精华内容
下载资源
问答
  • 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Course(C#,Cname,T#) C# --课程编号,Cname --课程名称,T# --教师编号--3.教师Teacher(T#,Tname) T# ...

    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 nvarchar(10))
    insert into Student values('01' , N'
    赵雷' , '1990-01-01' , N'')
    insert into Student values('02' , N'
    钱电' , '1990-12-21' , N'')
    insert into Student values('03' , N'
    孙风' , '1990-05-20' , N'')
    insert into Student values('04' , N'
    李云' , '1990-08-06' , N'')
    insert into Student values('05' , N'
    周梅' , '1991-12-01' , N'')
    insert into Student values('06' , N'
    吴兰' , '1992-03-01' , N'')
    insert into Student values('07' , N'
    郑竹' , '1989-07-01' , N'')
    insert into Student values('08' , N'
    王菊' , '1990-01-20' , N'')
    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')
    create 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('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.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 on 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 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) < 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.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)) >= 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 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.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(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
    select count(Tname) ["
    "姓老师的数量] from Teacher where Tname like N'%'
    --
    方法2
    select count(Tname) ["
    "姓老师的数量] from Teacher where left(Tname,1) = N''

    --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.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.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 Student.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.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
    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 sc.S# from sc , course , 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 student.S# in (select S# from SC where score < 60 group by S# having count(1) >= 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 and sc.C# = '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.S# 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.S# = b.S#
    left join Course c on b.C# = c.C#
    group by a.S# , a.Sname
    order by
    平均分 desc
    --17.2 SQL 2000
    动态
    declare @sql nvarchar(4000)
    set @sql = 'select a.S# ' + 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.S# = b.S# left join Course c on b.C# = c.C#
    group by a.S# , 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.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) + 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.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] ,
             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
    ) t1
    order by px
    --24.2
    查询学生的平均成绩并进行排名,sql 2005rank,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.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.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 = (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
    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
    select 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''
    select count(Ssex) as
    女生人数 from Student where Ssex = N''
    select sum(case when Ssex = N'
    ' then 1 else 0 end) [男生人数],sum(case when Ssex = N'' then 1 else 0 end) [女生人数] from student
    select case when Ssex = N'
    ' 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'' then N'男生人数' else N'女生人数' end

    --29、查询名字中含有""字的学生信息
    select * from student where sname like N'%%'
    select * from student where charindex(N'
    ' , 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.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
    order 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
    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 < 60

    --35、查询所有学生的课程及分数情况;
    select Student.* , Course.Cname , SC.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C#
    order by Student.S# , SC.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 Student.* , Course.Cname , SC.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C# and SC.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 < 60
    order by Student.S# , SC.C#

    --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.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C# and SC.C# = '01' and SC.score >= 80
    order by Student.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
    where Student.S# = SC.S# and SC.C# = Course.C# and Course.T# = Teacher.T# 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.C# , SC.score 
    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' and
    SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C# = Course.C# and Course.T# = Teacher.T# 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 C# , score from SC group by C# , score having count(1) > 1) n
    where m.C#= n.C# and m.score = n.score order by m.C# , m.score , m.S#
    --
    方法2
    select m.* from SC m where exists (select 1 from (select C# , score from SC group by C# , score having count(1) > 1) n
    where m.C#= n.C# and m.score = n.score) order by m.C# , m.score , m.S#

    --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 C# = T.C# order by score desc) order by t.C# , 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.C# , Course.Cname , count(*) [
    学生人数]
    from Course , SC
    where Course.C# = SC.C#
    group by  Course.C# , Course.Cname
    having count(*) >= 5
    order by [
    学生人数] desc , Course.C#

    --44、检索至少选修两门课程的学生学号
    select student.S# , student.Sname
    from student , SC
    where student.S# = SC.S#
    group by student.S# , student.Sname
    having count(1) >= 2
    order by student.S#

