问题及描述： --1.学生表
Student(Sid,Sname,Sage,Ssex) --Sid 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别 --2.课程表
Course(Cid,Cname,Tid) --Cid --课程编号,Cname 课程名称,Tid 教师编号 --3.教师表
Teacher(Tid,Tname) --Tid 教师编号,Tname 教师姓名 --4.成绩表
SC(Sid,Cid,score) --Sid 学生编号,Cid 课程编号,score 分数 */
--创建测试数据
create table Student(Sid varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(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‘ , ‘1992-03-01‘ , 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(Cid varchar(10),Cname nvarchar(10),Tid 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‘)
create table Teacher(Tid 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(Sid varchar(10),Cid varchar(10),score decimal(18,1)) insert into SC values(‘01‘ , ‘01‘ , 80) insert into SC values(‘01‘ , ‘02‘ , 90) insert into SC values(‘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 SC values(‘05‘ , ‘01‘ , 76) insert into SC values(‘05‘ , ‘02‘ , 87) insert into SC values(‘06‘ , ‘01‘ , 31) 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.score [课程‘02‘的分数] from Student a , SC b , SC c where a.Sid = b.Sid and a.Sid = c.Sid and b.Cid = ‘01‘ and c.Cid = ‘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.Sid = b.Sid and b.Cid = ‘01‘
left join SC c on a.Sid = c.Sid and c.Cid = ‘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.Sid = b.Sid and a.Sid = c.Sid and b.Cid = ‘01‘ and c.Cid = ‘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 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.Sid = b.Sid and b.Cid = ‘01‘ left join SC c on a.Sid = c.Sid and c.Cid = ‘02‘ where isnull(b.score,0) < c.score
--3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select a.Sid , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score from Student a , sc b where a.Sid = b.Sid group by a.Sid , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60 order by a.Sid
--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.Sid , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score from Student a , sc b where a.Sid = b.Sid group by a.Sid , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60 order by a.Sid
--4.2、查询在sc表中不存在成绩的学生信息的SQL语句。
select a.Sid , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0) avg_score
from Student a left join sc b on a.Sid = b.Sid group by a.Sid , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 order by a.Sid
--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.Sid [学生编号], a.Sname [学生姓名], count(b.Cid) 选课总数, sum(score) [所有课程的总成绩]
from Student a , SC b where a.Sid = b.Sid group by a.Sid,a.Sname order by a.Sid
--5.2、查询所有(包括有成绩和无成绩)的SQL。
select a.Sid [学生编号], a.Sname [学生姓名], count(b.Cid) 选课总数, sum(score) [所有课程的总成绩]
from Student a left join SC b on a.Sid = b.Sid group by a.Sid,a.Sname order by a.Sid
--6、查询"李"姓老师的数量 --方法1
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‘
--7、查询学过"张三"老师授课的同学的信息
select distinct Student.* from Student , SC , Course , Teacher
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and
Teacher.Tname = N‘张三‘ order by Student.Sid
--8、查询没学过"张三"老师授课的同学的信息
select m.* from Student m where Sid not in (select distinct SC.Sid from SC , Course , Teacher where SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and Teacher.Tname = N‘张三‘) order by m.Sid
--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.Sid = SC.Sid and SC.Cid = ‘01‘ and exists (Select 1 from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid = ‘02‘) order by Student.Sid
--方法2
select Student.* from Student , SC where Student.Sid = SC.Sid and SC.Cid = ‘02‘ and exists (Select 1 from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid = ‘01‘) order by Student.Sid
--方法3
select m.* from Student m where Sid in (
select Sid from (
select distinct Sid from SC where Cid = ‘01‘ union all
select distinct Sid from SC where Cid = ‘02‘ ) t group by Sid having count(1) = 2 )
order by m.Sid
--10、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.Sid = SC.Sid and SC.Cid = ‘01‘ and not exists (Select 1 from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid = ‘02‘) order by Student.Sid
--方法2
select Student.* from Student , SC where Student.Sid = SC.Sid and SC.Cid = ‘01‘ and Student.Sid not in (Select SC_2.Sid from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid = ‘02‘) order by Student.Sid
--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.Sid = SC.Sid
