说明：以下五十个语句都按照测试数据进行过测试，最好每次只单独运行一个语句。
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))
insertintoTeachervalues(’01’, N’张三’)
insertintoTeachervalues(’02’, N’李四’)
insertintoTeachervalues(’03’, N’王五’)
createtableSC(S#varchar(10),C#varchar(10),scoredecimal(18,1))
insertintoSCvalues(’01’,’01’,80)
insertintoSCvalues(’01’,’02’,90)
insertintoSCvalues(’01’,’03’,99)
insertintoSCvalues(’02’,’01’,70)
insertintoSCvalues(’02’,’02’,60)
insertintoSCvalues(’02’,’03’,80)
insertintoSCvalues(’03’,’01’,80)
insertintoSCvalues(’03’,’02’,80)
insertintoSCvalues(’03’,’03’,80)
insertintoSCvalues(’04’,’01’,50)
insertintoSCvalues(’04’,’02’,30)
insertintoSCvalues(’04’,’03’,20)
insertintoSCvalues(’05’,’01’,76)
insertintoSCvalues(’05’,’02’,87)
insertintoSCvalues(’06’,’01’,31)
insertintoSCvalues(’06’,’03’,34)
insertintoSCvalues(’07’,’02’,89)
insertintoSCvalues(’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)
注：
LEFT JOIN 关键字从左表（table1）返回所有的行，即使右表（table2）中没有匹配。如果右表中没有匹配，则结果为 NULL。
对于left join，不管on后面跟什么条件，左表的数据全部查出来，因此要想过滤需把条件放到where后面
对于inner join，满足on后面的条件表的数据才能查出，可以起到过滤作用。也可以把条件放到where后 面。
–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
selecta.*, b.score[课程’01’的分数],c.score[课程’02’的分数]fromStudent a , SC b , SC c
wherea.S#=b.S#anda.S#=c.S#andb.C#=’01’andc.C#=’02’andb.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
selecta.*, b.score[课程”01″的分数],c.score[课程”02″的分数]fromStudent a
leftjoinSC bona.S#=b.S#andb.C#=’01’
leftjoinSC cona.S#=c.S#andc.C#=’02’
whereisnull(b.score,0)
–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
orderby a.S#
注：CAST函数用于将某种数据类型的表达式显式转换为另一种数据类型 (CAST (expression AS data_type))
group by:根据(by)一定的规则进行分组(Group),它的作用是通过一定的规则将一个数据集划分成若干个小的区域，然后针对若干个小区域进行数据处理
聚合函数，例如SUM, COUNT, MAX, AVG等。这些函数和其它函数的根本区别就是它们一般作用在多条记录上。
当出现了 聚合函数的时候 一般情况下 我们就会用到group by来进行分组，而且group by出现的字段一般都要在查询中出现
–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。
selecta.S# , a.Sname ,cast(avg(b.score)asdecimal(18,2)) avg_score
fromStudent a , sc b
wherea.S#=b.S#
groupbya.S# , a.Sname
havingcast(avg(b.score)asdecimal(18,2))<60
orderbya.S#
–4.2、查询在sc表中不存在成绩的学生信息的SQL语句。
selecta.S# , a.Sname ,isnull(cast(avg(b.score)asdecimal(18,2)),0) avg_score
fromStudent aleftjoinsc b
ona.S#=b.S#
groupbya.S# , a.Sname
havingisnull(cast(avg(b.score)asdecimal(18,2)),0)<60
orderbya.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。
selecta.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]
fromStudent a , SC b
wherea.S#=b.S#
groupbya.S#,a.Sname
orderbya.S#
–5.2、查询所有(包括有成绩和无成绩)的SQL。
selecta.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]
fromStudent aleftjoinSC b
ona.S#=b.S#
groupbya.S#,a.Sname
orderbya.S#
–6、查询”李”姓老师的数量
–方法1
selectcount(Tname)[“李”姓老师的数量]fromTeacherwhereTnamelikeN’李%’
–方法2
selectcount(Tname)[“李”姓老师的数量]fromTeacherwhereleft(Tname,1)=N’李’
/*
“李”姓老师的数量
———–
1
*/
–7、查询学过”张三”老师授课的同学的信息
selectdistinctStudent.*fromStudent , SC , Course , Teacher
whereStudent.S#=SC.S#andSC.C#=Course.C#andCourse.T#=Teacher.T#andTeacher.Tname=N’张三’
orderbyStudent.S#
–8、查询没学过”张三”老师授课的同学的信息
selectm.*fromStudent mwhereS#notin(selectdistinctSC.S#fromSC ,
Course , TeacherwhereSC.C#=Course.C#andCourse.T#=Teacher.T#andTeacher.Tname=N’张三’)orderbym.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
selectStudent.*fromStudent , SCwhereStudent.S#=SC.S#andSC.C#=’01’andexists(Select1fromSC SC_2whereSC_2.S#=SC.S#andSC_2.C#=’02’)orderbyStudent.S#
–方法2
selectStudent.*fromStudent , SCwhereStudent.S#=SC.S#andSC.C#=’02’andexists(Select1fromSC SC_2whereSC_2.S#=SC.S#andSC_2.C#=’01’)orderbyStudent.S#
–方法3
selectm.*fromStudent mwhereS#in
(
selectS#from
(
selectdistinctS#fromSCwhereC#=’01’
unionall
selectdistinctS#fromSCwhereC#=’02’
) tgroupbyS#havingcount(1)=2
)
orderbym.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
selectStudent.*fromStudent , SCwhereStudent.S#=SC.S#andSC.C#=’01’andnotexists(Select1fromSC SC_2whereSC_2.S#=SC.S#andSC_2.C#=’02’)orderbyStudent.S#
–方法2
selectStudent.*fromStudent , SCwhereStudent.S#=SC.S#andSC.C#=’01’andStudent.S#notin(SelectSC_2.S#fromSC
