我贴 Student(S#,Sname,Sage,Ssex) 学生表 Course(C#,Cname,T#) 课程表 SC(S#,C#,score) 成绩表 Teacher(T#,Tname) 教师表 问题: 1、查询“001”课程比“002”课程成绩高的所有学生的学号; select a.S# from (select s#,score from SC where C#='001') a,(select s#,score from SC where C#='002') b where a.score>b.score and a.s#=b.s#; 2、查询平均成绩大于60分的同学的学号和平均成绩; select S#,avg(score) from sc group by S# having avg(score) >60; 3、查询所有同学的学号、姓名、选课数、总成绩; select Student.S#,Student.Sname,count(SC.C#),sum(score) from Student left Outer join SC on Student.S#=SC.S# group by Student.S#,Sname 4、查询姓“李”的老师的个数; select count(distinct(Tname)) from Teacher where Tname like '李%'; 5、查询没学过“叶平”老师课的同学的学号、姓名; select Student.S#,Student.Sname from Student where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平'); 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名; select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002'); 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名; select S#,Sname from Student where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平')); 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名; Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2 from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score; 9、查询所有课程成绩小于60分的同学的学号、姓名; select S#,Sname from Student where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60); 10、查询没有学全所有课的同学的学号、姓名; select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course); 11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名; select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='1001'; 12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名; select distinct SC.S#,Sname from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='001'); 13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩; update SC set score=(select avg(SC_2.score) from SC SC_2 where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名; select S# from SC where C# in (select C# from SC where S#='1002') group by S# having count(*)=(select count(*) from SC where S#='1002'); 15、删除学习“叶平”老师课的SC表记录; Delect SC from course ,Teacher where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平'; 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“003”课程的同学学号、2、 号课的平均成绩; Insert SC select S#,'002',(Select avg(score) from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002'); 17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示: 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分 SELECT S# as 学生ID ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语 ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩 FROM SC AS t GROUP BY S# ORDER BY avg(t.score) 18、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 L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分 FROM SC L ,SC AS R WHERE L.C# = R.C# and L.score = (SELECT MAX(IL.score) FROM SC AS IL,Student AS IM WHERE L.C# = IL.C# and IM.S#=IL.S# GROUP BY IL.C#) AND R.Score = (SELECT MIN(IR.score) FROM SC AS IR WHERE R.C# = IR.C# GROUP BY IR.C# ); 19、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 t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩 ,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数 FROM SC T,Course where t.C#=course.C# GROUP BY t.C# ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004) SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分 ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数 ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数 FROM SC 21、查询不同老师所教不同课程平均分从高到低显示 SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩 FROM SC AS T,Course AS C ,Teacher AS Z where T.C#=C.C# and C.T#=Z.T# GROUP BY C.C# ORDER BY AVG(Score) DESC 22、查询如下课程成绩第 3 名到第 6 名的学生成绩单:企业管理(001),马克思(002),UML (003),数据库(004) [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩 SELECT DISTINCT top 3 SC.S# As 学生学号, Student.Sname AS 学生姓名 , T1.score AS 企业管理, T2.score AS 马克思, T3.score AS UML, T4.score AS 数据库, ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分 FROM Student,SC LEFT JOIN SC AS T1 ON SC.S# = T1.S# AND T1.C# = '001' LEFT JOIN SC AS T2 ON SC.S# = T2.S# AND T2.C# = '002' LEFT JOIN SC AS T3 ON SC.S# = T3.S# AND T3.C# = '003' LEFT JOIN SC AS T4 ON SC.S# = T4.S# AND T4.C# = '004' WHERE student.S#=SC.S# and ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) NOT IN (SELECT DISTINCT TOP 15 WITH TIES ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) FROM sc LEFT JOIN sc AS T1 ON sc.S# = T1.S# AND T1.C# = 'k1' LEFT JOIN sc AS T2 ON sc.S# = T2.S# AND T2.C# = 'k2' LEFT JOIN sc AS T3 ON sc.S# = T3.S# AND T3.C# = 'k3' LEFT JOIN sc AS T4 ON sc.S# = T4.S# AND T4.C# = 'k4' ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 23、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] SELECT SC.C# as 课程ID, Cname as 课程名称 ,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 where SC.C#=Course.C# GROUP BY SC.C#,Cname; 24、查询学生平均成绩及其名次 SELECT 1+(SELECT COUNT( distinct 平均成绩) FROM (SELECT S#,AVG(score) AS 平均成绩 FROM SC GROUP BY S# ) AS T1 WHERE 平均成绩 > T2.平均成绩) as 名次, S# as 学生学号,平均成绩 FROM (SELECT S#,AVG(score) 平均成绩 FROM SC GROUP BY S# ) AS T2 ORDER BY 平均成绩 desc;
