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' , '19-07-01' , 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' , )

--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(*) 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 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(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 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'

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,a.Cname 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#,a.Cname

--方法二

select a.C# 课程ID,a.Cname 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], 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（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],a.Cname[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

on a.C#=b.C# group by a.C#,a.Cname

--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 a.C#,b.Cname,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#,b.Cname,(

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] , 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 b.C#[课程号],b.Cname[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#,b.Cname,(

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#,b.Cname,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 , ( ---//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# , m.Cname , 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

n 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# , 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 ,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 c.Cname=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 Course.Cname = 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.* , 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 a.Sname[姓名],c.Cname[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.* , 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.s

core

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.* , 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 b.C#,a.Cname,count(*) [学生数]

from Course as a ,SC as b where a.C#=b.C# group by b.C#,a.Cname

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

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 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.* , 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 a.*,c.Cname,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.* , 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

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# , 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 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