    --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 S# in
    (select S# from sc group by S# 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.S# not in
    (
      select distinct m.S# from
      (
        select S# , C# from student , course
      ) m where not exists (select 1 from sc n where n.S# = m.S# and n.C# = m.C#)
    )
    --
    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.S# from
      (
        select S# , C# from student , course
      ) m where not exists (select 1 from sc n where n.S# = m.S# and n.C# = m.C#)
    ) k where k.S# = t.S#
    )

    --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

    drop table  Student,Course,Teacher,SC

    展开全文
  • color:rgb(39,158,176)">sql语句多查询(学生表/课程/教师/成绩 )作者:海豚湾孬蛋 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...
    
    

    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 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(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')
    create 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('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.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 on 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 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) < 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.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)) >= 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 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.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(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
    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.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.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 Student.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、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.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
    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 sc.S# from sc , course , 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 student.S# in (select S# from SC where score < 60 group by S# having count(1) >= 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 and sc.C# = '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.S# 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.S# = b.S#
    left join Course c on b.C# = c.C#
    group by a.S# , a.Sname
    order by 平均分 desc
    --17.2 SQL 2000 动态
    declare @sql nvarchar(4000)
    set @sql = 'select a.S# ' + 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.S# = b.S# left join Course c on b.C# = c.C#
    group by a.S# , 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.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) + 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.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] ,
             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
    ) 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.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
    ) 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.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 = (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
    select 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 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.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
    order 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
    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 < 60

    --35、查询所有学生的课程及分数情况;
    select Student.* , Course.Cname , SC.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C#
    order by Student.S# , SC.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 Student.* , Course.Cname , SC.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C# and SC.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 < 60
    order by Student.S# , SC.C#

    --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.C# , SC.score 
    from Student, SC , Course
    where Student.S# = SC.S# and SC.C# = Course.C# and SC.C# = '01' and SC.score >= 80
    order by Student.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
    where Student.S# = SC.S# and SC.C# = Course.C# and Course.T# = Teacher.T# 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.C# , SC.score 
    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' and
    SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C# = Course.C# and Course.T# = Teacher.T# 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 C# , score from SC group by C# , score having count(1) > 1) n
    where m.C#= n.C# and m.score = n.score order by m.C# , m.score , m.S#
    --方法2
    select m.* from SC m where exists (select 1 from (select C# , score from SC group by C# , score having count(1) > 1) n
    where m.C#= n.C# and m.score = n.score) order by m.C# , m.score , m.S#

    --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 C# = T.C# order by score desc) order by t.C# , 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.C# , Course.Cname , count(*) [学生人数]
    from Course , SC
    where Course.C# = SC.C#
    group by  Course.C# , Course.Cname
    having count(*) >= 5
    order by [学生人数] desc , Course.C#

    --44、检索至少选修两门课程的学生学号
    select student.S# , student.Sname
    from student , SC
    where student.S# = SC.S#
    group by student.S# , student.Sname
    having count(1) >= 2
    order by student.S#

    --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 S# in
    (select S# from sc group by S# 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.S# not in
    (
      select distinct m.S# from
      (
        select S# , C# from student , course
      ) m where not exists (select 1 from sc n where n.S# = m.S# and n.C# = m.C#)
    )
    --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.S# from
      (
        select S# , C# from student , course
      ) m where not exists (select 1 from sc n where n.S# = m.S# and n.C# = m.C#)
    ) k where k.S# = t.S#
    )

    --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

    drop table  Student,Course,Teacher,SC

    展开全文
  • //学生类封装 ,(链表) 排序成绩 无动态数组 ,动态开辟空间 以及静态函数,数据成员的使用#include #include #include #include using namespace std; #define MAXNO 200 clas

     //学生类封装  ,(链表) 排序成绩 无动态数组 ,动态开辟空间 以及静态函数,数据成员的使用

    风骚的全指针结构体走位

     
    #include <iostream>
    #include <string>
    using namespace std;
    
