SQL Server(四)——查询练习（45道习题）

题目：设有一数据库，包括四个表：学生表（Student）、课程表（Course）、成绩表（Score）以及教师信息表（Teacher）。

             四个表的结构分别如表1-1的表（一）~表（四）所示，数据如表1-2的表（一）~表（四）所示。用SQL语句创建四个表并完成相关题目。

表1-1数据库的表结构

表（一）Student (学生表)                        

表（二）Course（课程表）

表（三）Score(成绩表)

表（四）Teacher(教师表)

表1-2数据库中的数据

表（一）Student

表（二）Course

表（三）Score

表（四）Teacher

查询问题：

1、 查询Student表中的所有记录的Sname、Ssex和Class列。

2、 查询教师所有的单位即不重复的Depart列。

3、 查询Student表的所有记录。

4、 查询Score表中成绩在60到80之间的所有记录。

5、 查询Score表中成绩为85，86或88的记录。

6、 查询Student表中“95031”班或性别为“女”的同学记录。

7、 以Class降序查询Student表的所有记录。

8、 以Cno升序、Degree降序查询Score表的所有记录。

9、 查询“95031”班的学生人数。

10、 查询Score表中的最高分的学生学号和课程号。（子查询或者排序）

11、 查询每门课的平均成绩。

12、查询Score表中至少有5名学生选修的并以3开头的课程的平均分数。

13、查询分数大于70，小于90的Sno列。

14、查询所有学生的Sname、Cno和Degree列。

15、查询所有学生的Sno、Cname和Degree列。

16、查询所有学生的Sname、Cname和Degree列

17、 查询“95033”班学生的平均分。

18、 假设使用如下命令建立了一个grade表：

create table grade(low  int(3),upp  int(3),rank  char(1))

insert into grade values(90,100,’A’)

insert into grade values(80,89,’B’)

insert into grade values(70,79,’C’)

insert into grade values(60,69,’D’)

insert into grade values(0,59,’E’)

现查询所有同学的Sno、Cno和rank列。

19、  查询选修“3-105”课程的成绩高于“109”号同学成绩的所有同学的记录。

20、查询score中选学多门课程的同学中分数为非最高分成绩的记录。

21、查询成绩高于学号为“109”、课程号为“3-105”的成绩的所有记录。

22、查询和学号为108的同学同年出生的所有学生的Sno、Sname和Sbirthday列。

23、查询“张旭“教师任课的学生成绩。

24、查询选修某课程的同学人数多于5人的教师姓名。

25、查询95033班和95031班全体学生的记录。

26、  查询存在有85分以上成绩的课程Cno.

27、查询出“计算机系“教师所教课程的成绩表。

28、查询“计算机系”与“电子工程系“不同职称的教师的Tname和Prof。

29、查询选修编号为“3-105“课程且成绩至少高于选修编号为“3-245”的同学的Cno、Sno和Degree,并按Degree从高到低次序排序。

30、查询选修编号为“3-105”且成绩高于选修编号为“3-245”课程的同学的Cno、Sno和Degree.

31、 查询所有教师和同学的name、sex和birthday.

32、查询所有“女”教师和“女”同学的name、sex和birthday.

33、 查询成绩比该课程平均成绩低的同学的成绩表。

34、 查询所有任课教师的Tname和Depart.

35 、 查询所有未讲课的教师的Tname和Depart.

36、查询至少有2名男生的班号。

37、查询Student表中不姓“王”的同学记录。

38、查询Student表中每个学生的姓名和年龄。

39、查询Student表中最大和最小的Sbirthday日期值。

40、以班号和年龄从大到小的顺序查询Student表中的全部记录。

41、查询“男”教师及其所上的课程。

42、查询最高分同学的Sno、Cno和Degree列。

43、查询和“李军”同性别的所有同学的Sname.

44、查询和“李军”同性别并同班的同学Sname.

45、查询所有选修“计算机导论”课程的“男”同学的成绩表。

建表程序：

create database xueshenguse xueshenggocreate table Student(Sno varchar(20) not null primary key,Sname varchar(20) not null,Ssex varchar(20) not null,Sbirthday datetime,Class varchar(20))gocreate table Teacher(Tno varchar(20) not null primary key,Tname varchar(20) not null,Tsex varchar(20) not null,Tbrithday datetime,Prof varchar(20),Depart varchar(20) not null)gocreate table Course(Cno varchar(20) not null primary key,Cname varchar(20) not null,Tno varchar(20) not null references Teacher(Tno))gocreate table Score(Sno varchar(20) not null references Student(Sno),Cno varchar(20) not null references Course(Cno),Degree Decimal(4,1))alter table Score add constraint pk primary key(Sno, Cno)

