2022年SQL查询练习及答案 .pdf

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1、问题及描述：-1.学生表Student(S#,Sname,Sage,Ssex)-S#学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别-2.课程表Course(C#,Cname,T#)-C#-课程编号,Cname 课程名称,T#教师编号-3.教师表Teacher(T#,Tname)-T#教师编号,Tname 教师姓名-4.成绩表SC(S#,C#,score)-S#学生编号,C#课程编号,score 分数*/-创建测试数据create table Student(S#varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarc

2、har(10)insert into Student values(01,N 赵雷 ,1990-01-01,N 男)insert into Student values(02,N 钱电 ,1990-12-21,N 男)insert into Student values(03,N 孙风 ,1990-05-20,N 男)insert into Student values(04,N 李云 ,1990-08-06,N 男)insert into Student values(05,N 周梅 ,1991-12-01,N 女)insert into Student values(06,N 吴兰 ,19

3、92-03-01,N 女)insert into Student values(07,N 郑竹 ,1989-07-01,N 女)insert into Student values(08,N 王菊 ,1990-01-20,N 女)create table Course(C#varchar(10),Cname nvarchar(10),T#varchar(10)insert into Course values(01,N 语文 ,02)insert into Course values(02,N 数学 ,01)insert into Course values(03,N 英语 ,03)creat

4、e table Teacher(T#varchar(10),Tname nvarchar(10)insert into Teacher values(01,N 张三)insert into Teacher values(02,N 李四)insert into Teacher values(03,N 王五)create table SC(S#varchar(10),C#varchar(10),score decimal(18,1)insert into SC values(01,01,80)insert into SC values(01,02,90)insert into SC values(

5、01,03,99)insert into SC values(02,01,70)insert into SC values(02,02,60)insert into SC values(02,03,80)insert into SC values(03,01,80)insert into SC values(03,02,80)insert into SC values(03,03,80)insert into SC values(04,01,50)insert into SC values(04,02,30)insert into SC values(04,03,20)insert into

6、SC values(05,01,76)insert into SC values(05,02,87)insert into SC values(06,01,31)名师资料总结-精品资料欢迎下载-名师精心整理-第 1 页，共 10 页 -insert into SC values(06,03,34)insert into SC values(07,02,89)insert into SC values(07,03,98)go-1、查询 01课程比 02课程成绩高的学生的信息及课程分数-1.1、查询同时存在01课程和 02 课程的情况select a.*,b.score 课程 01的分数,c.sc

7、ore 课程 02的分数 from Student a,SC b,SC c where a.S#=b.S#and a.S#=c.S#and b.C#=01 and c.C#=02 and b.score c.score-1.2、查询同时存在01 课程和 02 课程的情况和存在01 课程但可能不存在02课程的情况(不存在时显示为null)(以下存在相同内容时不再解释)select a.*,b.score 课程 01的分数,c.score 课程 02 的分数 from Student a left join SC b on a.S#=b.S#and b.C#=01 left join SC c o

8、n a.S#=c.S#and c.C#=02 where b.score isnull(c.score,0)-2、查询 01课程比 02课程成绩低的学生的信息及课程分数-2.1、查询同时存在01课程和 02 课程的情况select a.*,b.score 课程 01的分数,c.score 课程 02的分数 from Student a,SC b,SC c where a.S#=b.S#and a.S#=c.S#and b.C#=01 and c.C#=02 and b.score c.score-2.2、查询同时存在01课程和 02 课程的情况和不存在01课程但存在 02课程的情况select

9、 a.*,b.score 课程 01的分数,c.score 课程 02的分数 from Student a left join SC b on a.S#=b.S#and b.C#=01 left join SC c on a.S#=c.S#and c.C#=02 where isnull(b.score,0)=60 order by a.S#-4、查询平均成绩小于60 分的同学的学生编号和学生姓名和平均成绩-4.1、查询在sc 表存在成绩的学生信息的SQL语句。select a.S#,a.Sname,cast(avg(b.score)as decimal(18,2)avg_score from