    
    
    class CStudent;
    typedef struct stunode{
    	CStudent *stuPoint;
    	struct stunode * next;
    }*Linklist;//指针结构体 是对对象的一个映射链表
    
    class CStudent
    {
    public:
    	CStudent();
    	CStudent(string name,string num,float score);
    	virtual ~CStudent();
    public:
    
    	bool init(string name,string num,float score);//对学生进行初始化
    	static bool output();//输出
    	static bool insertsort();//插入法排序
    private:
    	 /*struct stunode{
    		CStudent *stuPoint;
    		struct stunode * next;
    	}*Linklist; zhe li de wenti shi xia mian buzhi zen me ding yi*/
    	static Linklist linklist;//生成链表的头节点,,,,,,这里仅仅是对静态数据成员进行引用性声明
    
    private:
    	static int count;//记录总学生人数的个数
    	float score	;//学生的分数
    	string name,num;//名字和学号
    };
    
    
    int CStudent::count = 0;
    Linklist CStudent::linklist = NULL;//静态数据成员必须在命名空间作用域的某个地方使用类名限定定义性声明
    CStudent::CStudent()
    {
    /*	static struct stunode * last = NULL;
    	count++;
    	linklist = new (struct stunode *)();
    	linklist->next = last;
    	linklist->stuPoint = this;
    	last = linklist;
    	//zhe li de wenti shi buneng diaoyong linklist??? 在构造函数中不能使用this不能赋值给linklist->stupoint???
    */
    	//stuPointList[count] = this;
    }
    
    CStudent::~CStudent()
    {
    }
    
    CStudent::CStudent(string name,string num,float score):name(name),num(num),score(score)
    {
    }
    bool CStudent::insertsort()
    {
    Linklist p = linklist,q=linklist,newhead,comp,last;//用来取 主链的 头节点,下次要取源的节点和生成新链表的 头节点,以及要比较的节点和记录要插入节点的前驱
    	for(int i=1;i<=count;i++){
    		q = q->next;// 把下次要从源中取的点 先记下
    		if(i == 1){newhead = p;}
    		else{
    			comp = newhead;
    			//找到要插入的位置 comp 为与之想比较的节点
    			while( comp!= NULL &&p->stuPoint->score < comp->stuPoint->score  )
    			{
    				last = comp;
    				comp = comp->next;
    			}
    			p->next = comp;
    			if(comp == newhead)
    			{
    				newhead = p;
    			}
    			else{
    				last->next = p;
    			}
    		}
    		last = p;
    		p = q;//待插入节点在 源 中取节点
    		last->next = NULL;
    	}
    	linklist = newhead;//新链表的头结点变了
    	return true;
    }
    bool CStudent::init(string name1,string num1,float score1)//:name(name),num(num),score(score)
    {
    	name = name1;
    	num = num1;
    	score = score1;
    
    	count++;
    	Linklist temp;
    	temp = new (struct stunode )();
    	temp->next = linklist;
    	temp->stuPoint = this;//将每次产生的学生对象的地址 与 映射链表 对接上
    	linklist = temp;
    
    	return true;
    }
    bool CStudent::output()
    {
    	struct stunode *temp = linklist;
    	cout<<"num\t"<<"name\t"<<"score"<<endl;;
    	while(temp != NULL)
    	{
    		cout<<temp->stuPoint->num<<"\t";
    		cout<<temp->stuPoint->name<<"\t";
    		cout<<temp->stuPoint->score<<endl;
    
    		temp = temp->next;
    	}
    	return true;
    }
    int main()
    {
    	freopen("in.txt","r",stdin);
    	CStudent *list;
    	string name,num;
    	float score;
    	while( 1 )
    	{
    		cin>>name>>num>>score;//脭脷 string脰脨虏脜脛脺cin  string.h虏禄脛脺cin
    		if(name == "0")
    			break;
    		list = new (CStudent);//能否在这里进行初始化 ?
    		list->init(name,num,score);
    	}
    	cout<<"before sort:"<<endl;
    	CStudent::output();
    	CStudent::insertsort();
    	cout<<"after insertsort:"<<endl;
    	CStudent::output();
    
    	return 0;
    }/*
     * main.cpp
     *
     *  Created on: Jul 21, 2011
     *      Author: root
     */
    
