Sql高级查询练习题(有答案!)

--create database practice
--on primary
--(name='practice_data',
--filename='E:\sqlspace\practice_data.mdf'
--)
--use practice

--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

use practice
select * from SC
select * from Teacher
select * from Student
select * from Course



--1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

select a.*,b.score as '01课程分数',c.score as '02课程分数'
from student as a
left join SC as b on a.S#=b.S# and b.C#='01'
left join SC as c on c.S#=a.S# and c.C#='02'
where b.score>ISNULL(c.score,0)


--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
select a.*,c.score as '01课程分数',b.score as '02课程分数'
from Student as a
left join SC as c on c.S#=a.S# and c.C#='01'
left join SC as b on b.S#=a.S# and b.C#='02'
where c.score
--2.1、查询同时存在"01"课程和"02"课程的情况
select a.*,b.score as'01课程分数',c.score as '02课程分数'
from student as a,SC as b ,SC as c
where b.C#='01' and c.C#='02'and b.S#=a.S# and c.S#=a.S# and b.score


----------------------------------------------
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
--分析：即如果没有01的分数，有02的分数就认为是01的分数比02的分数低，就要用isnull 函数，判断是否为空，如果空则用0分替代
--所以查询时要用left做外链接,把所有的人都查进来

select a.*,b.score as '01课程分数' ,c.score as '02课程分数'
from Student as a
left join SC as b on a.S#=b.S# and b.C#='01'
left join SC as c on c.S#=a.S# and c.C#='02'
where ISNULL(b.score,0)-------------------------------------------------------------------

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(c.score) as decimal(5,2))
from Student as a
join SC as c on a.S#=c.S#
group by a.S#,a.Sname
having CAST(avg(c.score) as decimal(5,2))>=60
order by a.S#

----------------------------------------------------------
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
select a.S# 学号,a.Sname 姓名,CAST(AVG(b.score) as decimal(5,2)) 平均分
from Student as a ,SC as b
where a.S#=b.S#
group by a.S#,a.Sname
having CAST(AVG(b.score) as decimal(5,2))<60

----------------------------------------------------


--4.1、查询在sc表存在成绩的学生信息的SQL语句。
select a.*
from student as a
where a.S# in( select distinct(b.S#) from SC as b )



---------------------------------------------------
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.*
from student as a
where a.S# not in (select b.S# from SC as b )



--------------------------------------------------
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
select a.S# 学生编号,a.Sname 姓名,COUNT(b.C#) 选课总数, 总成绩=isnull(convert(varchar(20),sum(b.score)),'空')---聚合函数里面的内容可以不放分组里面
from Student as a
join SC as b
on a.S#=b.S#
group by a.S#,b.S#,a.Sname
----------------------------------------

use practice
--5.1、查询所有有成绩的SQL。
select a.S# 学号,a.Sname 姓名,count(b.C#) 选课总数,SUM(b.score) 总成绩
from Student as a ,SC as b
where a.S#=b.S#
group by a.S#,a.Sname






------------------------------------
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# 学号,a.Sname 姓名,count(b.c#) 选课总数,sum(b.score) 总分
from Student as a left join
SC as b on a.S#=b.S#
group by a.S#,a.Sname






--------------------
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
select count(*) 李老师的数量
from Teacher
where left(Tname,1)=N'李'
--where Tname like'李%'

-------------------
--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'
/*
"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
select a.*
from student as a
where a.S# in (select distinct(b.S#) from SC as b ,Course as c ,Teacher as d where b.C#=c.C# and c.T#=c.T# and d.Tname='张三')

-----------------------------------------------------------------------------------------------



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 a.*
from student as a
where a.S# not in (se

lect distinct(b.S#) from SC as b ,Course as c ,Teacher as d where b.C#=c.C# and c.T#=c.T# and d.Tname='张三')
--------------------------------------------------------------
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
select a.*
from Student as a
join SC as b on b.S#=a.S#
join SC as c on c.S#=a.S#
where c.C#='01' and b.C#='02'