10、 Student a,sc b where a.S#=b.S#group by a.S#,a.Sname having cast(avg(b.score)as decimal(18,2)60 order by a.S#-4.2、查询在sc 表中不存在成绩的学生信息的SQL语句。select a.S#,a.Sname,isnull(cast(avg(b.score)as decimal(18,2),0)avg_score from Student a left join sc b on a.S#=b.S#group by a.S#,a.Sname having isnull(cast(avg(b

11、.score)as decimal(18,2),0)60 order by a.S#-5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩-5.1、查询所有有成绩的SQL。select a.S#学生编号,a.Sname 学生姓名,count(b.C#)选课总数,sum(score)所有课程的总成绩 from Student a,SC b where a.S#=b.S#group by a.S#,a.Sname order by a.S#-5.2、查询所有(包括有成绩和无成绩)的 SQL。select a.S#学生编号,a.Sname 学生姓名,count(b.C#)选课总数,sum

12、(score)所有课程的总成绩 from Student a left join SC b on a.S#=b.S#group by a.S#,a.Sname order by a.S#-6、查询 李 姓老师的数量-方法 1 名师资料总结-精品资料欢迎下载-名师精心整理-第 2 页，共 10 页 -select count(Tname)李姓老师的数量 from Teacher where Tname like N 李%-方法 2 select count(Tname)李姓老师的数量 from Teacher where left(Tname,1)=N李/*李姓老师的数量-1*/-7、查询学过

13、张三 老师授课的同学的信息select distinct Student.*from Student,SC,Course,Teacher where Student.S#=SC.S#and SC.C#=Course.C#and Course.T#=Teacher.T#and Teacher.Tname=N张三 order by Student.S#-8、查询没学过 张三 老师授课的同学的信息select m.*from Student m where S#not in(select distinct SC.S#from SC,Course,Teacher where SC.C#=Course.

14、C#and Course.T#=Teacher.T#and Teacher.Tname=N张三)order by m.S#-9、查询学过编号为01 并且也学过编号为02的课程的同学的信息-方法 1 select Student.*from Student,SC where Student.S#=SC.S#and SC.C#=01 and exists(Select 1 from SC SC_2 where SC_2.S#=SC.S#and SC_2.C#=02)order by Student.S#-方法 2 select Student.*from Student,SC where Stud

15、ent.S#=SC.S#and SC.C#=02 and exists(Select 1 from SC SC_2 where SC_2.S#=SC.S#and SC_2.C#=01)order by Student.S#-方法 3 select m.*from Student m where S#in(select S#from(select distinct S#from SC where C#=01 union all select distinct S#from SC where C#=02)t group by S#having count(1)=2)order by m.S#-10

16、、查询学过编号为01 但是没有学过编号为02的课程的同学的信息-方法 1 select Student.*from Student,SC where Student.S#=SC.S#and SC.C#=01 and not exists(Select 1 from SC SC_2 where SC_2.S#=SC.S#and SC_2.C#=02)order by Student.S#-方法 2 select Student.*from Student,SC where Student.S#=SC.S#and SC.C#=01 and Student.S#not in(Select SC_2.

17、S#from SC SC_2 where SC_2.S#=SC.S#and SC_2.C#=02)order by Student.S#-11、查询没有学全所有课程的同学的信息-11.1、select Student.*from Student,SC where Student.S#=SC.S#group by Student.S#,Student.Sname,Student.Sage,Student.Ssex having count(C#)(select count(C#)from Course)-11.2 select Student.*from Student left join SC

18、 on Student.S#=SC.S#group by Student.S#,Student.Sname,Student.Sage,Student.Ssex having count(C#)(select count(C#)from Course)-12、查询至少有一门课与学号为01 的同学所学相同的同学的信息select distinct Student.*from Student,SC where Student.S#=SC.S#and SC.C#in(select C#from SC where S#=01)and Student.S#01-13、查询和 01 号的同学学习的课程完全相

19、同的其他同学的信息select Student.*from Student where S#in(select distinct SC.S#from SC where S#01 and SC.C#in(select distinct C#from SC where S#=01)group by SC.S#having count(1)=(select count(1)from SC where S#=01)-14、查询没学过张三 老师讲授的任一门课程的学生姓名select student.*from student where student.S#not in(select distinct