    
    
    

    /*in.txt 附上测试文件
    a  1 99
    b  2 100
    c  3 101
    d  4 102
    e  5 123
    0  0 0
    */
    
    
    
    
    
    
    
    
    
    
    
    
    

    展开全文
  • mysql经典面试题之学生成绩

    千次阅读 2019-09-24 11:25:32
    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 Course(CID,Cname,TID) --CID --课程编号,Cname 课程名称,TID 教师编号 3.教师 Teacher(TID,...

    需要数据库表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

    添加测试数据1.学生表

    create table Student(SID varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10));

    insert into Student values('01' , '赵雷' , '1990-01-01' , '男');

    insert into Student values('02' , '钱电' , '1990-12-21' , '男');

    insert into Student values('03' , '孙风' , '1990-05-20' , '男');

    insert into Student values('04' , '李云' , '1990-08-06' , '男');

    insert into Student values('05' , '周梅' , '1991-12-01' , '女');

    insert into Student values('06' , '吴兰' , '1992-03-01' , '女');

    insert into Student values('07' , '郑竹' , '1989-07-01' , '女');

    insert into Student values('08' , '王菊' , '1990-01-20' , '女');

    2.课程表

    create table Course(CID varchar(10),Cname nvarchar(10),TID varchar(10));

    insert into Course values('01' , '语文' , '02');

    insert into Course values('02' , '数学' , '01');

    insert into Course values('03' , '英语' , '03');

    3.教师表

    create table Teacher(TID varchar(10),Tname nvarchar(10));

    insert into Teacher values('01' , '张三');

    insert into Teacher values('02' , '李四');

    insert into Teacher values('03' , '王五');

    4.成绩表

    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);

    --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 '李%'

    --方法2

    select count(Tname) 李姓老师的数量 from Teacher where left(Tname,1) = '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 = '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 = '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 = '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 '语文' then b.score else null end) 语文 ,

    max(case c.Cname when '数学' then b.score else null end) 数学 ,

    max(case c.Cname when '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 ' + '学生编号' + ' , a.Sname ' + '学生姓名'

    select @sql = @sql + ',max(case c.Cname when '''+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)) ' + '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'

    exec(@sql)

    --18、查询各科成绩最高分、最低分和平均分:以如下形式显示:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率

    --及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90

    --方法1

    select m.CID 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.score) as decimal(18,2)) 平均分 ,

    cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 及格率 ,

    cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 中等率 ,

    cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优良率 ,

    cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优秀率

    from Course m , SC n

    where m.CID = n.CID

    group by m.CID , m.Cname

    order by m.CID

    --方法2

    select m.CID 课程编号 , m.Cname 课程名称 ,

    (select max(score) from SC where CID = m.CID) 最高分 ,

    (select min(score) from SC where CID = m.CID) 最低分 ,

    (select cast(avg(score) as decimal(18,2)) from SC where CID = m.CID) 平均分 ,

    cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 及格率,

    cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 中等率 ,

    cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优良率 ,

    cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 优秀率

    from Course m

    order by m.CID

    --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

    select t.* , px = (select count(1) from SC where CID = t.CID and score > t.score) + 1 from sc t order by t.cid , px

    --Score重复时合并名次

    select t.* , px = (select count(distinct score) from SC where CID = t.CID and score >= t.score) from sc t order by t.cid , 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() over(partition by cid order by score desc) from sc t order by t.CID , px

