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--1.学生表 Student(S,Sname,Sage,Ssex) --S 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别 --2.课程表 Course(C,Cname,T) --C --课程编号,Cname 课程名称,T 教师编号 --3.教师表 Teacher(T,Tname) --T 教师编号,Tname 教师姓名 --4.成绩表 SC(S,C,score) --S 学生编号,C 课程编号,score 分数 */ --创建测试数据 create table Student(S varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10)) insert into Student values(‘01‘ , N‘赵雷‘ , ‘1990-01-01‘ , N‘男‘) insert into Student values(‘02‘ , N‘钱电‘ , ‘1990-12-21‘ , N‘男‘) insert into Student values(‘03‘ , N‘孙风‘ , ‘1990-05-20‘ , N‘男‘) insert into Student values(‘04‘ , N‘李云‘ , ‘1990-08-06‘ , N‘男‘) insert into Student values(‘05‘ , N‘周梅‘ , ‘1991-12-01‘ , N‘女‘) insert into Student values(‘06‘ , N‘吴兰‘ , ‘1992-03-01‘ , N‘女‘) insert into Student values(‘07‘ , N‘郑竹‘ , ‘1989-07-01‘ , N‘女‘) insert into Student values(‘08‘ , N‘王菊‘ , ‘1990-01-20‘ , N‘女‘) create table Course(C varchar(10),Cname nvarchar(10),T varchar(10)) insert into Course values(‘01‘ , N‘语文‘ , ‘02‘) insert into Course values(‘02‘ , N‘数学‘ , ‘01‘) insert into Course values(‘03‘ , N‘英语‘ , ‘03‘) create table Teacher(T varchar(10),Tname nvarchar(10)) insert into Teacher values(‘01‘ , N‘张三‘) insert into Teacher values(‘02‘ , N‘李四‘) insert into Teacher values(‘03‘ , N‘王五‘) create table SC(S varchar(10),C varchar(10),score decimal(18,1)) insert into SC values(‘01‘ , ‘01‘ , 80) insert into SC values(‘01‘ , ‘02‘ , 90) insert into SC values(‘01‘ , ‘03‘ , 99) insert into SC values(‘02‘ , ‘01‘ , 70) insert into SC values(‘02‘ , ‘02‘ , 60) insert into SC values(‘02‘ , ‘03‘ , 80) insert into SC values(‘03‘ , ‘01‘ , 80) insert into SC values(‘03‘ , ‘02‘ , 80) insert into SC values(‘03‘ , ‘03‘ , 80) insert into SC values(‘04‘ , ‘01‘ , 50) insert into SC values(‘04‘ , ‘02‘ , 30) insert into SC values(‘04‘ , ‘03‘ , 20) insert into SC values(‘05‘ , ‘01‘ , 76) insert into SC values(‘05‘ , ‘02‘ , 87) insert into SC values(‘06‘ , ‘01‘ , 31) insert into SC values(‘06‘ , ‘03‘ , 34) insert into SC values(‘07‘ , ‘02‘ , 89) insert into SC values(‘07‘ , ‘03‘ , 98)
生成如下表：
--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
select student.*,course.cname,sc1.score from SC as sc1 inner join sc as sc2 on sc1.s=sc2.s AND sc1.c=‘01‘and sc2.c=‘02‘ left join student on sc1.s=student.s left join course on sc1.c=course.c where sc1.score>sc2.score
--1.2、查询同时存在"01"课程和"02"课程的情况
select * from Student where Student.Sname in (select Student.Sname from SC as sc1 inner join SC as sc2 on sc1.s=sc2.s AND sc1.c=‘01‘and sc2.c=‘02‘ inner join student on Student.S=sc1.S)--in中只能包括字段
--1.3、查询"01"课程分数高于"02"课程的学生情况
1 select a.* , b.score [课程"01"的分数],c.score [课程"02"的分数] from Student a 2 left join SC b on a.S = b.S and b.C = ‘01‘ 3 left join SC c on a.S = c.S and c.C = ‘02‘ 4 where b.score > c.score 5 --where b.score > isnull(c.score,0)--为null时候当0比较
--2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
--2.1、查询同时存在"01"课程和"02"课程的情况
select a.* , b.score [课程‘01‘的分数],c.score [课程‘02‘的分数] from Student a , SC b , SC c
where a.S = b.S and a.S = c.S and b.C = ‘01‘ and c.C = ‘02‘ and b.score < c.score
--2.2、查询同时存在"01"课程和"02"课程的情况和不存在"01"课程但存在"02"课程的情况
select a.* , b.score [课程"01"的分数],c.score [课程"02"的分数] from Student a
left join SC b on a.S = b.S and b.C = ‘01‘
left join SC c on a.S = c.S and c.C = ‘02‘
where isnull(b.score,0) < c.score