SC_2whereSC_2.S#=SC.S#andSC_2.C#=’02’)orderbyStudent.S#
–11、查询没有学全所有课程的同学的信息
–11.1、
selectStudent.*
fromStudent , SC
whereStudent.S#=SC.S#
groupbyStudent.S# , Student.Sname , Student.Sage ,
Student.Ssexhavingcount(C#)<(selectcount(C#)fromCourse)
–11.2
selectStudent.*
fromStudentleftjoinSC
onStudent.S#=SC.S#
groupbyStudent.S# , Student.Sname , Student.Sage ,
Student.Ssexhavingcount(C#)<(selectcount(C#)fromCourse)
–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
selectdistinctStudent.*fromStudent , SCwhereStudent.S#=SC.S#andSC.C#in(selectC#fromSCwhereS#=’01’)andStudent.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
selectStudent.*fromStudentwhereS#in
(selectdistinctSC.S#fromSCwhereS#<>’01’andSC.C#in(selectdistinctC#fromSCwhereS#=’01’)
groupbySC.S#havingcount(1)=(selectcount(1)fromSCwhereS#=’01’))
–14、查询没学过”张三”老师讲授的任一门课程的学生姓名
selectstudent.*fromstudentwherestudent.S#notin
(selectdistinctsc.S#fromsc ,
course , teacherwheresc.C#=course.C#andcourse.T#=teacher.T#andteacher.tname=N’张三’)
orderbystudent.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
selectstudent.S# , student.sname ,cast(avg(score)asdecimal(18,2)) avg_scorefromstudent , sc
wherestudent.S#=SC.S#andstudent.S#in(selectS#fromSCwherescore<60groupbyS#havingcount(1)>=2)
groupbystudent.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
selectstudent.*, sc.C# , sc.scorefromstudent , sc
wherestudent.S#=SC.S#andsc.score<60andsc.C#=’01’
orderbysc.scoredesc
–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
selecta.S#学生编号, a.Sname学生姓名,
max(casec.CnamewhenN’语文’thenb.scoreelsenullend)[语文],
max(casec.CnamewhenN’数学’thenb.scoreelsenullend)[数学],
max(casec.CnamewhenN’英语’thenb.scoreelsenullend)[英语],
cast(avg(b.score)asdecimal(18,2))平均分
fromStudent a
leftjoinSC bona.S#=b.S#
leftjoinCourse conb.C#=c.C#
groupbya.S# , a.Sname
orderby平均分desc
–17.2 SQL 2000动态
declare@sqlnvarchar(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(selectdistinctCnamefromCourse)ast
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)》。
二；
–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
selectm.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)asdecimal(18,2))[平均分],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=60)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[及格率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=70andscore<80)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[中等率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=80andscore<90)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[优良率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=90)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[优秀率(%)]
fromCourse m , SC n
wherem.C#=n.C#
groupbym.C# , m.Cname
orderbym.C#
–方法2
selectm.C#[课程编号], m.Cname[课程名称],
(selectmax(score)fromSCwhereC#=m.C#)[最高分],
(selectmin(score)fromSCwhereC#=m.C#)[最低分],
(selectcast(avg(score)asdecimal(18,2))fromSCwhereC#=m.C#)[平均分],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=60)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[及格率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=70andscore<80)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[中等率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=80andscore<90)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[优良率(%)],
cast((selectcount(1)fromSCwhereC#=m.C#andscore>=90)*100.0/(selectcount(1)fromSCwhereC#=m.C#)asdecimal(18,2))[优秀率(%)]
fromCourse m
orderbym.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
selectt.*, px=(selectcount(1)fromSCwhereC#=t.C#andscore>t.score)+1fromsc torderbyt.c# , px
–Score重复时合并名次
selectt.*, px=(selectcount(distinctscore)fromSCwhereC#=t.C#andscore>=t.score)fromsc torderbyt.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)
selectt.*, px=rank()over(partitionbyc#orderbyscoredesc)fromsc torderbyt.C# , px
–Score重复时合并名次(DENSE_RANK完成)
selectt.*, px=DENSE_RANK(
三；
–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