group by Student.Sid , Student.Sname , Student.Sage , Student.Ssex having count(Cid) < (select count(Cid) from Course) --11.2 select Student.* from Student left join SC on Student.Sid = SC.Sid
group by Student.Sid , Student.Sname , Student.Sage , Student.Ssex having count(Cid) < (select count(Cid) 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.Sid = SC.Sid and SC.Cid in (select Cid from SC where Sid = ‘01‘) and Student.Sid <> ‘01‘
--13、查询和"01"号的同学学习的课程完全相同的其他同学的信息
select Student.* from Student where Sid in
(select distinct SC.Sid from SC where Sid <> ‘01‘ and SC.Cid in (select distinct Cid from SC where Sid = ‘01‘)
group by SC.Sid having count(1) = (select count(1) from SC where Sid=‘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.Sid not in
(select distinct sc.Sid from sc , course , teacher where sc.Cid = course.Cid and course.Tid = teacher.Tid and teacher.tname = N‘张三‘) order by student.Sid
--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.Sid , student.sname , cast(avg(score) as decimal(18,2)) avg_score from student , sc where student.Sid = SC.Sid and student.Sid in (select Sid from SC where score < 60 group by Sid having count(1) >= 2)
group by student.Sid , 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.Cid , sc.score from student , sc
where student.Sid = SC.Sid and sc.score < 60 and sc.Cid = ‘01‘ order by sc.score desc
--17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
--17.1 SQL 2000 静态
select a.Sid 学生编号 , a.Sname 学生姓名 ,
max(case c.Cname when N‘语文‘ then b.score else null end) [语文], max(case c.Cname when N‘数学‘ then b.score else null end) [数学], max(case c.Cname when N‘英语‘ then b.score else null end) [英语], cast(avg(b.score) as decimal(18,2)) 平均分 from Student a
left join SC b on a.Sid = b.Sid left join Course c on b.Cid = c.Cid group by a.Sid , a.Sname order by 平均分 desc
--17.2 SQL 2000 动态
declare @sql nvarchar(4000)
set @sql = ‘select a.Sid ‘ + N‘学生编号‘ + ‘ , a.Sname ‘ + N‘学生姓名‘
select @sql = @sql + ‘,max(case c.Cname when N‘‘‘+Cname+‘‘‘ then b.score else null end)
[‘+Cname+‘]‘
from (select distinct Cname from Course) as t
set @sql = @sql + ‘ , cast(avg(b.score) as decimal(18,2)) ‘ + N‘平均分‘ + ‘ from Student a left join SC b on a.Sid = b.Sid left join Course c on b.Cid = c.Cid group by a.Sid , a.Sname order by ‘ + N‘平均分‘ + ‘ desc‘ exec(@sql)
--24、查询学生平均成绩及其名次
--24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，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 t1.* , px = (select count(1) from (
select m.Sid [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.Sid = n.Sid group by m.Sid , m.Sname
) t2 where 平均成绩 > t1.平均成绩) + 1 from (
select m.Sid [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.Sid = n.Sid group by m.Sid , m.Sname ) t1 order by px
select t1.* , px = (select count(distinct 平均成绩) from (
select m.Sid [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.Sid = n.Sid group by m.Sid , m.Sname
) t2 where 平均成绩 >= t1.平均成绩) from (
select m.Sid [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.Sid = n.Sid group by m.Sid , m.Sname ) t1 order by px
--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.Sid [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.Sid = n.Sid group by m.Sid , m.Sname ) t order by px
select t.* , px = DENSE_RANK() over(order by [平均成绩] desc) from (
select m.Sid [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.Sid = n.Sid group by m.Sid , 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.Cid , n.score from Student m, SC n where m.Sid = n.Sid and n.score in
(select top 3 score from sc where Cid = n.Cid order by score desc) order by n.Cid , 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 = (select count(distinct score) from SC where Cid = t.Cid and score >= t.score) from sc t) m where px between 1 and 3 order by m.cid , m.px --sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t) m where px between 1 and 3 order by m.Cid , m.px
--26、查询每门课程被选修的学生数
select cid , count(Sid)[学生数] from sc group by Cid
--27、查询出只有两门课程的全部学生的学号和姓名
select Student.Sid , Student.Sname from Student , SC where Student.Sid = SC.Sid
group by Student.Sid , Student.Sname having count(SC.Cid) = 2 order by Student.Sid
--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 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 * from 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 datediff(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.Cid , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score from Course m, SC n where m.Cid = n.Cid
