學號、姓名；8、查詢課程編號“c002”的成績比課程編號“c001”課程低的所有同學的學號、姓名；9、查詢所有課程成績小于60 分的同學的學號、姓名；10、查詢沒有學全所有課的同學的學號、姓名；11、查詢至少有一門課與學號為“s001”的同學所學相同的同學的學號和姓名；12、查詢至少學過學號為“s001”同學所有一門課的其他同學學號和姓名；13、把“SC表中“諶燕老師教的課的成績都更改為此課程的平均成績；14、查詢和“s001”號的同學學習的課程完全相同的其他同學學號和姓名；15、刪除學習“諶燕老師課的SC 表記錄；16、向SC 表中插入一些記

9、錄，這些記錄要求符合以下條件：沒有上過編號“c002”課程的同學學號、“c002”號課的平均成績；17、查詢各科成績最高和最低的分：以如下形式顯示：課程ID，最高分，最低分18、按各科平均成績從低到高和及格率的百分數(shù)從高到低順序19、查詢不同老師所教不同課程平均分從高到低顯示20、統(tǒng)計列印各科成績,各分數(shù)段人數(shù):課程ID,課程名稱,100-85,85-70,70-60, b.score;*select * from sc awhere a o=c001and exists(select * from sc b where b o=c002 and a.scoreb.scoreand a.sno

10、 = b.sno)*2. 查詢平均成績大于60 分的同學的學號和平均成績；*select sno,avg(score) from sc group by sno having avg(score)60;*3. 3、查詢所有同學的學號、姓名、選課數(shù)、總成績；*select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno錯錯誤示范 select st.*,count(sc o),sum(sc.score)from student st,sc

11、where st.sno=sc.snogroup by st.sno select count(sc o) sum (sc.score) 不是單數(shù)分組，要先group，不然和 select stu.sno關聯(lián)不起來*4. 4、查詢姓“劉的老師的個數(shù)；*select count(*) from teacher where tname like 劉%;*5. 5、查詢沒學過“諶燕老師課的同學的學號、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=諶燕)t where c.tno=t.tno) bwhere s o = b o ) *select * from student st where st.sno not in(select distinct sno from sc s join course c on s o=c ojoin teacher t on c.tno=t.tno where tname=諶燕)*6. 6、查詢學過“c001”并且也學過編號“c002”課程的同學的學號、姓名；*select st.* fr

13、om student stjoin sc a on st.sno=a.snojoin sc b on a.sno=b.snowhere a o=c001 and b o=c002 ;*7. 7、查詢學過“諶燕老師所教的所有課的同學的學號、姓名；*select st.* from student st join sc s on st.sno=s.snojoin course c on s o=c ojoin teacher t on c.tno=t.tnowhere t.tname=諶燕*8. 8、查詢課程編號“c002”的成績比課程編號“c001”課程低的所有同學的學號、姓名；*select

14、* from student stjoin sc a on st.sno=a.snojoin sc b on st.sno=b.snowhere a o=c002 and b o=c001 and a.score b.score*9. 9、查詢所有課程成績小于60 分的同學的學號、姓名*select st.*,s.score from student stjoin sc s on st.sno=s.snojoin course c on s o=c owhere s.score 60*10. 查詢沒有學全所有課的同學的學號、姓名*select stu.sno,stu.sname,count(s

15、c o) from student stuleft join sc on stu.sno=sc.snogroup by stu.sno,stu.snamehaving count(sc o)(select count(distinct cno)from course)=select * from student where sno in(select sno from (select stu.sno,c o from student stucross join course cminusselect sno,cno from sc)=*11. 查詢至少有一門課與學號為“s001”的同學所學相同

16、的同學的學號和姓名*select st.* from student st,(select distinct a.sno from (select * from sc) a,(select * from sc where sc.sno=s001) bwhere a o=b o) hwhere st.sno=h.sno and st.snos001*12. 查詢至少學過學號為“s001”同學所有一門課的其他同學學號和姓名；*select * from scleft join student ston st.sno=sc.sno where sc.snos001and sc o in(select