---方法二
select a.*
from Student as a
join SC as c on c.S#=a.S#
where c.C#='01' and a.S# in (select distinct(b.S#) from SC as b where b.C#='02')
--方法三
select a.*
from student as a
join SC as b on b.S#=a.S#
where b.C#='01' and exists(select * from SC as c where c.S#=a.S# and c.C#='02')
---方法三
select a.*
from student as a
where a.S# in (
select s# from
(
select distinct s# from SC where c#='01'
union all ----合并结果集，不会消除相同的结果
select distinct s# from sc where c#='02'
) b group by s# having count(*)>=2 --选中的课程数大于等于二

)

-----------------------------------------------------
--方法一
select a.*
from Student as a ,SC as b
where a.S#=b.S# and b.C#='01' and exists(select * from SC as c where c.S#=b.S# and c.C#='02')
----方法二
select a.*
from Student as a ,SC as b
where a.S#=b.S# and b.C#='01' and a.S# in (select S# from SC where C#='02')
--方法三
select a.*
from Student as a
where a.S# in
(
select S# from
(
select distinct S# from SC where C#='01'
union all
select distinct S# from SC where C#='02'

) b group by S# having COUNT(*)=2
)

----------------------------------------------------------------------------------------------------------
--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
select a.*
from student as a ,SC as b
where b.S#=a.S# and b.C#='01'--内链接，学过‘01'的才筛选
and not exists(select * from SC as c where c.S#=a.S# and c.C#='02')



select a.*
from student as a ,SC as b
where b.S#=a.S# and b.C#='01'--内链接，学过‘01'的才筛选
and b.S# not in (select c.S# from SC as c where c.S#=b.S# and c.C#='02')

----

------------------------------------------------------
--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
--要把所有的学生都考虑进来，一科都没有报的也要，注意这种特殊的分组
select a.*
from Student as a
left join SC as b
on a.S#=b.S#
group by a.S#,a.Sage,a.Sname,a.Ssex having COUNT(b.C#)<(select count(distinct(C#)) from Course )

----------------------------------------------------------------------------
--11.1、
--――把null忽略了，不合理
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)
--11.3
--=------------复杂高级查询，问 老师,递归查询，查第一个学生的第一个课程，然后，查第二个学生的第二个课程，........
select c.*
from student c
where exists
(
select *
from course b
where not exists( select *
from sc a
where a.c#=b.c# and c.s#=a.s#
)

)

--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
--切记不能是’01‘学生本人
select distinct a.*
from Student as a ,SC as b
where b.S#=a.S# and b.C# in( select distinct(c.C#) from SC as c where c.S#='01') and b.S#<>'01'



-------------------------------------------------------
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 a.*
from Student as a
where a.S# in
(select distinct(b.S#) from SC as b where b.S#<>'01' and b.C# in(select C# from SC where S#='01')
group by b.S# having COUNT(*)=(select count(*) from SC where S#='01'))
--------------------------------------------------------


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(*) = (select count(*) 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 a.Sname 姓名
from student as a
where a.S# not in (
select distinct(b.S#) from SC as b ,Teacher as c ,Course as d
where c.T#=d.T#

and c.Tname='张三' and b.C#=d.C#
)


--------------------------------
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 a.S#,Sname, cast(AVG(b.score) as decimal(5,2)) as 平均成绩
from Student as a
join SC as b
on a.S#=b.S#
where b.score<60
group by a.S#,a.Sname
having count(*)>=2




--------------------------------------------------------------------------------------------------
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 a.*,b.C#,b.score
from Student as a ,SC as b
where a.S#=b.S# and b.C#='01' and b.score<60
order by b.score desc


-----------------------------------------
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
select a.Sname 姓名,a.S# 学号 ,
max(case ame when N'语文' then b.score else null end) [语文],
max (case ame when N'数学' then b.score else null end ) [数学],
max(case ame when N'英语' then b.score else null end )[英语],
cast(avg(b.score) as decimal(5,2) ) 平均分
from Student as a
left join SC as b on b.S#=a.S#
left join Course as c on c.C#=b.C#
group by a.S#,a.Sname
order by avg(b.score) desc