25、查询各科成绩前三名的记录:(不考虑成绩并列情况) SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1 WHERE score IN (SELECT TOP 3 score FROM SC WHERE t1.C#= C# ORDER BY score DESC ) ORDER BY t1.C#;
26、查询每门课程被选修的学生数 select c#,count(S#) from sc group by C#; 27、查询出只选修了一门课程的全部学生的学号和姓名 select SC.S#,Student.Sname,count(C#) AS 选课数 from SC ,Student where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1; 28、查询男生、女生人数 Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男'; Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女'; 29、查询姓“张”的学生名单 SELECT Sname FROM Student WHERE Sname like '张%'; 30、查询同名同性学生名单,并统计同名人数 select Sname,count(*) from Student group by Sname having count(*)>1;; 31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime) select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age from student where CONVERT(char(11),DATEPART(year,Sage))='1981'; 32、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 C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ; 33、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 Sname,SC.S# ,avg(score) from Student,SC where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>85; 34、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,isnull(score,0) from Student,SC,Course where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60; 35、查询所有学生的选课情况; SELECT SC.S#,SC.C#,Sname,Cname FROM SC,Student,Course where SC.S#=Student.S# and SC.C#=Course.C# ; 36、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; SELECT distinct student.S#,student.Sname,SC.C#,SC.score FROM student,Sc WHERE SC.score>=70 AND SC.S#=student.S#; 37、查询不及格的课程,并按课程号从大到小排列 select c# from sc where scor e <60 order by C# ; 38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名; select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003'; 39、求选了课程的学生人数 select count(*) from sc; 40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩 select Student.Sname,score from Student,SC,Course C,Teacher where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# ); 41、查询各个课程及相应的选修人数 select count(*) from sc group by C#; 42、查询不同课程成绩相同的学生的学号、课程号、学生成绩 select distinct A.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ; 43、查询每门功成绩最好的前两名 SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1 WHERE score IN (SELECT TOP 2 score FROM SC WHERE t1.C#= C# ORDER BY score DESC ) ORDER BY t1.C#; 44、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,查询结果按人数降序排列,若人数相同,按课程号升序排列 select C# as 课程号,count(*) as 人数 from sc group by C# order by count(*) desc,c# 45、检索至少选修两门课程的学生学号 select S# from sc group by s# having count(*) > = 2 46、查询全部学生都选修的课程的课程号和课程名 select C#,Cname from Course where C# in (select c# from sc group by c#) 47、查询没学过“叶平”老师讲授的任一门课程的学生姓名 select Sname from Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平'); 48、查询两门以上不及格课程的同学的学号及其平均成绩 select S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#; 49、检索“004”课程分数小于60,按分数降序排列的同学学号 select S# from SC where C#='004'and score <60 order by score desc; 50、删除“002”同学的“001”课程的成绩 delete from Sc where S#='001'and C#='001'; 够了吧
SQL语句练习题参考答案 1、 select Sname,Ssex,Class from Student; 2、 select distinct depart from teacher; 3、 select Sno as '学号',Sname as '姓名',Ssex as '性别',Sbirthday as'出生日期',Class as'班号'from student; 或 select Sno as 学号,Sname as 姓名,Ssex as 性别,Sbirthday as 出生日期,Class as 班号 from student; 4、 select * from score where degree between 60 and 80; 或select * from score where degree>=60 and degree<=80; 5、 select * from score where degree in (85,86,88); 6、 select * from student where class='95031'or Ssex='女'; 7、 select * from student order by class desc; 8、 select * from score order by cno asc ,degree desc; 或select * from score order by cno ,degree desc; 9、 select count(*) as CNT from student where class='95031'; 10、select Sno as '学号',cno as '课程号', degree as '最高分' from score where degree=(select max(degree) from score) 11、select avg(degree)as 课程平均分 from score where cno='3-105'; 12、select cno,avg(degree) from score where cno like'3%'group by cno having count(*) >5; 13、select Sno from score group by Sno having min(degree)>70 and max(degree)<90; 14、select student.Sname,score.Cno,score.degree from student,score where student.Sno=score.Sno; 15、select x.Sno,y.Cname,x.degree from score x,course y where x.Cno=y.Cno; 16、select x.Sname,y.Cname,z.degree from student x,course y,score z where x.Sno=z.Sno and z.Cno=y.Cno; 17、select y.Cno,avg(y.degree) from student x,score y where x.Sno=y.Sno and