--3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select Student.S,student.Sname,CAST( avg(sc.score) as decimal(18,2) )as ‘average‘ from student,SC where sc.S=student.S group by Student.S,Student.Sname--group by后面必须包含select之后所有非聚合函数？？ having CAST( avg(sc.score) as decimal(18,2) )>=60--此处比较大小必须是聚合函数 order by Student.S
--4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
--4.1、查询在sc表存在成绩的学生信息的SQL语句。
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60
order by a.S
--4.2、查询在sc表中不存在成绩或平均分不及格的学生信息的SQL语句。
select a.S , a.Sname , isnull (cast(avg(b.score) as decimal(18,2)),0) avg_score from Student a , sc b where a.S = b.S group by a.S , a.Sname having isnull (cast(avg(b.score) as decimal(18,2)),0)=0 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 Student.S ‘学生编号‘,student.Sname ‘学生编号‘,count(sc.C) as ‘选课总数‘,CAST( sum(sc.score) as decimal(18,2) )as ‘总分‘ from student,SC where sc.S=student.S group by Student.S,Student.Sname--group by后面必须包含select之后所有非聚合函数？？ order by Student.S
--5.1、查询所有有成绩的SQL。
select a.S [学生编号], a.Sname [学生姓名], count(b.C) 选课总数, sum(score) [所有课程的总成绩]
from Student a , SC b
where a.S = b.S
group by a.S,a.Sname
order by a.S
--5.2、查询所有(包括有成绩和无成绩)的SQL。
select a.S [学生编号], a.Sname [学生姓名], count(b.C) 选课总数, sum(score) [所有课程的总成绩]
from Student a left join SC b --right join是指选取有成绩的学生
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(tname) from Teacher where Tname like N‘%三%‘ select count(Tname) ["李"姓老师的数量] from Teacher where right(Tname,1)=N‘三‘
--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 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‘--排除一个课程可能会有两个老师任课的情况
order by Student.S
--8、查询没学过"张三"老师授课的同学的信息
select m.* from Student m where S not in (select distinct SC.S from SC , Course , Teacher where SC.C = Course.C and Course.T = Teacher.T and Teacher.Tname = N‘张三‘) order by m.S
--9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
--方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C = ‘01‘ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C = ‘02‘) order by Student.S
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----exists引导的子句有结果集返回，那么exists这个条件就算成立了,大家注意返回的字段始终为1，如果改成“select 2 from grade where ...”，那么返回的字段就是2，这个数字没有意义。所以exists子句不在乎返回什么，而是在乎是不是有结果集返回。而 exists 与 in 最大的区别在于 in引导的子句只能返回一个字段
--slect 1 from 详解移步 http://www.cnblogs.com/han1028/archive/2009/10/09/1579672.html
select Student.* from Student , SC where Student.S = SC.S and SC.C = ‘01‘ and Student.S in (select Student.S from Student,SC where Student.S = SC.S and SC.C = ‘02‘)--in引导的子句只能返回一个字段
--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 * from Student where Student.S in
( select S from (select s from SC where C=‘01‘ union all select s from SC where C=‘02‘) T
group by s having sum(1)=2 ) --此处用count也是OK的 order by Student.S
select m.* from Student m where S in
(
select S from
(
select distinct S from SC where C = ‘01‘
union all
select distinct S from SC where C = ‘02‘
) t group by S having count(1) = 2
)
order by m.S
--10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
--方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C = ‘01‘ and not exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C = ‘02‘) order by Student.S
--方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C = ‘01‘ and Student.S not in (Select SC_2.S from SC SC_2 where SC_2.S = SC.S and SC_2.C = ‘02‘) order by Student.S
--11、查询没有学全所有课程的同学的信息
--11.1、