-----------------------------------
--17.1 SQL 2000 静态
select a.S# 学生编号 , a.Sname 学生姓名 ,
max(case ame when N'语文' then b.score else null end) [语文],
max(case ame when N'数学' then b.score else null end) [数学],
max(case ame 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 ame 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'
print @sql
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
select a.C# 课程ID,ame 课程名,max(b.score) 最高分,min(b.score) 最低分,cast(avg(b.score) as decimal(5,2)) 平均分,
cast((select count(*) from SC as c where c.score>=60 and c.C#=a.C#)*100.0/(select count(*) from sc where C#=a.C#) as decimal(5,2) )[及格率(%)],
cast((select count(*) from sc where score>=70 and score<80 and C#=a.C#)*100.0/(select count(*) from sc where a.C#=C#) as decimal(5,2)) [中等率(%)],
cast((select count(*) from sc where score>=80 and score<90 and C#=a.C#)*100.0/(select count(*) from sc where a.C#=C#) as decimal(5,2))[优良率(%)],
cast((select count(*) from sc where score>=90 and C#=a.C#)*100.0/(select count(*) from sc where C#=a.C#) as decimal(5,2)) [优秀率(%)]
from Course as a
left join SC as b on b.C#=a.C#
group by a.C#,ame

--方法二
select a.C# 课程ID,ame 课程名,
(select max(score) from SC where C#=a.C#) as 最高分,
(select min(score) from SC where C#=a.C#) as 最低分,
(select avg(score) from sc where C#=a.C#) as 平均分,
cast((select count(*) from sc where C#=a.C# and score>=60 )*100.0/(select count(*) from sc where C#=a.C# )as decimal(5,2)) [合格率(%)],
cast((select count(*) from sc where C#=a.C# and score>=70 and score<80 )*100.0/(select count(*) from sc where C#=a.C# )as decimal(5,2)) [中等率(%)],
cast((select count(*) from sc where C#=a.C# and score>=80 and score<90 )*100.0/(select count(*) from sc where C#=a.C# )as decimal(5,2)) [优良率(%)],
cast((select count(*) from sc where C#=a.C# and score>=90 )*100.0/(select count(*) from sc where C#=a.C# )as decimal(5,2)) [优秀率(%)]
from Course as a
order by a.C#

------------------------------------------------------------------------------
--方法1
select m.C# [课程编号], ame [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# ,

ame
order by m.C#
--方法2
select m.C# [课程编号], ame [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（超）
select a.* ,px=(select count(*) from sc where C#=a.c# and score>a.score)+1
--- a.score是主体，首先找出和它课程号一样的所有的分数，然后进行比较，大于当前分数就记1，自己的排名就应该是2，所以应该在最后的查询结果后面加1
from SC as a
order by a.C# ,px

-----------------------------------------------------------------------------
--19.1 sql 2000用子查询完成
--Score重复时保留名次空缺--高级查询,问老师
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
---解析：子查询中count(1) 是记同科目中，有多少人的分数比你高，然后加1就是你在这个科目中的名次


--Score重复时合并名次---驱除重复项，不保留空缺名次
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=(
select count(distinct(o.总成绩)) from (
select a.S#,a.Sname,ISNULL(sum(b.score),0) 总成绩
from Student as a left join SC as b on a.S#=b.S#
group by a.S#,a.Sname
) o where o.总成绩>=t.总成绩

)

from
(
select a.S#,a.Sname,ISNULL(sum(b.score),0) 总成绩
from Student as a left join SC as b on a.S#=b.S#
group by a.S#,a.Sname

) t

order by t.总成绩 desc



-------------------------------------------------------------------------

select a.*,px=rank()over(partition by c# order by score desc)
from sc as a order by a.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
select a.S# 学号, b.Sname 姓名,sum(a.score) 总分
from sc as a
join Student as b on a.S#

=b.S#
group by a.S#,b.Sname
order by sum(a.score) desc

--------------------------------------------------------------------

-------------------------------------------------------------------------------
--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。
-- 总分重复时