x.class='95033'group by y.cno; 18、select Sno,Cno,rank from score,grade where degree between low and upp order by rank; 19、select x.Cno,x.Sno,x.degree from score x,score y where x.cno='3-105' and x.degree>y.degree and y.sno='109'and y.cno='3-105'; 20、 1,查询成绩非本科最高 select * from score b where degree <(select max(degree) from score a where a.cno=b.cno); 2,查询成绩非本科最高并且选2门以上的学生的成绩: 21、select x.cno,x.Sno,x.degree from score x,score y where x.degree>y.degree and y.sno='109'and y.cno='3-105'; select cno,sno,degree from score where degree >(select degree from score where sno='109' and cno='3-105') 22、select sno,sname,sbirthday from student where to_char(sbirthday,'yyyy')=(select to_char(sbirthday,'yyyy') from student where sno='108'); 23、select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where x.tno=y.tno and y.tname='张旭'); 24、select tname from teacher where tno in(select x.tno from course x,score y where x.cno=y.cno group by x.tno having count(x.tno)>5); 25、select * from student where class in('95033','95031'); 26、select distinct cno from score where degree in (select degree from score where degree>85); 27、select * from score where cno in(select x.cno from course x,teacher y where y.tno=x.tno and y.depart='计算机系'); 28、select tname,prof from teacher where depart='计算机系' and prof not in (select prof from teacher where depart='电子工程系'); 29、select * from score where cno='3-105' and degree>any (select degree from score where cno='3-245')order by degree desc; 30、select * from score where cno='3-105' and degree>all(select degree from score where cno='3-245'); 31、select tname,tsex,tbirthday from teacher union select sname,ssex,sbirthday from student; 32、select tname,tsex,tbirthday from teacher where tsex='女' union select sname,ssex,sbirthday from student where ssex='女'; 33、select * from score a where degree<(select avg(degree) from score b where a.cno=b.cno); 34、select tname,depart from teacher a where exists (select * from course b where a.tno=b.tno); 35、select tname,depart from teacher a where not exists (select * from course b where a.tno=b.tno); 36、select class from student where ssex='男'group by class having count(*)>=2; 37、select * from student where sname not like'王_'; 38、select sname as 姓名,(to_char(sysdate,'yyyy')-to_char(sbirthday,'yyyy')) as 年龄 from student 39、select sname,sbirthday as 最大 from student where sbirthday =(select min (sbirthday) from student) union select sname,sbirthday as 最小 from student where sbirthday =(select max(sbirthday) from student) 40、select class,sname,sbirthday from student order by class desc,sbirthday; 41、select x.tname,y.cname from teacher x,course y where x.tno=y.tno and x.tsex='男'; 42、select * from score where degree=(select max(degree)from score); 43、select sname from student where ssex=(select ssex from student where sname='李军'); 44、select sname from student where ssex=(select ssex from student where sname='李军') and class=(select class from student where sname='李军'); 45、select * from score where sno in(select sno from student where ssex='男') and cno=(select cno from course where cname='计算机导论');
问题描述: 为管理岗位业务培训信息,建立3个表: S (S#,SN,SD,SA) S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄 C (C#,CN ) C#,CN 分别代表课程编号、课程名称 SC ( S#,C#,G ) S#,C#,G 分别代表学号、所选修的课程编号、学习成绩要求实现如下6个处理: 1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名 2. 使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位 3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位 4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位 5. 查询选修了课程的学员人数 6. 查询选修课程超过5门的学员学号和所属单位1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名 --实现代码: SELECT SN,SD FROM S WHERE [S#] IN( SELECT [S#] FROM C,SC WHERE C.[C#]=SC.[C#] AND CN=N'税收基础') 2. 使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位 --实现代码: SELECT S.SN,S.SD FROM S,SC WHERE S.[S#]=SC.[S#] AND SC.[C#]='C2'3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位 --实现代码: SELECT SN,SD FROM S WHERE [S#] NOT IN( SELECT [S#] FROM SC WHERE [C#]='C5')4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位 --实现代码: SELECT SN,SD FROM S WHERE [S#] IN( SELECT [S#] FROM SC RIGHT JOIN C ON SC.[C#]=C.[C#] GROUP BY [S#] HAVING COUNT(*)=COUNT(DISTINCT [S#]))5. 查询选修了课程的学员人数 --实现代码: SELECT 学员人数=COUNT(DISTINCT [S#]) FROM SC6. 查询选修课程超过5门的学员学号和所属单位 --实现代码: SELECT SN,SD FROM S WHERE [S#] IN( SELECT [S#] FROM SC GROUP BY [S#] HAVING COUNT(DISTINCT [C#])>5)