    --Score重复时合并名次(DENSE_RANK完成)

    select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t order by t.CID , 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.SID 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.SID = n.SID

    group by m.SID , 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.SID 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.SID = n.SID

    group by m.SID , m.Sname

    ) t2 where 总成绩 > t1.总成绩) + 1 from

    (

    select m.SID 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.SID = n.SID

    group by m.SID , m.Sname

    ) t1

    order by px

    select t1.* , px = (select count(distinct 总成绩) from

    (

    select m.SID 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.SID = n.SID

    group by m.SID , m.Sname

    ) t2 where 总成绩 >= t1.总成绩) from

    (

    select m.SID 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.SID = n.SID

    group by m.SID , 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.SID 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.SID = n.SID

    group by m.SID , m.Sname

    ) t

    order by px

    select t.* , px = DENSE_RANK() over(order by 总成绩 desc) from

    (

    select m.SID 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.SID = n.SID

    group by m.SID , m.Sname

    ) t

    order 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.TID , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score

    from Teacher m , Course n , SC o

    where m.TID = n.TID and n.CID = o.CID

    group by m.TID , m.Tname

    order by avg_score 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重复时保留名次空缺

    select * from (select t.* , px = (select count(1) from SC where CID = t.CID and score > t.score) + 1 from sc t) m where px between 2 and 3 order by m.cid , m.px

    --Score重复时合并名次

    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 2 and 3 order by m.cid , 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 cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , m.px

    --Score重复时合并名次(DENSE_RANK完成)

    select * from (select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , 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.CID 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 < 85 then 1 else 0 end) 70-85 ,

    sum(case when score >= 60 and score < 70 then 1 else 0 end) 60-70 ,

    sum(case when score < 60 then 1 else 0 end) 0-60

    from sc , Course

    where SC.CID = Course.CID

    group by Course.CID , Course.Cname

    order by Course.CID

    --纵向显示1(显示存在的分数段)

    select m.CID 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 < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end) ,

    count(1) 数量

    from Course m , sc n

    where m.CID = n.CID

    group by m.CID , m.Cname , (

    case when n.score >= 85 then '85-100'

    when n.score >= 70 and n.score < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end)

    order by m.CID , m.Cname , 分数段

    --纵向显示2(显示存在的分数段,不存在的分数段用0显示)

    select m.CID 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 < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end) ,

    count(1) 数量

    from Course m , sc n

    where m.CID = n.CID

    group by all m.CID , m.Cname , (

    case when n.score >= 85 then '85-100'

    when n.score >= 70 and n.score < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end)

    order by m.CID , 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.CID 课程编号, m.Cname 课程名称,

    (select count(1) from SC where CID = m.CID and score < 60) 0-60 ,

    cast((select count(1) from SC where CID = m.CID and score < 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,

    (select count(1) from SC where CID = m.CID and score >= 60 and score < 70) 60-70 ,

    cast((select count(1) from SC where CID = m.CID and score >= 60 and score < 70)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,

    (select count(1) from SC where CID = m.CID and score >= 70 and score < 85) 70-85 ,

    cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比 ,

    (select count(1) from SC where CID = m.CID and score >= 85) 85-100 ,

    cast((select count(1) from SC where CID = m.CID and score >= 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2)) 百分比

    from Course m

    order by m.CID

    --纵向显示1(显示存在的分数段)

    select m.CID 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 < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end) ,

    count(1) 数量 ,

    cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2)) 百分比

    from Course m , sc n

    where m.CID = n.CID

    group by m.CID , m.Cname , (

    case when n.score >= 85 then '85-100'

    when n.score >= 70 and n.score < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end)

    order by m.CID , m.Cname , 分数段

    --纵向显示2(显示存在的分数段,不存在的分数段用0显示)

    select m.CID 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 < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end) ,

    count(1) 数量 ,

    cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2)) 百分比

    from Course m , sc n

    where m.CID = n.CID

    group by all m.CID , m.Cname , (

    case when n.score >= 85 then '85-100'