------------------------------------------------------------------------
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。
--不留空名
select t.*,rank()over(order by t.总成绩 desc)[名次]
from (
select a.S#,a.Sname,ISNULL(sum(b.score),0) [总成绩]
from Student as a left join SC as b
on a.S#=b.S#
group by a.S#,a.Sname

) t

------------------------------------------------------------------------------------
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 c.T#,c.Tname, cast(avg(a.score) as decimal(5,2)) 平均分
from SC a ,Course b,Teacher c where a.C#=b.C# and b.T#=c.T#
group by c.T#,c.Tname
order by avg(a.score) desc

-------------------------------------------------------------------------
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
select *
from ( select C#,score,px=rank()over(partition by c# order by score desc)
from SC

) as a
where a.px between 2 and 3
order by a.C#,a.px

----------------------------------------------
select a.*
from (
select b.C#,b.score,px=RANK()over(partition by b.c# order by b.score desc )
from sc as b
) a
where a.px between 2 and 3 order by a.C#,a.px

------------------------------------------------------------------------------------
--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

select a.C#[课程号],ame[课程名] ,
(select count(*) from sc where C#=a.c# and score>=0 and score<60) [0-60所占人数],
cast(((select count(*) from sc where C#=a.c# and score>=0 and score<60 )*100/(select count(*) from sc where c#=a.C#)) as decimal(5,2))[0-60百分比(%)],
(select count(*) from sc where C#=a.c# and score>=60 and score<70 )[60-70所占人数],
cast(((select count(*) from sc where C#=a.c# and score>=60 and score<70 )*100/(select count(*) from sc where c#=a.C#)) as decimal(5,2))[60-70百分比(%)],
(select count(*) from sc where C#=a.c# and score>=70 and score<85 )[70-85所占人数],
cast(((select count(*) from sc where C#=a.c# and score>=70 and score<85 )*100/(select count(*) from sc where c#=a.C#)) as decimal(5,2))[70-85百分比(%)],
(select count(*) from sc where C#=a.c# and score>=85)[85-100所占人数],
cast(((select count(*) from sc where C#=a.c# and score>=85 )*100/(select count(*) from sc where c#=a.C#)) as decimal(5,2))[85-100百分比(%)]
from Course as a left join SC as b o

n a.C#=b.C# group by a.C#,ame



--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# , ame
order by Course.C#
--纵向显示1(显示存在的分数段)
select a.C#,ame,分数段=(
case when a.score >=85 then '100-85'
when a.score<85 and a.score>70 then '85-70'
when a.score>=60 and a.score<70 then '70-60'
else '60-0'
end
),count(1)人数---核心是分组，分好组，就计数
from SC as a,Course as b where a.C#=b.C#
group by a.C#,ame,(
case when a.score >=85 then '100-85'
when a.score<85 and a.score>70 then '85-70'
when a.score>=60 and a.score<70 then '70-60'
else '60-0'
end
)

-------------------------------------------------------------------------------
select m.C# [课程编号] , ame [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# , ame , (
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# , ame , 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] , ame [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# , ame , (
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# , ame , 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, ame 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 S

C 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 b.C#[课程号],ame[课程名],分数段=(
case when a.score >=85 then '100-85'
when a.score>=70and a.score<85 then '70-85'
when a.score>=60 and a.score<70 then '60-70'
else '60-0'
end
),count(1) 数量,cast(count(1)*100/(select count(*) from sc where C#=b.C#) as decimal(5,2))[百分比(%)]
from sc as a ,Course as b
where a.C#=b.C#
group by all b.C#,ame,(
case when a.score >=85 then '100-85'
when a.score>=70and a.score<85 then '70-85'
when a.score>=60 and a.score<70 then '60-70'
else '60-0'
end
)
order by b.c#,ame,分数段

--------------------------------------------------------------
select m.C# [课程编号] , ame [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# , ame , (
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# , ame , 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] , ame [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# , ame , ( ---//包含所有的分组结果
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# , ame , 分数段