问题描述: 已知关系模式: S (SNO,SNAME) 学生关系。SNO 为学号,SNAME 为姓名 C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号,CNAME 为课程名,CTEACHER 为任课教师 SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩要求实现如下5个处理: 1. 找出没有选修过“李明”老师讲授课程的所有学生姓名 2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩 3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名 4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号 5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩1. 找出没有选修过“李明”老师讲授课程的所有学生姓名 --实现代码: SELECT SNAME FROM S WHERE NOT EXISTS( SELECT * FROM SC,C WHERE SC.CNO=C.CNO AND C.CTEACHER='李明' AND SC.SNO=S.SNO)2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩 --实现代码: SELECT S.SNO,S.SNAME,AVG_SCGRADE=AVG(SC.SCGRADE) FROM S,SC,( SELECT SNO FROM SC WHERE SCGRADE<60 GROUP BY SNO HAVING COUNT(DISTINCT CNO)>=2 )A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO GROUP BY S.SNO,S.SNAME3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名 --实现代码: SELECT S.SNO,S.SNAME FROM S,( SELECT SC.SNO FROM SC,C WHERE SC.CNO=C.CNO AND C.CNAME IN('1','2') GROUP BY SNO HAVING COUNT(DISTINCT CNO)=2 )SC WHERE S.SNO=SC.SNO 4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号 --实现代码: SELECT S.SNO,S.SNAME FROM S,SC SC1,SC SC2 WHERE SC1.CNO='1' AND SC2.SNO='2' AND SC1.CNO=S.CNO AND SC1.SCGRADE>SC2.SCGRADE5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩 --实现代码: SELECT SC1.SNO,[1号课成绩]=SC1.SCGRADE,[2号课成绩]=SC2.SCGRADE FROM SC SC1,SC SC2 WHERE SC1.CNO='1' AND SC2.CNO='2' AND SC1.SNO=SC2.SNO AND SC1.SCGRADE>SC2.SCGRADE
问题描述: 本题用到下面三个关系表: CARD 借书卡。 CNO 卡号,NAME 姓名,CLASS 班级 BOOKS 图书。 BNO 书号,BNAME 书名,AUTHOR 作者,PRICE 单价,QUANTITY 库存册数 BORROW 借书记录。 CNO 借书卡号,BNO 书号,RDATE 还书日期 备注:限定每人每种书只能借一本;库存册数随借书、还书而改变。 要求实现如下15个处理: 1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束。 2. 找出借书超过5本的读者,输出借书卡号及所借图书册数。 3. 查询借阅了"水浒"一书的读者,输出姓名及班级。 4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期。 5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者。 6. 查询现有图书中价格最高的图书,输出书名及作者。 7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出。 8. 将"C01"班同学所借图书的还期都延长一周。 9. 从BOOKS表中删除当前无人借阅的图书记录。 10.如果经常按书名查询图书信息,请建立合适的索引。 11.在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表)。 12.建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名)。 13.查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出。 14.假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句。 15.对CARD表做如下修改: a. 将NAME最大列宽增加到10个字符(假定原为6个字符)。 b. 为该表增加1列NAME(系名),可变长,最大20个字符。 1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束 --实现代码: CREATE TABLE BORROW( CNO int FOREIGN KEY REFERENCES CARD(CNO), BNO int FOREIGN KEY REFERENCES BOOKS(BNO), RDATE datetime, PRIMARY KEY(CNO,BNO)) 2. 找出借书超过5本的读者,输出借书卡号及所借图书册数 --实现代码: SELECT CNO,借图书册数=COUNT(*) FROM BORROW GROUP BY CNO HAVING COUNT(*)>53. 查询借阅了"水浒"一书的读者,输出姓名及班级 --实现代码: SELECT * FROM CARD c WHERE EXISTS( SELECT * FROM BORROW a,BOOKS b WHERE a.BNO=b.BNO AND b.BNAME=N'水浒' AND a.CNO=c.CNO) 4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期 --实现代码: SELECT * FROM BORROW WHERE RDATE<GETDATE() 5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者 --实现代码: SELECT BNO,BNAME,AUTHOR FROM BOOKS WHERE BNAME LIKE N'%网络%' 6. 查询现有图书中价格最高的图书,输出书名及作者 --实现代码: SELECT BNO,BNAME,AUTHOR FROM BOOKS WHERE PRICE=( SELECT MAX(PRICE) FROM BOOKS) 7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出 --实现代码: SELECT a.CNO FROM BORROW a,BOOKS b WHERE a.BNO=b.BNO AND b.BNAME=N'计算方法' AND NOT EXISTS( SELECT * FROM BORROW aa,BOOKS bb WHERE aa.BNO=bb.BNO AND bb.BNAME=N'计算方法习题集' AND aa.CNO=a.CNO) ORDER BY a.CNO DESC 8. 将"C01"班同学所借图书的还期都延长一周 --实现代码: UPDATE b SET RDATE=DATEADD(Day,7,b.RDATE) FROM CARD a,BORROW b WHERE a.CNO=b.CNO AND a.CLASS=N'C01' 9. 从BOOKS表中删除当前无人借阅的图书记录 --实现代码: DELETE A FROM BOOKS a WHERE NOT EXISTS( SELECT * FROM BORROW WHERE BNO=a.BNO) 10. 如果经常按书名查询图书信息,请建立合适的索引 --实现代码: CREATE CLUSTERED INDEX IDX_BOOKS_BNAME ON BOOKS(BNAME)11. 在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表) --实现代码: CREATE TRIGGER TR_SAVE ON BORROW FOR INSERT,UPDATE AS IF @@ROWCOUNT>0 INSERT BORROW_SAVE SELECT i.* FROM INSERTED i,BOOKS b WHERE i.BNO=b.BNO AND b.BNAME=N'数据库技术及应用' 12. 建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名) --实现代码: CREATE VIEW V_VIEW AS SELECT a.NAME,b.BNAME FROM BORROW ab,CARD a,BOOKS b WHERE ab.CNO=a.CNO AND ab.BNO=b.BNO AND a.CLASS=N'力01'13. 查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出 --实现代码: SELECT a.CNO FROM BORROW a,BOOKS b WHERE a.BNO=b.BNO AND b.BNAME IN(N'计算方法',N'组合数学') GROUP BY a.CNO HAVING COUNT(*)=2 ORDER BY a.CNO DESC 14. 假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句 --实现代码: ALTER TABLE BOOKS ADD PRIMARY KEY(BNO) 15.1 将NAME最大列宽增加到10个字符(假定原为6个字符) --实现代码: ALTER TABLE CARD ALTER COLUMN NAME varchar(10) 15.2 为该表增加1列NAME(系名),可变长,最大20个字符 --实现代码: ALTER TABLE CARD ADD 系名 varchar(20)
Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表 问题:
1、查询“001”课程比“002”课程成绩高的所有学生的学号;
select a.S# from (select s#,score from SC where C#='001') a,(select s#,score
from SC where C#='002') b
where a.score>b.score and a.s#=b.s#;
2、查询平均成绩大于60分的同学的学号和平均成绩;
select S#,avg(score)
from sc