20、sc.S#from sc,course,名师资料总结-精品资料欢迎下载-名师精心整理-第 3 页，共 10 页 -teacher where sc.C#=course.C#and course.T#=teacher.T#and teacher.tname=N张三)order by student.S#-15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩select student.S#,student.sname,cast(avg(score)as decimal(18,2)avg_score from student,sc where student.S#=SC.S#and st

21、udent.S#in(select S#from SC where score=2)group by student.S#,student.sname-16、检索 01课程分数小于60，按分数降序排列的学生信息select student.*,sc.C#,sc.score from student,sc where student.S#=SC.S#and sc.score=60，中等为：70-80，优良为：80-90，优秀为：=90-方法 1 select m.C#课程编号,m.Cname 课程名称,max(n.score)最高分,min(n.score)最低分,cast(avg(n.scor

22、e)as decimal(18,2)平均分,cast(select count(1)from SC where C#=m.C#and score=60)*100.0/(select count(1)from SC where C#=m.C#)as decimal(18,2)及格率(%),cast(select count(1)from SC where C#=m.C#and score=70 and score=80 and score=90)*100.0/(select count(1)from SC where C#=m.C#)as decimal(18,2)优秀率(%)from Cour

23、se m,SC n where m.C#=n.C#group by m.C#,m.Cname order by m.C#-方法 2 select m.C#课程编号,m.Cname 课程名称,(select max(score)from SC where C#=m.C#)最高分,(select min(score)from SC where C#=m.C#)最低分,(select cast(avg(score)as decimal(18,2)from SC where C#=m.C#)平均分,cast(select count(1)from SC where C#=m.C#and score=6

24、0)*100.0/(select count(1)from SC where C#=m.C#)as decimal(18,2)及格率(%),cast(select count(1)from SC where C#=m.C#and score=70 and score=80 and score=90)*100.0/(select count(1)from SC where C#=m.C#)as decimal(18,2)优秀率(%)from Course m order by m.C#-19、按各科成绩进行排序，并显示排名-19.1 sql 2000 用子查询完成-Score重复时保留名次空缺s

25、elect t.*,px=(select count(1)from SC where C#=t.C#and score t.score)+1 from sc t order by t.c#,px-Score重复时合并名次select t.*,px=(select count(distinct score)from SC where C#=t.C#and score=t.score)from sc t order by t.c#,px-19.2 sql 2005 用 rank,DENSE_RANK 完成-Score重复时保留名次空缺(rank 完成)select t.*,px=rank()ove

26、r(partition by c#order by score desc)from sc t order by t.C#,px-Score重复时合并名次(DENSE_RANK 完成)select t.*,px=DENSE_RANK()over(partition by c#order by score desc)from sc t order by t.C#,px-20、查询学生的总成绩并进行排名-20.1 查询学生的总成绩select m.S#学生编号 ,m.Sname 学生姓名 ,isnull(sum(score),0)总成绩 from Student m left join SC n o

27、n m.S#=n.S#group by m.S#,m.Sname order by 总成绩 desc-20.2 查询学生的总成绩并进行排名，sql 2000 用子查询完成，分总分重复时保留名次空缺和不保留名次空缺两种。select t1.*,px=(select count(1)from(select m.S#学生编号 ,m.Sname 学生姓名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t2 where 总成绩 t1.总成绩)+1 from(select m

28、.S#学 生 编 号 ,m.Sname 学 生 姓 名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t1 order by px select t1.*,px=(select count(distinct 总成绩)from(select m.S#学 生 编 号 ,m.Sname 学 生 姓 名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sn

29、ame)t2 where 总成绩=t1.总成绩)from(select m.S#学生编号 ,m.Sname 学生姓名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t1 order by px-20.3 查询学生的总成绩并进行排名，sql 2005 用 rank,DENSE_RANK 完成，分总分重复时保留名次空缺和不保留名次空缺两种。select t.*,px=rank()over(order by 总成绩 desc)from(select m.S#学生编号 ,