--24、查询学生平均成绩及其名次
select t.*,
px=(

select count(*)
from
( select a.S#[学生编号],a.Sname[学生姓名],isnull(cast(avg(b.score) as decimal(5,2)),0)[平均分]from Student as a left join SC as b on a.S#=b.S# group by a.S#,a.Sname) d
where d.平均分>t.平均分
)+1
from( select a.S#[学生编号],a.Sname[学生姓名], isnull(cast(avg(b.score) as decimal(5,2)),0)[平均分] from Student as a left join SC as b on a.S#=b.S# group by a.S#,a.Sname) t
order by px
-------- 函数

select t.*,px=(Dense_rank()over(order by t.平均分 desc))
from
( select a.S#[学生编号],a.Sname[学生姓名], isnull(cast(avg(b.score) as decimal(5,2)),0)[平均分] from Student as a left join SC as b on a.S#=b.S# group by a.S#,a.Sname) t
order by px

---------------------------------------------------------------------
--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
select t.*
from (select a.*,px=(dense_rank()over(partition

by C# order by score desc ))
from sc as a
) t where t.px<=3 order by t.C#


------------------------------------
--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 a.C# [课程号],count(S#) 选修人数
from SC as a group by a.C#
--------------------------------------
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 a.S#,a.Sname
from Student as a ,SC as b where a.S#=b.S#
group by a.S#,a.Sname having count(b.s#)=2


----------------------------------------
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 a.Ssex [性别],count(*) [人数]
from student as a
group by a.Ssex
---版本2
select sum(case when a.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 as a
---版本3
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
--------------------------
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 CHARINDEX('风',sname)>0
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 distinct(a.Sname),count(*)[同名人数]from Student as a group by a.Sname having count(*)>1
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 DATEPART(yy,sage)=1990
select * from Student where DATEDIFF(yy,Sage,'1990-01-01')=0
select * from Student where convert(varchar(4),Sage,120)='1990'
select * from Student where DATENAME(yy,sage)='1990'
--------------------------------------------------------------------
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 a.C#[课程ID],isnull(cast(avg(a.score) as decimal(5,2)),0)[平均分数]
from sc as a group by a.C# order by isnull(cast(avg(a.score) as decimal(5,2)),0),a.C#
---------------------------------
select m.C# , ame , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n
where m.C# = n.C#
group by m.C# , ame
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 ,isnull(cast(avg(b.score) as decimal(5,2)),0)[平均成绩]
from student as a ,sc as b where a.S#=b.S#
group by a.S#,a.Sname having isnull(cast(avg(b.score) as decimal(5,2)),0)>=85

------------------------------------------
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 a.Sname,b.score
from Student as a ,SC as b,Course as c
where a.S#=b.S# and c.C#=b.C# and ame=N'数学'
and b.score<60


-----------------------------------------------------
select sname , score
from Student , SC , Course
where SC.S# = Student.S# and SC.C# = Course.C# and ame = N'数学' and score < 60

--35、查询所有学生的课程及分数情况；

select a.*,b.C#[课程号],b.score
from student as a ,SC as b where a.S#=b.S#



----------------------------------------------------
select Student.* , ame , 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 a.Sname[姓名],ame[课程名],b.score[分数]
from Student as a ,SC as b ,Course as c where a.S#=b.S# and b.C#=c.C# and b.score>70

-------------------------------------
select Student.* , ame , 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.* , ame , SC.C# , SC.sco

re
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 a.S#[学号],a.Sname[姓名]
from student as a ,sc as b where a.S#=b.S# and b.score>=80 and b.C#='01'

--------------------------------------------------------------------
select Student.* , ame , 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 b.C#,ame,count(*) [学生数]
from Course as a ,SC as b where a.C#=b.C# group by b.C#,ame

---------------------------------------------
select Course.C# , ame , count(*) [学生人数]
from Course , SC
where Course.C# = SC.C#
group by Course.C# , ame
order by Course.C# , ame

--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 a.S#,b.score
from student as a ,SC as b ,Course as c,Teacher as t
where a.S#=b.S# and b.C#=c.C# and t.T#=c.T# and t.Tname=N'张三'