group by S# having avg(score) >60;
3、查询所有同学的学号、姓名、选课数、总成绩;
select Student.S#,Student.Sname,count(SC.C#),sum(score)
from Student left Outer join SC on Student.S#=SC.S#
group by Student.S#,Sname
4、查询姓“李”的老师的个数;
select count(distinct(Tname))
from Teacher
where Tname like '李%';
5、查询没学过“叶平”老师课的同学的学号、姓名;
select Student.S#,Student.Sname
from Student
where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平');
6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002');
7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
select S#,Sname
from Student
where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平'));
8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2
from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score;
9、查询所有课程成绩小于60分的同学的学号、姓名;
select S#,Sname
from Student
where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);
10、查询没有学全所有课的同学的学号、姓名;
select Student.S#,Student.Sname
from Student,SC
where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course);
11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;
select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='1001';
12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;
select distinct SC.S#,Sname
from Student,SC
where Student.S#=SC.S# and C# in (select C# from SC where S#='001');
13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;
update SC set score=(select avg(SC_2.score)
from SC SC_2
where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平');
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;
select S# from SC where C# in (select C# from SC where S#='1002')
group by S# having count(*)=(select count(*) from SC where S#='1002');
15、删除学习“叶平”老师课的SC表记录;
Delect SC
from course ,Teacher
where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平';
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“003”课程的同学学号、2、
号课的平均成绩;
Insert SC select S#,'002',(Select avg(score)
from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002');
17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示: 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
SELECT S# as 学生ID
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语
,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
FROM SC AS t
GROUP BY S#
ORDER BY avg(t.score)
18、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 L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分
FROM SC L ,SC AS R
WHERE L.C# = R.C# and
L.score = (SELECT MAX(IL.score)
FROM SC AS IL,Student AS IM
WHERE L.C# = IL.C# and IM.S#=IL.S#
GROUP BY IL.C#)
AND
R.Score = (SELECT MIN(IR.score)
FROM SC AS IR
WHERE R.C# = IR.C#
GROUP BY IR.C#
);
19、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 t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩
,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
FROM SC T,Course
where t.C#=course.C#
GROUP BY t.C#
ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC
20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004)
SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分
,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数
,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分
,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数
,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分
,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数
,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分
,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数
FROM SC
21、查询不同老师所教不同课程平均分从高到低显示
SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C ,Teacher AS Z
where T.C#=C.C# and C.T#=Z.T#
GROUP BY C.C#
ORDER BY AVG(Score) DESC
22、查询如下课程成绩第 3 名到第 6 名的学生成绩单:企业管理(001),马克思(002),UML (003),数据库(004)
[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
SELECT DISTINCT top 3
SC.S# As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC LEFT JOIN SC AS T1
ON SC.S# = T1.S# AND T1.C# = '001'
LEFT JOIN SC AS T2
ON SC.S# = T2.S# AND T2.C# = '002'
LEFT JOIN SC AS T3
ON SC.S# = T3.S# AND T3.C# = '003'
LEFT JOIN SC AS T4
ON SC.S# = T4.S# AND T4.C# = '004'
WHERE student.S#=SC.S# and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.S# = T1.S# AND T1.C# = 'k1'
LEFT JOIN sc AS T2
ON sc.S# = T2.S# AND T2.C# = 'k2'
LEFT JOIN sc AS T3
ON sc.S# = T3.S# AND T3.C# = 'k3'
LEFT JOIN sc AS T4
ON sc.S# = T4.S# AND T4.C# = 'k4'
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 23、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]
SELECT SC.C# as 课程ID, Cname as 课程名称
,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
where SC.C#=Course.C#
GROUP BY SC.C#,Cname; 24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT S#,AVG(score) AS 平均成绩
FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩
FROM SC
GROUP BY S#
) AS T2
ORDER BY 平均成绩 desc;
25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