select student.S,student.Sname,COUNT(SC.C) as ‘所学课程科目‘ from SC,Student where SC.S=Student.S--left join也是一样的 group by student.S,student.Sname having COUNT(SC.C)<(select COUNT(course.C) from Course)
--11.2
select Student.*
from Student left join SC
on Student.S = SC.S
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having count(C) < (select count(C) from Course)
--12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息
select distinct Student.* from Student , SC where Student.S = SC.S and SC.C in (select C from SC where S = ‘01‘) and Student.S <> ‘01‘
（引号需要加0，不要引号为int 不要加0）
--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 * from student where student.S in (select distinct s from SC where C in (select c from sc where S=‘01‘ ) and s<>1 group by sc.S having count(1)=(select count(1) from sc where S=‘01‘))--这里使用count（1）更加好，因为前面需要使用in
--14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select * from student where student.s not in (select S from sc,Teacher,Course where SC.C=Course.C and Teacher.T=Course.T and Teacher.Tname=N‘张三‘)--张三前要加N，不加不行？？
--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 sc.S,Student.Sname,cast(avg(score) as decimal(18,2) ) N‘平均分‘,sum(case when score<60 then 1 else 0 end) from SC inner join Student on sc.S=Student.S group by sc.S,Student.Sname having sum(case when score<60 then 1 else 0 end)>=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 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 sc.S as ‘学号‘,Student.Sname as ‘姓名‘, sum(case when c=‘01‘then score else 0 end) as ‘语文‘, sum(case when c=‘02‘then score else 0 end) as ‘数学‘, sum(case when c=‘03‘then score else 0 end) as ‘英语‘, cast(avg(score) as decimal(18,2)) as ‘平均分‘ from sc left join Student on sc.S=Student.S group by sc.S,Student.Sname order by cast(avg(score) as decimal(18,2)) desc --按照总分排名，降序
select Student.S as ‘学号‘,Student.Sname as ‘姓名‘, sum(case when c=‘01‘then score else 0 end) as ‘语文‘, sum(case when c=‘02‘then score else 0 end) as ‘数学‘, sum(case when c=‘03‘then score else 0 end) as ‘英语‘, cast(avg(score) as decimal(18,2)) as ‘平均分‘ from sc right join Student on sc.S=Student.S--考虑没有分数的学生，所以这里使用right join group by Student.S ,Student.Sname order by cast(avg(score) as decimal(18,2)) desc --按照总分排名，降序
--17.1 SQL 2000 静态
select a.S 学生编号 , a.Sname 学生姓名 ,
max(case c.Cname when N‘语文‘ then b.score else null end) [语文],
max(case c.Cname when N‘数学‘ then b.score else null end) [数学],
max(case c.Cname when N‘英语‘ then b.score else null end) [英语],
cast(avg(b.score) as decimal(18,2)) 平均分
from Student a
left join SC b on a.S = b.S
left join Course c on b.C = c.C
group by a.S , a.Sname
order by 平均分 desc
--17.2 SQL 2000 动态
declare @sql nvarchar(4000)
set @sql = ‘select a.S ‘ + N‘学生编号‘ + ‘ , a.Sname ‘ + N‘学生姓名‘
select @sql = @sql + ‘,max(case c.Cname when N‘‘‘+Cname+‘‘‘ then b.score else null end) [‘+Cname+‘]‘
from (select distinct Cname from Course) as t
set @sql = @sql + ‘ , cast(avg(b.score) as decimal(18,2)) ‘ + N‘平均分‘ + ‘ from Student a left join SC b on a.S = b.S left join Course c on b.C = c.C
group by a.S , a.Sname order by ‘ + N‘平均分‘ + ‘ desc‘
exec(@sql)
--17.3 有关sql 2005的动静态写法参见我的文章《普通行列转换(version 2.0)》或《普通行列转换(version 3.0)》。
SQL code