------------------------------------------------------------
select top 1 Student.* , ame , 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 a.*,ame,c.C#,b.score
from Student as a ,SC as b,Course as c ,Teacher as t where a.S#=b.S# and b.C#=c.C# and t.T#=c.T# and t.Tname=N'张三' and b.score =( select top 1 b.score
from student as a ,SC as b ,Course as c,Teacher as t
where a.S#=b.S# and b.C#=c.C# and t.T#=c.T# and t.Tname=N'张三')



-------------------------------------------------------------------
select Student.* , ame , 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
select t.S#,t.C#,t.score
from ( select a.S#,b.C#,b.score from Student as a ,SC as b where a.S#=b.S# ) t,(select a.S#,b.C#,b.score from Student as a ,SC as b where a.S#=b.S# )y
where t.C#<>y.C# and t.score=y.score group by t.S#,t.C#,t.score



----------------------------------------------------------------------------------不理解
--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.C#,t.S#,t.score,t.px[名次]
from (select *,px=dense_RANK()over(partition by a.c# order by a.score desc ) from SC as a ) t where px<=2 order by t.C#,t.px


-------------------------------------------------
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 a.C#[课程号],count(*) [选修人数]
from sc as a
group by a.C#
having count(*) >5
order by COUNT(*) desc,a.C# asc


-------------------------------------------------------------------------------------------------------------------------------
select Course.C# , ame , count(*) [学生人数]
from Course , SC
where Course.C# = SC.C#
group by Course.C# , ame
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 a.S#[学号],count(a.s#)[选修的课程数]
from sc as a
group by a.S# having count(a.S#)>=2
------------------------------------------
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
---方法一，计数
select a.S#[学生号],b.Sname[学生姓名],count(a.S#)[选修课程数]
from SC as a ,student as b where a.S#=b.S#
group by a.S#,b.Sname
having count(a.S#)=(select count(distinct(b.C#)) from Course as b)
--方法二，取反
select a.*
from Student as a
where a.S# not in(
select distinct t.S#
from (
select S#,C# from Student ,Course
) t where not exists( select 1 from sc where C#=t.C# and S#=t.S#)
)
------------------------------------------
--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 --当学号不在这里面时
(
select distinct m.S# from---查询不同的学号,条件是有科目没有选修的
(
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#)--难点,就是没有选修所有课程的同学
)




--方法3 使用双重否定来完成
select t.* from student t
where not exists ---取反,下面的条件是所有学生对应没有选修所有课程的学生,就不能被查出来，反之考完了的学生救能被

查出来
(select 1 from
(
select distinct m.S# from
(
select S# , C# from student , course
)--整个结果集就是所有学生对应的所有课程，类似全链接
m where not exists (select 1 from sc n where n.S# = m.S# and n.C# = m.C#)--当有一个学生没有选修对应的课程时会被查出来
) --整个结果集就是没有考完的学生号
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 a.Sname,年龄=datediff(YY,Sage,GETDATE())
from student as a

----------------------------
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 * ,[年龄]=case when( datepart(MM,Sage)then datediff(yy , sage , getdate())
else datediff(yy , sage , getdate())-1
end
from Student
---------------------------------------------------------------------------------------------------------------
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 a.*
from Student as a
where DATEPART(WEEK,Sage)=DATEPART(week,getdate())

-- where DATEDIFF(WEEK,DATENAME(yy,GETDATE())+RIGHT(convert(varchar(20),opendate,120),6),getdate())=0
-------------------------------------------------------------------------------------
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 a.*
from Student as a
where DATEPART(WEEK,Sage)=DATEPART(week,getdate())+1
-------------------------------------------------------------------------------------
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 a.*
from Student as a
where DATEPART(MONTH,Sage)=DATEPART(MONTH,getdate())
-------------------------------------------------------------------------------------
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 a.*
from Student as a
where DATEPART(MONTH,Sage)=DATEPART(MONTH,getdate())+1
-------------------------------------------------------------------------------------

select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),sage,120),6),getdate()) = -1


drop table Student,Course,Teacher,SC