select c#,count(S#) from sc group by C#;
27、查询出只选修了一门课程的全部学生的学号和姓名
select SC.S#,Student.Sname,count(C#) AS 选课数
from SC ,Student
where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
28、查询男生、女生人数
Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男';
Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女';
29、查询姓“张”的学生名单
SELECT Sname FROM Student WHERE Sname like '张%';
30、查询同名同性学生名单,并统计同名人数
select Sname,count(*) from Student group by Sname having count(*)>1;;
31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
from student
where CONVERT(char(11),DATEPART(year,Sage))='1981';
32、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 C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ;
33、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 Sname,SC.S# ,avg(score)
from Student,SC
where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>85;
34、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,isnull(score,0)
from Student,SC,Course
where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60;
35、查询所有学生的选课情况;
SELECT SC.S#,SC.C#,Sname,Cname
FROM SC,Student,Course
where SC.S#=Student.S# and SC.C#=Course.C# ;
36、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;
SELECT distinct student.S#,student.Sname,SC.C#,SC.score
FROM student,Sc
WHERE SC.score>=70 AND SC.S#=student.S#;
37、查询不及格的课程,并按课程号从大到小排列
select c# from sc where scor e <60 order by C# ;
38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
39、求选了课程的学生人数
select count(*) from sc;
40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
select Student.Sname,score
from Student,SC,Course C,Teacher
where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# );
41、查询各个课程及相应的选修人数
select count(*) from sc group by C#;
42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
select distinct A.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
43、查询每门功成绩最好的前两名
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
44、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,查询结果按人数降序排列,若人数相同,按课程号升序排列
select C# as 课程号,count(*) as 人数
from sc
group by C#
order by count(*) desc,c#
45、检索至少选修两门课程的学生学号
select S#
from sc
group by s#
having count(*) > = 2
46、查询全部学生都选修的课程的课程号和课程名
select C#,Cname
from Course
where C# in (select c# from sc group by c#)
47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
select Sname from Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平');
48、查询两门以上不及格课程的同学的学号及其平均成绩
select S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#;
49、检索“004”课程分数小于60,按分数降序排列的同学学号
select S# from SC where C#='004'and score <60 order by score desc;
50、删除“002”同学的“001”课程的成绩
delete from Sc where S#='001'and C#='001';
够了吧
1、 select Sname,Ssex,Class from Student;
2、 select distinct depart from teacher;
3、 select Sno as '学号',Sname as '姓名',Ssex as '性别',Sbirthday as'出生日期',Class as'班号'from student;
或
select Sno as 学号,Sname as 姓名,Ssex as 性别,Sbirthday as 出生日期,Class as 班号 from student;
4、 select * from score where degree between 60 and 80;
或select * from score where degree>=60 and degree<=80;
5、 select * from score where degree in (85,86,88);
6、 select * from student where class='95031'or Ssex='女';
7、 select * from student order by class desc;
8、 select * from score order by cno asc ,degree desc;
或select * from score order by cno ,degree desc;
9、 select count(*) as CNT from student where class='95031';
10、select Sno as '学号',cno as '课程号', degree as '最高分' from score
where degree=(select max(degree) from score)
11、select avg(degree)as 课程平均分 from score where cno='3-105';
12、select cno,avg(degree) from score where cno like'3%'group by cno having count(*) >5;
13、select Sno from score group by Sno having min(degree)>70 and max(degree)<90;
14、select student.Sname,score.Cno,score.degree from student,score where student.Sno=score.Sno;
15、select x.Sno,y.Cname,x.degree from score x,course y where x.Cno=y.Cno;
16、select x.Sname,y.Cname,z.degree from student x,course y,score z where x.Sno=z.Sno and z.Cno=y.Cno;
17、select y.Cno,avg(y.degree) from student x,score y where x.Sno=y.Sno and x.class='95033'group by y.cno;
18、select Sno,Cno,rank from score,grade where degree between low and upp order by rank;
19、select x.Cno,x.Sno,x.degree from score x,score y
where x.cno='3-105' and x.degree>y.degree and y.sno='109'and y.cno='3-105';
20、
1,查询成绩非本科最高 select * from score b where degree <(select max(degree) from score a where a.cno=b.cno);
2,查询成绩非本科最高并且选2门以上的学生的成绩:
21、select x.cno,x.Sno,x.degree from score x,score y where x.degree>y.degree and y.sno='109'and y.cno='3-105';
select cno,sno,degree from score where degree >(select degree from score where sno='109' and cno='3-105')
22、select sno,sname,sbirthday from student where to_char(sbirthday,'yyyy')=(select to_char(sbirthday,'yyyy') from student where sno='108');