30、m.Sname 学生姓名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t order by px select t.*,px=DENSE_RANK()over(order by 总 成 绩 desc)from(select m.S#学 生 编 号 ,m.Sname 学生姓名 ,isnull(sum(score),0)总成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t orde

31、r by px-21、查询不同老师所教不同课程平均分从高到低显示select m.T#,m.Tname,cast(avg(o.score)as decimal(18,2)avg_score from Teacher m,Course n,SC o where m.T#=n.T#and n.C#=o.C#group by m.T#,m.Tname order by avg_score 名师资料总结-精品资料欢迎下载-名师精心整理-第 5 页，共 10 页 -desc-22、查询所有课程的成绩第2 名到第 3 名的学生信息及该课程成绩-22.1 sql 2000 用子查询完成-Score重复时保留

32、名次空缺select*from(select t.*,px=(select count(1)from SC where C#=t.C#and score t.score)+1 from sc t)m where px between 2 and 3 order by m.c#,m.px-Score重复时合并名次select*from(select t.*,px=(select count(distinct score)from SC where C#=t.C#and score=t.score)from sc t)m where px between 2 and 3 order by m.c#

33、,m.px-22.2 sql 2005 用 rank,DENSE_RANK 完成-Score重复时保留名次空缺(rank 完成)select*from(select t.*,px=rank()over(partition by c#order by score desc)from sc t)m where px between 2 and 3 order by m.C#,m.px-Score重复时合并名次(DENSE_RANK 完成)select*from(select t.*,px=DENSE_RANK()over(partition by c#order by score desc)fro

34、m sc t)m where px between 2 and 3 order by m.C#,m.px-23、统计各科成绩各分数段人数：课程编号,课程名称,100-85,85-70,70-60,0-60 及所占百分比-23.1 统计各科成绩各分数段人数：课程编号,课程名称,100-85,85-70,70-60,0-60-横向显示select Course.C#课程编号 ,Cname as 课程名称 ,sum(case when score=85 then 1 else 0 end)85-100,sum(case when score=70 and score=60 and score 70

35、then 1 else 0 end)60-70,sum(case when score=85 then 85-100 when n.score=70 and n.score=60 and n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score 70 then 60-70 else 0-60

36、end)order by m.C#,m.Cname,分数段-23.2 统计各科成绩各分数段人数：课程编号,课程名称,100-85,85-70,70-60,60 及所占百分比-横向显示select m.C#课程编号,m.Cname 课程名称,名师资料总结-精品资料欢迎下载-名师精心整理-第 6 页，共 10 页 -(select count(1)from SC where C#=m.C#and score 60)0-60,cast(select count(1)from SC where C#=m.C#and score=60 and score=60 and score=70 and scor

37、e=70 and score=85)85-100,cast(select count(1)from SC where C#=m.C#and score=85)*100.0/(select count(1)from SC where C#=m.C#)as decimal(18,2)百分比(%)from Course m order by m.C#-纵向显示1(显示存在的分数段)select m.C#课程编号 ,m.Cname 课程名称 ,分数段=(case when n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score

38、=85 then 85-100 when n.score=70 and n.score=60 and n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score=85 then 85-100 when n.score=70 and n.score=60 and n.score t1.平均成绩)+1 from(select m.S#学生编号,m.Sname 学生姓名,isnull(cast(avg(score)as decimal(18,2),0)平均成绩 from Student m left join SC n on m

39、.S#=n.S#group by m.S#,m.Sname)t1 order by px select t1.*,px=(select count(distinct 平均成绩)from(select m.S#学生编号 ,m.Sname 学生姓名 ,isnull(cast(avg(score)as decimal(18,2),0)平均成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t2 where 平均成绩=t1.平均成 绩)from(select m.S#学生 编号 ,m.Sname 学 生姓名 ,isnul

40、l(cast(avg(score)as decimal(18,2),0)平均成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t1 order by px 名师资料总结-精品资料欢迎下载-名师精心整理-第 7 页，共 10 页 -24.2 查询学生的平均成绩并进行排名，sql 2005 用 rank,DENSE_RANK 完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。select t.*,px=rank()over(order by 平均成绩 desc)from(select m.S#学生编号 ,m.