--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 Course.C as ‘课程代码‘ ,Course.Cname as ‘课程名称‘,max(score) as ‘最高分‘,MIN(score) as ‘最低分‘,cast(avg(score) as decimal(18,2))as ‘平均分‘, cast(sum(case when score>=60 then 1 else 0 end)*100.0/count(sc.C) as decimal(18,2)) as ‘及格率(%)‘,--转换为保留小数两位数 cast(sum(case when score>=70 and score<80 then 1 else 0 end)*100.0/count(sc.C) as decimal(18,2)) as ‘中等率(%)‘, cast(sum(case when score>=80 and score<90 then 1 else 0 end)*100.0/count(sc.C) as decimal(18,2)) as ‘优良率(%)‘, cast(sum(case when score>=90 then 1 else 0 end)*100/count(sc.C) as decimal(18,2)) as ‘优秀率(%)‘ from SC right join Course on sc.C=Course.C group by Course.C,Course.Cname order by Course.C asc
--上面乘以100小数点后面的数为零，乘以100.0则会显示小数点
--方法1
select m.C [课程编号], m.Cname [课程名称], max(n.score) [最高分], min(n.score) [最低分], cast(avg(n.score) as decimal(18,2)) [平均分], cast((select count(1) from SC where C = m.C and score >= 60)*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [及格率(%)], cast((select count(1) from SC where C = m.C and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [中等率(%)], cast((select count(1) from SC where C = m.C and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [优良率(%)], cast((select count(1) from SC where C = m.C and score >= 90)*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [优秀率(%)] from Course m , SC n where m.C = n.C group by m.C , m.Cname order by m.C
--方法2
select m.C [课程编号], m.Cname [课程名称],
(select max(score) from SC where C = m.C) [最高分],
(select min(score) from SC where C = m.C) [最低分],
(select cast(avg(score) as decimal(18,2)) from SC where C = m.C) [平均分],
cast((select count(1) from SC where C = m.C and score >= 60)*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [及格率(%)],
cast((select count(1) from SC where C = m.C and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [中等率(%)],
cast((select count(1) from SC where C = m.C and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [优良率(%)],
cast((select count(1) from SC where C = m.C and score >= 90)*100.0 / (select count(1) from SC where C = m.C) as decimal(18,2)) [优秀率(%)]
from Course m
order by m.C
--19、按各科成绩进行排序，并显示排名
--19.1 sql 2000用子查询完成
--Score重复时保留名次空缺
select t.* , px = (select count(1) from SC where C = t.C and score > t.score) + 1 from sc t order by t.C , px
--Score重复时合并名次
select t.* , px = (select count(distinct score) from SC where C = t.C and score >= t.score) from sc t order by t.C , px
--19.2 sql 2005用rank,DENSE_RANK完成
--Score重复时保留名次空缺(rank完成)
select t.* , px = rank() over(partition by C order by score desc) from sc t order by t.C , px
--Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by C order by score desc) from sc t order by t.C , px
select rank() over(partition by c order by score desc) as t,* from sc --排列相同时跳一个序号 select ROW_NUMBER() over(partition by c order by score desc) as t,* from sc select dense_rank() over(partition by c order by score desc) as t,* from sc--序号连续 select ntile(6) over(partition by c order by score desc) as t,* from sc
--20、查询学生的总成绩并进行排名
--20.1 查询学生的总成绩
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
order by [总成绩] desc
--20.2 查询学生的总成绩并进行排名，sql 2000用子查询完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t1.* , px = (select count(1) from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t1
order by px
select t1.* , px = (select count(distinct 总成绩) from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t2 where 总成绩 >= t1.总成绩) from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t1
order by px
--20.3 查询学生的总成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t.* , px = rank() over(order by [总成绩] desc) from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t
order by px