    when n.score >= 70 and n.score < 85 then '70-85'

    when n.score >= 60 and n.score < 70 then '60-70'

    else '0-60'

    end)

    order by m.CID , m.Cname , 分数段

    --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

    转载于:https://www.cnblogs.com/chaojiyingxiong/p/9380820.html

    展开全文
  • 对象数组 —— 学生信息

    千次阅读 2020-10-26 12:43:51
    本关任务:编写一个能管理多条学生信息的程序。 相关知识 为了完成本关任务,你需要掌握构造函数与析构函数的调用和对象数组的使用。 构造函数与析构函数的调用 构造函数不能直接调用,只能通过声明一个对象或者使用...
  • 用C语言链表编写学生成绩管理系统

    万次阅读 多人点赞 2018-09-06 20:04:12
    一、设计题目:学生成绩管理系统 二、目的与要求 每位学生记录包含有学号、姓名、性别、出生日期、三门功课的成绩(高等数学、大学英语、C语言)、总分和平均分 系统菜单: (1)录入学生记录 (2)添加学生...
  • 静态测试动态测试相关知识点

    千次阅读 2015-09-27 11:41:46
    目 录 知识总结................................. 5 第一章........................................ 5 第二章软件测试基础............................. ...第四章软件测试的分类.............
  • 注:本文为个人学习的测试题,此题目如果可以自己独立完成基本上算入门了,题目来源于上官可编程陈老师出的的一个c语测试题,有想学习c语了解嵌入式的朋友可以抖音搜索上官可编程。 链表A,每个节点存放一个新的链表...
  • 实验11-2-1 建立学生信息链表 (20分)

    千次阅读 2020-02-03 14:07:17
    本题要求实现一个将输入的学生成绩组织成单向链表的简单函数。 函数接口定义: void input(); 该函数利用scanf从输入中获取学生的信息,并将其组织成单向链表。链表节点结构定义如下: struct stud_node { int num; ...
  • 建议使用动态内存分配来实现。 输入格式: 输入第一行首先给出一个正整数N,表示学生的个数。接下来一行给出N个学生的成绩,数字间以空格分隔。 输出格式: 按照以下格式输出: average = 平均成绩 max = 最高...
  • Student(S#,Sname,Sage,Ssex) 学生表 Course(C#,Cname,T#) 课程 SC(S#,C#,score) 成绩 Teacher(T#,Tname) 教师
  • 有 n 个学生站成一排,每个学生有一个能力值,牛牛想从这 n 个学生中按照顺序选取 k 名学生,要求相邻两个学生的位置编号的差不超过 d,使得这 k 个学生的能力值的乘积最大,你能返回最大的乘积吗? 输入描述: ...
  • 第17周报告2:动态链表

    千次阅读 2011-12-19 15:06:36
    任务:数据来自score.txt,在程序中建立一个动态链表:每读入一个同学的数据,计算总分,分配结点的存储空间并赋值,并建立起前后相链的关系。在建立链表的同时,要进行统计,以便于求出所有同学总分的平均成绩。...
  • 链表《5》使用链表实现学生成绩管理系统

    万次阅读 多人点赞 2014-05-05 18:45:29
    上次我使用动态数组和结果
  • 使用数组(或者动态数组)完成一个简单的学生信息管理系统,要求包含学生添加,删除,查询,修改等功能;学生信息包含: 学号 姓名 性别 专业 年龄 成绩 首先创建一个学生类 package Exp; public class ...
  • Deep Knowledge Tracing and Dynamic Student Classification for ...i)捕获学生的学习能力,并以固定的时间间隔将学生动态分配给具有相似能力的不同群体, ii)并且将这些信息与称为深度知识追踪的递归神经网
  • C++第2周项目4——动态链表初试