17、 cno from scwhere sno=s001)*13. 把“SC表中“諶燕老師教的課的成績都更改為此課程的平均成績*update sc c set score=(select avg(c.score) from course a,teacher bwhere a.tno=b.tnoand b.tname=諶燕and a o=c ogroup by c o)where cno in(select cno from course a,teacher bwhere a.tno=b.tnoand b.tname=諶燕)*不會做14.(原來是錯的我改正來了) 查詢和“s001”號的同學學習的課程

18、完全相同的其他同學學號和姓名*15. 刪除學習“諶燕老師課的SC 表記錄；*delete from scwhere sc o in(select cno from course cleft join teacher t on c.tno=t.tnowhere t.tname=諶燕 )*16. 16、向SC 表中插入一些記錄，這些記錄要求符合以下條件：沒有上過編號“c002”課程的同學學號、“c002”號課的平均成績；*insert into sc (sno,cno,score)select distinct st.sno,sc o,(select avg(score)from sc where

19、 cno=c002) from student st,scwhere not exists(select * from sc where cno=c002 and sc.sno=st.sno) and sc o=c002;*17. 17、查詢各科成績最高和最低的分：以如下形式顯示：課程ID，最高分，最低分*select cno ,max(score),min(score) from sc group by cno;*18. 18、按各科平均成績從低到高和及格率的百分數(shù)從高到低順序*select cno,avg(score),round(sum(case when score=60 then 1

20、 else 0 end)/count(*),2)*100|% as 及格率from sc group by cnoorder by avg(score) , 及格率 desc*19. 19、查詢不同老師所教不同課程平均分從高到低顯示*select max(t.tno),max(t.tname),max(c o),max(c ame),c o,avg(score) from sc , course c,teacher t where sc o=c o and c.tno=t.tnogroup by c oorder by avg(score) desc*20. 20、統(tǒng)計列印各科成績,各分數(shù)段人

21、數(shù):課程ID,課程名稱,100-85,85-70,70-60, 60*答案是錯的select sc o,c ame,sum(case when 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)

22、AS 85 and score70 and score60 andthen 1 else 0 end) 60到70,sum(case when score 60 then 1 else 0 end) 60以下from sc, course cwhere sc o=c o group by sc o ,c ame;*21. 查詢各科成績前三名的記錄:(不考慮成績并列情況)*select * from (select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)where rn1;*

23、27. 27、1981 年出生的學生名單(注：Student 表中Sage 列的類型是number)*select sno,sname,sage,ssex from student t where to_char(sysdate,yyyy)-sage =1988*28. 28、查詢每門課程的平均成績，結果按平均成績升序排列，平均成績相同時，按課程號降序排列*select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;*29. 29、查詢平均成績大于85 的所有學生的學號、姓名和平均成績*select st

24、.sno,st.sname,avg(score) from student stleft join sc on sc.sno=st.snogroup by st.sno,st.sname having avg(score)85;*30. 30、查詢課程名稱為“數(shù)據(jù)庫，且分數(shù)低于60 的學生姓名和分數(shù)*select sname,score from student st,sc,course cwhere st.sno=sc.sno and sc o=c o and c ame=Oracle and sc.score70*33. 33、查詢不及格的課程，并按課程號從大到小排列*select sc.

25、sno,c ame,sc.score from sc,course cwhere sc o=c o and sc.score80;*35. 35、求選了課程的學生人數(shù)*select count(distinct sno) from sc;*y有問題36、查詢選修“諶燕老師所授課程的學生中，成績最高的學生姓名及其成績*select st.sname,score from student st,sc ,course c,teacher twherest.sno=sc.sno and sc o=c o and c.tno=t.tnoand t.tname=諶燕 and sc.score=(select max(score)from sc where sc o=c o)*37. 37、查詢各個課程及相應的選修人數(shù)*select cno,count(sno) from sc group by cno;*38. 38、查詢不同課程成績相同的學生的學號、課程號、學生成績*select a.* from sc a ,sc b where a.score=b.score and a ob o*