23、select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where x.tno=y.tno and y.tname='张旭');
24、select tname from teacher where tno in(select x.tno from course x,score y where x.cno=y.cno group by x.tno having count(x.tno)>5);
25、select * from student where class in('95033','95031');
26、select distinct cno from score where degree in (select degree from score where degree>85);
27、select * from score where cno in(select x.cno from course x,teacher y where y.tno=x.tno and y.depart='计算机系');
28、select tname,prof from teacher where depart='计算机系' and prof not in (select prof from teacher where depart='电子工程系');
29、select * from score where cno='3-105' and degree>any (select degree from score where cno='3-245')order by degree desc;
30、select * from score where cno='3-105' and degree>all(select degree from score where cno='3-245');
31、select tname,tsex,tbirthday from teacher
union select sname,ssex,sbirthday from student;
32、select tname,tsex,tbirthday from teacher where tsex='女'
union select sname,ssex,sbirthday from student where ssex='女';
33、select * from score a where degree<(select avg(degree)
from score b where a.cno=b.cno);
34、select tname,depart from teacher a where exists
(select * from course b where a.tno=b.tno);
35、select tname,depart from teacher a where not exists
(select * from course b where a.tno=b.tno);
36、select class from student where ssex='男'group by class having count(*)>=2;
37、select * from student where sname not like'王_';
38、select sname as 姓名,(to_char(sysdate,'yyyy')-to_char(sbirthday,'yyyy')) as 年龄 from student
39、select sname,sbirthday as 最大 from student where sbirthday =(select min (sbirthday) from student)
union select sname,sbirthday as 最小 from student where sbirthday =(select max(sbirthday) from student)
40、select class,sname,sbirthday from student order by class desc,sbirthday;
41、select x.tname,y.cname from teacher x,course y where x.tno=y.tno and x.tsex='男';
42、select * from score where degree=(select max(degree)from score);
43、select sname from student where ssex=(select ssex from student where sname='李军');
44、select sname from student where ssex=(select ssex from student where sname='李军') and class=(select class from student where sname='李军');
45、select * from score where sno in(select sno from student where ssex='男') and cno=(select cno from course
where cname='计算机导论');
为管理岗位业务培训信息,建立3个表:
S (S#,SN,SD,SA) S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄
C (C#,CN ) C#,CN 分别代表课程编号、课程名称
SC ( S#,C#,G ) S#,C#,G 分别代表学号、所选修的课程编号、学习成绩要求实现如下6个处理:
1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
2. 使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位
3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位
4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
5. 查询选修了课程的学员人数
6. 查询选修课程超过5门的学员学号和所属单位1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM C,SC
WHERE C.[C#]=SC.[C#]
AND CN=N'税收基础')
2. 使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位
--实现代码:
SELECT S.SN,S.SD FROM S,SC
WHERE S.[S#]=SC.[S#]
AND SC.[C#]='C2'3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] NOT IN(
SELECT [S#] FROM SC
WHERE [C#]='C5')4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM SC
RIGHT JOIN C ON SC.[C#]=C.[C#]
GROUP BY [S#]
HAVING COUNT(*)=COUNT(DISTINCT [S#]))5. 查询选修了课程的学员人数
--实现代码:
SELECT 学员人数=COUNT(DISTINCT [S#]) FROM SC6. 查询选修课程超过5门的学员学号和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM SC
GROUP BY [S#]
HAVING COUNT(DISTINCT [C#])>5)
已知关系模式:
S (SNO,SNAME) 学生关系。SNO 为学号,SNAME 为姓名
C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号,CNAME 为课程名,CTEACHER 为任课教师
SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩要求实现如下5个处理:
1. 找出没有选修过“李明”老师讲授课程的所有学生姓名
2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩
3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名
4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号
5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩1. 找出没有选修过“李明”老师讲授课程的所有学生姓名
--实现代码:
SELECT SNAME FROM S
WHERE NOT EXISTS(
SELECT * FROM SC,C
WHERE SC.CNO=C.CNO
AND C.CTEACHER='李明'
AND SC.SNO=S.SNO)2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩
--实现代码:
SELECT S.SNO,S.SNAME,AVG_SCGRADE=AVG(SC.SCGRADE)
FROM S,SC,(
SELECT SNO
FROM SC
WHERE SCGRADE<60
GROUP BY SNO
HAVING COUNT(DISTINCT CNO)>=2
)A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO
GROUP BY S.SNO,S.SNAME3. 列出既学过“1”号课程,又学过“2”号课程的所有学生姓名
--实现代码:
SELECT S.SNO,S.SNAME
FROM S,(
SELECT SC.SNO
FROM SC,C
WHERE SC.CNO=C.CNO
AND C.CNAME IN('1','2')
GROUP BY SNO
HAVING COUNT(DISTINCT CNO)=2
)SC WHERE S.SNO=SC.SNO 4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号
--实现代码:
SELECT S.SNO,S.SNAME
FROM S,SC SC1,SC SC2
WHERE SC1.CNO='1'