数据输入程序未附，自己在设计中添加记好了

题目答案：

1、select Sname , Ssex , Class from Student 2、select distinct Depart from Teacher ——去重查询3、select * from Student 4、select * from Score where Degree between 60 and 80 ——范围查询，between关键字5、select * from Score where Degree=85 or Degree=86 or Degree=88  select * from Score where Degree in(85,86,88)6、select * from Student where Ssex='女' or Class=95031 ——条件查询7、select * from Student order by Class desc8、select * from Score order by Cno asc,Degree desc ——排序查询9、select count(*) from Student where Class=9503110、select Sno,Cno from score where Degree=(select MAX(Degree) from score)——运用子查询11、select cno ,AVG(Degree) from Score group by Cno12、select Cno,avg(Degree) from score group by Cno having count(Sno)>=5 and Cno like '3%' ——分组查询与模糊查询  (师)select Cno,avg(Degree) from score where Cno like '3%' group by Cno having count(Sno)>=5 ——聚合函数 count13、select Sno from Score where Degree between 70 and 9014、select Sname,Cno,degree from student,score where student.Sno=score.Sno  (师)select Sname,Cno,degree from score join student on score.sno=student.sno ——连接查询，join on15、select Sno,Cname,degree from course,score where course.cno=score.cno  (师)select Sname,Cno,degree from score join course on score.cno=course.cno16、select Sname,Cname,degree from student, course,score where student.Sno=score.Sno and course.cno=score.cno  (师)select Sname,Cno,degree from score join student on score.sno=student.sno join course on score.cno=course.cno17、select avg(degree) from score where sno in (select sno from student where class='95033') ——离散查询18、select Sno,Cno,rank from score,grade where score.degree between low and upp19、select * from score where cno='3-105' and degree>(select degree from score where sno='109'and cno='3-105')——第一种理解  select * from score where cno='3-105' and degree>(select max（degree） from score where sno='109')——理解二20、select * from score b where degree <(select max(degree) from score a where a.cno=b.cno)  select * from score where degree not in (select max(degree) from score group by cno having count(sno)>1）（师）select * from score where sno in (select sno from score group by sno having count(*)>1) and degree <(select max(degree) from score )   select * from score a where sno in (select sno from score group by sno having count(*)>1) and degree <(select max(degree) from score b where b.cno=a.cno )21、select * from score where cno='3-105' and degree>(select degree from score where sno='109'and cno='3-105')22、select sno,sname,sbirthday from student where datediff(year,sbirthday,getdate())=(select datediff(year,sbirthday,getdate()) from student where sno='108')   select * from student where year(sbirthdy)=(select year(sbirthday) from student where sno='108')23、select cno,sno,degree from score where cno=(select cno from course ,teacher where course.tno=teacher.tno and teacher.tname='张旭')  (师)select * from score where cno in(select cno from course where tno in(select tno from teacher where tname='张旭'))24、select tname from teacher where tno in (select tno from course where cno in (select cno from score group by cno having count(*)>5))25、select * from student where class in('95033','95031')或者select * from student where class ='95033'or class='95031'26、select distinct cno from score where degree>8527、select cno,sno,degree from score where cno in (select cno from course ,teacher where course.tno=teacher.tno and teacher.depart='计算机系')（师）select * from score where cno in(select cno from course where tno in(select tno from teacher where 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')  union  select tname,prof from teacher where depart='电子工程系' and prof not in (select prof from teacher where depart='计算机系')    select tname,prof from teacher a where prof not in (select prof from teacher b where b.depart != a.depart)29、select * from score where cno='3-105' and degree>any (select degree from score where cno='3-245')order by degree desc——any其中任何一个、all所有30、select * from score where cno='3-105' and degree>(select max(degree) from score where cno='3-245')31、select tname,tsex,tbirthday from teacher  union  select sname,ssex,sbirthday from student32、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);  select tname,depart from teacher,course where Teacher.tno=Course.tno  （师）select tname,depart from teacher where tno in( select tno from course where cno in( select distinct cno from score))35、select tname,depart from teacher a where not exists(select * from course b where a.tno=b.tno);  （师）select tname,depart from teacher where tno not in( select tno from course where cno in( select distinct cno from score))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,datediff(year,sbirthday,getdate()) as 年龄 from student  select sname as 姓名,year(getdate())-year(sbirthday) as 年龄 from student39、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)  （师）select min (sbirthday),max(sbirthday) from student40、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';  select tname,cname from teacher ,course where teacher.tno=course.tno and tsex='男'42、select * from score where degree=(select max(degree)from score);(师)selsct top 1 * from score order by degree desc——只适用于有一个最高分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');

答案仅代表自己意见，若有错误和不同做法，希望看到的告知，相互学习，十分感谢！

原标题：SQL Server(四)——查询练习（45道习题）

关键词：sql