41、Sname 学生姓名,isnull(cast(avg(score)as decimal(18,2),0)平均成绩 from Student m left join SC n on m.S#=n.S#group by m.S#,m.Sname)t order by px select t.*,px=DENSE_RANK()over(order by 平均成绩 desc)from(select m.S#学生编号,m.Sname 学生姓名 ,isnull(cast(avg(score)as decimal(18,2),0)平均成绩 from Student m left join SC n on m

42、.S#=n.S#group by m.S#,m.Sname)t order by px-25、查询各科成绩前三名的记录-25.1 分数重复时保留名次空缺select m.*,n.C#,n.score from Student m,SC n where m.S#=n.S#and n.score in(select top 3 score from sc where C#=n.C#order by score desc)order by n.C#,n.score desc-25.2 分数重复时不保留名次空缺，合并名次-sql 2000 用子查询实现select*from(select t.*,px

43、=(select count(distinct score)from SC where C#=t.C#and score=t.score)from sc t)m where px between 1 and 3 order by m.c#,m.px-sql 2005 用DENSE_RANK 实现select*from(select t.*,px=DENSE_RANK()over(partition by c#order by score desc)from sc t)m where px between 1 and 3 order by m.C#,m.px-26、查询每门课程被选修的学生数se

44、lect c#,count(S#)学生数 from sc group by C#-27、查询出只有两门课程的全部学生的学号和姓名select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S#group by Student.S#,Student.Sname having count(SC.C#)=2 order by Student.S#-28、查询男生、女生人数select count(Ssex)as 男生人数from Student where Ssex=N男 select count(Ssex)as 女生人数

45、from Student where Ssex=N 女 select sum(case when Ssex=N男 then 1 else 0 end)男生人数,sum(case when Ssex=N 女 then 1 else 0 end)女生人数 from student select case when Ssex=N男 then N男生人数 else N女生人数 end 男女情况 ,count(1)人数 from student group by case when Ssex=N男 then N 男生人数 else N女生人数 end-29、查询名字中含有风字的学生信息select*fr

46、om student where sname like N%风%select*from student where charindex(N风,sname)0-30、查询同名同性学生名单，并统计同名人数select Sname 学生姓名,count(*)人数 from Student group by Sname having count(*)1-31、查询 1990 年出生的学生名单(注：Student 表中 Sage列的类型是datetime)select*from Student where year(sage)=1990 select*from Student where datedif

47、f(yy,sage,1990-01-01)=0 select*from Student where datepart(yy,sage)=1990 select*from Student where convert(varchar(4),sage,120)=1990-32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列select m.C#,m.Cname,cast(avg(n.score)as decimal(18,2)avg_score from Course m,SC n where m.C#=n.C#group by m.C#,m.Cname orde

48、r by avg_score desc,m.C#asc-33、查询平均成绩大于等于85 的所有学生的学号、姓名和平均成绩名师资料总结-精品资料欢迎下载-名师精心整理-第 8 页，共 10 页 -select a.S#,a.Sname,cast(avg(b.score)as decimal(18,2)avg_score from Student a,sc b where a.S#=b.S#group by a.S#,a.Sname having cast(avg(b.score)as decimal(18,2)=85 order by a.S#-34、查询课程名称为数学，且分数低于60 的学生姓

49、名和分数select sname,score from Student,SC,Course where SC.S#=Student.S#and SC.C#=Course.C#and Course.Cname=N 数学 and score=70 order by Student.S#,SC.C#-37、查询不及格的课程select Student.*,Course.Cname,SC.C#,SC.score from Student,SC,Course where Student.S#=SC.S#and SC.C#=Course.C#and SC.score=80 order by Student

50、.S#,SC.C#-39、求每门课程的学生人数select Course.C#,Course.Cname,count(*)学生人数 from Course,SC where Course.C#=SC.C#group by Course.C#,Course.Cname order by Course.C#,Course.Cname-40、查询选修 张三 老师所授课程的学生中，成绩最高的学生信息及其成绩-40.1 当最高分只有一个时select top 1 Student.*,Course.Cname,SC.C#,SC.score from Student,SC,Course,Teacher wh