    千次阅读 热门讨论 2013-03-09 22:28:30
    课程首页地址:http://blog.csdn.net/sxhelijian/article/details/7910565,本周题目链接:http://blog.csdn.net/sxhelijian/article/details/8635385【项目4-动态链表初试】数据依然来自score.txt,在程序中建立一...
  • C++第2周(春)项目6 动态链表初体验

    千次阅读 2014-03-08 17:40:24
     不要被貌似复杂的指针操作迷惑,这正是专业学生应该具备的思维基本功。涉及到链接如何建立的操作,在纸上画一画,想一想,道理就通了。  作为需要用脑作的实践,由画一画的形象思维,会形成头脑中不用画也很清楚...
  • 题目描述有 n 个学生站成一排,每个学生有一个能力值,牛牛想从这 n 个学生中按照顺序选取 k 名学生,要求相邻两个学生的位置编号的差不超过 d,使得这 k 个学生的能力值的乘积最大,你能返回最大的乘积吗?...
  • Python题目:学生信息管理系统-高级版(图形界面+MySQL数据库) 使用图形界面显示,选用list、tuple、dictionary或map等数据结构,操作数据库存储X个学生的三门课的成绩(机器学习、Python程序设计、研究生英语)...
  • 使用C++结合文件操作和链表实现学生成绩管理系统

    万次阅读 多人点赞 2014-08-30 12:56:05
    对于学生成绩管理系统,我是
  • “今日芯声”是读芯术推出的一档简读栏目,汇聚每日国内外最新最热的AI应用资讯,敬请关注。1、浙江一小学为学生佩戴头环监测走神,数据实时上传IT之家10月31日消息 根据新...
  • 我们经常会遇到所要制作的表格列数不固定的情况,比如第一列是科目名,后面的列是各个学生的情况(当然实际中一般都是第一列为学生,后面每列是科目,这里只是举个例子),而birt的视图界面在建立table时,会要求...
  • 动态规划,这个是算法里面一直比较难的,当我拿到这个题的时候,有点难以下手,虽然知道要用动态规划但是如何用,自己完全不知道,首先想到找出这个n个数中k个最大的相乘 ,但是很遗憾不对,①要求相邻两个学生之间...
  • jq 动态增加、删除tr行

    千次阅读 2017-08-30 22:14:57
    利用jquery给指定的table添加一行、删除一行 ////////添加一行、删除一行封装方法... * tab id * row 行数,如:0->第一行 1->第二行 -2->倒数第二行 -1->最后一行 * trHtml 添加行的html代码 * */
  • 提高学生设计实验、发现问题、分析问题和解决问题的能力,并学习撰写规范的科学研究报告。 2. 实验要求: 任选C高级程序语言编写源程序,在Linux操作系统下调试通过,测试正确。 3. 实验内容:  (1).用C或其
  • 第五章 学生成绩管理系统系统测试 5.1 概述 37 5.2 测试方法 37 5.2.1 界面测试 37 5.2.2 功能测试 37 5.2.3 功能测试边界测试\越界测试技术详述 38 5.2.4 状态测试技术 38 5.2.5 竞争条件测试技术 38 5.2.6 ...
  • 大四学生如何求Java工作?

    万次阅读 多人点赞 2020-03-29 17:17:00
    有读者给我留言:大四的学生如何求一份Java的工作?并且还特别强调了他非科班和文凭不出众的事实背景。我想这个问题具有代表性,所以写一些个人的建议,希望能帮到一些人。 定位 要想求得一份工作,首先需要对自己...
  • 记录一下:从此题开始正式慢慢研究一下动态规划。源代码来自牛客网; 题目:有 n 个学生站成一排,每个学生有一个能力值,牛牛想从这 n 个学生中按照顺序选取 k 名学生,要求相邻两个学生的位置编号的差不超过 d,...

空空如也

空空如也

1 2 3 4 5 ... 20
收藏数 46,822
精华内容 18,728
关键字:

学生动态监测表