AND SC2.SNO='2'
AND SC1.CNO=S.CNO
AND SC1.SCGRADE>SC2.SCGRADE5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩
--实现代码:
SELECT SC1.SNO,[1号课成绩]=SC1.SCGRADE,[2号课成绩]=SC2.SCGRADE
FROM SC SC1,SC SC2
WHERE SC1.CNO='1'
AND SC2.CNO='2'
AND SC1.SNO=SC2.SNO
AND SC1.SCGRADE>SC2.SCGRADE
本题用到下面三个关系表:
CARD 借书卡。 CNO 卡号,NAME 姓名,CLASS 班级
BOOKS 图书。 BNO 书号,BNAME 书名,AUTHOR 作者,PRICE 单价,QUANTITY 库存册数
BORROW 借书记录。 CNO 借书卡号,BNO 书号,RDATE 还书日期
备注:限定每人每种书只能借一本;库存册数随借书、还书而改变。
要求实现如下15个处理:
1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束。
2. 找出借书超过5本的读者,输出借书卡号及所借图书册数。
3. 查询借阅了"水浒"一书的读者,输出姓名及班级。
4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期。
5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者。
6. 查询现有图书中价格最高的图书,输出书名及作者。
7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出。
8. 将"C01"班同学所借图书的还期都延长一周。
9. 从BOOKS表中删除当前无人借阅的图书记录。
10.如果经常按书名查询图书信息,请建立合适的索引。
11.在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表)。
12.建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名)。
13.查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出。
14.假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句。
15.对CARD表做如下修改:
a. 将NAME最大列宽增加到10个字符(假定原为6个字符)。
b. 为该表增加1列NAME(系名),可变长,最大20个字符。
1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束
--实现代码:
CREATE TABLE BORROW(
CNO int FOREIGN KEY REFERENCES CARD(CNO),
BNO int FOREIGN KEY REFERENCES BOOKS(BNO),
RDATE datetime,
PRIMARY KEY(CNO,BNO)) 2. 找出借书超过5本的读者,输出借书卡号及所借图书册数
--实现代码:
SELECT CNO,借图书册数=COUNT(*)
FROM BORROW
GROUP BY CNO
HAVING COUNT(*)>53. 查询借阅了"水浒"一书的读者,输出姓名及班级
--实现代码:
SELECT * FROM CARD c
WHERE EXISTS(
SELECT * FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO
AND b.BNAME=N'水浒'
AND a.CNO=c.CNO) 4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期
--实现代码:
SELECT * FROM BORROW
WHERE RDATE<GETDATE() 5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者
--实现代码:
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE BNAME LIKE N'%网络%' 6. 查询现有图书中价格最高的图书,输出书名及作者
--实现代码:
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE PRICE=(
SELECT MAX(PRICE) FROM BOOKS) 7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出
--实现代码:
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO AND b.BNAME=N'计算方法'
AND NOT EXISTS(
SELECT * FROM BORROW aa,BOOKS bb
WHERE aa.BNO=bb.BNO
AND bb.BNAME=N'计算方法习题集'
AND aa.CNO=a.CNO)
ORDER BY a.CNO DESC 8. 将"C01"班同学所借图书的还期都延长一周
--实现代码:
UPDATE b SET RDATE=DATEADD(Day,7,b.RDATE)
FROM CARD a,BORROW b
WHERE a.CNO=b.CNO
AND a.CLASS=N'C01' 9. 从BOOKS表中删除当前无人借阅的图书记录
--实现代码:
DELETE A FROM BOOKS a
WHERE NOT EXISTS(
SELECT * FROM BORROW
WHERE BNO=a.BNO) 10. 如果经常按书名查询图书信息,请建立合适的索引
--实现代码:
CREATE CLUSTERED INDEX IDX_BOOKS_BNAME ON BOOKS(BNAME)11. 在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表)
--实现代码:
CREATE TRIGGER TR_SAVE ON BORROW
FOR INSERT,UPDATE
AS
IF @@ROWCOUNT>0
INSERT BORROW_SAVE SELECT i.*
FROM INSERTED i,BOOKS b
WHERE i.BNO=b.BNO
AND b.BNAME=N'数据库技术及应用' 12. 建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名)
--实现代码:
CREATE VIEW V_VIEW
AS
SELECT a.NAME,b.BNAME
FROM BORROW ab,CARD a,BOOKS b
WHERE ab.CNO=a.CNO
AND ab.BNO=b.BNO
AND a.CLASS=N'力01'13. 查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出
--实现代码:
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO
AND b.BNAME IN(N'计算方法',N'组合数学')
GROUP BY a.CNO
HAVING COUNT(*)=2
ORDER BY a.CNO DESC 14. 假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句
--实现代码:
ALTER TABLE BOOKS ADD PRIMARY KEY(BNO) 15.1 将NAME最大列宽增加到10个字符(假定原为6个字符)
--实现代码:
ALTER TABLE CARD ALTER COLUMN NAME varchar(10) 15.2 为该表增加1列NAME(系名),可变长,最大20个字符
--实现代码:
ALTER TABLE CARD ADD 系名 varchar(20)
CID CType CName
1 1 张三
2 1 李四
3 2 王二
CID(int 用户标识号):主键;
CName(varchar(50) 用户名):
CType(int 用户类型):1-一般用户;2-VIP用户; T_Order(订单表)
OID CID OName
1 1 订单1
2 1 订单2
3 1 订单3
4 2 订单4
5 2 订单5
6 2 订单6
7 2 订单7
8 3 订单8
9 3 订单9
OID(int 订单标识号):主键;
CID(int 用户标识号):外键;
OName(varchar(50) 订单名): T_Record(订单送达表)
OID OTime
1 2004-9-10
3 2004-11-10
5 2004-11-10
6 2004-11-20
8 2004-10-20
OID(int 订单标识号):主键,外键;
OTime(datetime 订单送达时间): 1. 用一条语句查询出订单数目至少有3个的用户的用户标识号、用户名、订单数:
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2. 用一条语句查询出所有用户的用户标识号、用户类型、用户名:(当用户类型为1时显示“一般用户”,当用户类型为2时显示“VIP用户”。)
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3. 用一条语句查询出所有订单的订单标识号、订单名、用户名、订单送达时间:(如果订单有送达时间,则显示订单送达时间,如果订单没有送达时间,则显示“未送达”。)
________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
4. 用一条语句查询出所有订单送达记录中2004年11月及以后的记录:
________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
5. 删除三个表中所有CID=3的数据:
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6. 用一条语句查询出“订单送达表”中送达时间最近的三条纪录:
________________________________________________________________________________
7. 向用户表中加入一条记录,CID=4,CType=1,CName=’赵六’:
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