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)
–17.3 有关sql 2005的动静态写法参见我的文章《普通行列转换(version 2.0)》或《普通行列转换(version 3.0)》。
–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.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.score) 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 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [中等率(%)],
cast((select count(1) from SC where C# = m.C# and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [优良率(%)],
cast((select count(1) from SC where C# = m.C# and score >= 90)*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [优秀率(%)]
from Course 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 >= 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 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [中等率(%)],
cast((select count(1) from SC where C# = m.C# and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [优良率(%)],
cast((select count(1) from SC where C# = m.C# 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
select 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() over(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 on 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.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.Sname
) 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] ,
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
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.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 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 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# , 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) from 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 < 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.C# = Course.C#
group by Course.C# , Course.Cname
order by Course.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 < 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.C# = n.C#
group by m.C# , 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.C# , 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.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 < 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.C# = n.C#
group by all m.C# , 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.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,
(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)*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [百分比(%)],
(select count(1) from SC where C# = m.C# and score >= 60 and score < 70) [60-70],
cast((select count(1) from SC where C# = m.C# and score >= 60 and score < 70)*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [百分比(%)],
(select count(1) from SC where C# = m.C# and score >= 70 and score < 85) [70-85],
cast((select count(1) from SC where C# = m.C# and score >= 70 and score < 85)*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) [百分比(%)],
(select count(1) from SC where C# = m.C# 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 < 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 C# = m.C#) as decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C# = n.C#
group by m.C# , 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.C# , 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.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 < 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 C# = m.C#) as decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C# = n.C#
group by all m.C# , 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.C# , 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.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