group by m.Cid , m.Cname order by avg_score desc, m.Cid 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
select a.Sid , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score from Student a , sc b where a.Sid = b.Sid group by a.Sid , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85 order by a.Sid
--34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select sname , score from Student , SC , Course
where SC.Sid = Student.Sid and SC.Cid = Course.Cid and Course.Cname = N‘数学‘ and score < 60
--35、查询所有学生的课程及分数情况；
select Student.* , Course.Cname , SC.Cid , SC.score from Student, SC , Course
where Student.Sid = SC.Sid and SC.Cid = Course.Cid order by Student.Sid , SC.Cid
--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 Student.* , Course.Cname , SC.Cid , SC.score from Student, SC , Course
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and SC.score >= 70 order by Student.Sid , SC.Cid
--37、查询不及格的课程
select Student.* , Course.Cname , SC.Cid , SC.score from Student, SC , Course
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and SC.score < 60 order by Student.Sid , SC.Cid
--38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名；
select Student.* , Course.Cname , SC.Cid , SC.score from Student, SC , Course
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and SC.Cid = ‘01‘ and SC.score >= 80 order by Student.Sid , SC.Cid
--39、求每门课程的学生人数
select Course.Cid , Course.Cname , count(*) [学生人数] from Course , SC where Course.Cid = SC.Cid
group by Course.Cid , Course.Cname order by Course.Cid , 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.Cid , SC.score from Student, SC , Course , Teacher
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and Teacher.Tname = N‘张三‘ order by SC.score desc
--40.2 当最高分出现多个时
select Student.* , Course.Cname , SC.Cid , SC.score from Student, SC , Course , Teacher
where Student.Sid = SC.Sid and SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and Teacher.Tname = N‘张三‘ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and Teacher.Tname = N‘张三‘)
--41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 --方法1
select m.* from SC m ,(select Cid , score from SC group by Cid , score having count(1) > 1) n where m.Cid= n.Cid and m.score = n.score order by m.Cid , m.score , m.Sid --方法2
select m.* from SC m where exists (select 1 from (select Cid , score from SC group by Cid , score having count(1) > 1) n
where m.Cid= n.Cid and m.score = n.score) order by m.Cid , m.score , m.Sid
--42、查询每门功成绩最好的前两名
select t.* from sc t where score in (select top 2 score from sc where Cid = T.Cid order by score desc) order by t.Cid , t.score desc
--43、统计每门课程的学生选修人数（超过5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
select Course.Cid , Course.Cname , count(*) [学生人数] from Course , SC where Course.Cid = SC.Cid
group by Course.Cid , Course.Cname having count(*) >= 5
order by [学生人数] desc , Course.Cid
--44、检索至少选修两门课程的学生学号 select student.Sid , student.Sname from student , SC where student.Sid = SC.Sid
group by student.Sid , student.Sname having count(1) >= 2 order by student.Sid
--45、查询选修了全部课程的学生信息 --方法1 根据数量来完成
select student.* from student where Sid in
(select Sid from sc group by Sid having count(1) = (select count(1) from course))
--方法2 使用双重否定来完成 select t.* from student t where t.Sid not in (
select distinct m.Sid from (
select Sid , Cid from student , course
) m where not exists (select 1 from sc n where n.Sid = m.Sid and n.Cid = m.Cid) )
--方法3 使用双重否定来完成
select t.* from student t where not exists(select 1 from (
select distinct m.Sid from (
select Sid , Cid from student , course
) m where not exists (select 1 from sc n where n.Sid = m.Sid and n.Cid = m.Cid) ) k where k.Sid = t.Sid )
--46、查询各学生的年龄 --46.1 只按照年份来算
select * , datediff(yy , sage , getdate()) [年龄] from student
--46.2 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一 select * , case when right(convert(varchar(10),getdate(),120),5) <
right(convert(varchar(10),sage,120),5) then datediff(yy , sage , getdate()) - 1 else datediff(yy , sage , getdate()) end [年龄] from student
--47、查询本周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0
--48、查询下周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1
--49、查询本月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = 0
--50、查询下月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1