select t.* , px = DENSE_RANK() over(order by [总成绩] desc) from
(
select m.S [学生编号] ,
m.Sname [学生姓名] ,
isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S
group by m.S , m.Sname
) t
order by px
--21、查询不同老师所教不同课程平均分从高到低显示
select Teacher.Tname,cast(avg(score) as decimal(18,2)) as ‘平均分‘ from Course inner join SC on sc.C=Course.C inner join Teacher on Teacher.T=Course.T group by sc.C,Teacher.Tname order by cast(avg(score) as decimal(18,2)) 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 *, rank() over (partition by c order by score ) as rank from SC) t where t.rank between 2 and 3 order by t.C
--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 Course.C [课程编号] , Cname as [课程名称] , cast(sum(case when score>=85 and score<100 then 1 else 0 end)*100.0/(sum(case when score>=0 then 1 else 0 end)) as decimal(18,2)) [100-85] ,cast(sum(case when score>=70 and score<85 then 1 else 0 end ) *100.0/(sum(case when score>=0 then 1 else 0 end)) as decimal(18,2)) [85-70] ,cast(sum(case when score>=60 and score<70 then 1 else 0 end) *100.0/(sum(case when score>=0 then 1 else 0 end)) as decimal(18,2)) [70-60] ,cast(sum(case when score<60 then 1 else 0 end) *100.0/(sum(case when score>=0 then 1 else 0 end)) as decimal(18,2)) [0-60] from sc , Course where SC.C = Course.C group by Course.C , Course.Cname order by Course.C
--23.1 统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]
--横向显示
select Course.C [课程编号] , Cname as [课程名称] , sum(case when score >= 85 then 1 else 0 end) [85-100], sum(case when score >= 70 and score < 85 then 1 else 0 end) [70-85], sum(case when score >= 60 and score < 70 then 1 else 0 end) [60-70], sum(case when score < 60 then 1 else 0 end) [0-60] from sc , Course where SC.C = Course.C group by Course.C , Course.Cname order by Course.C
--纵向显示1(显示存在的分数段)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
case when n.score >= 85 then ‘85-100‘
when n.score >= 70 and n.score < 85 then ‘70-85‘
when n.score >= 60 and n.score < 70 then ‘60-70‘
else ‘0-60‘
end) ,
count(1) 数量
from Course m , sc n
where m.C = n.C
group by m.C , m.Cname , (
case when n.score >= 85 then ‘85-100‘
when n.score >= 70 and n.score < 85 then ‘70-85‘
when n.score >= 60 and n.score < 70 then ‘60-70‘
else ‘0-60‘
end)
order by m.C , m.Cname , 分数段
--纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select course.C,course.Cname,‘nokia‘=(case when score<60 then ‘[0-60]‘when score>=60 and score<70 then ‘[60-70]‘ when score>=70 and score<85 then ‘[70-85]‘ when score>=85 and score<100 then ‘[85-100]‘ else null end ),count(1) from SC ,Course where Course.C=SC.C group by all course.C,course.Cname,case when score<60 then ‘[0-60]‘when score>=60 and score<70 then ‘[60-70]‘ when score>=70 and score<85 then ‘[70-85]‘--多一个单词all when score>=85 and score<100 then ‘[85-100]‘ else null end
--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 course.C,course.Cname, ‘nokia‘=(case when score<60 then ‘[0-60]‘when score>=60 and score<70 then ‘[60-70]‘ when score>=70 and score<85 then ‘[70-85]‘ when score>=85 and score<100 then ‘[85-100]‘ else null end), count(1) qty, cast(count(1)*100.0/(select count(1)from SC where c=Course.C)as decimal(18,2)) rate --后面以及groupby分好类 from SC ,Course where Course.C=SC.C group by all course.C,course.Cname,case when score<60 then ‘[0-60]‘when score>=60 and score<70 then ‘[60-70]‘ when score>=70 and score<85 then ‘[70-85]‘ when score>=85 and score<100 then ‘[85-100]‘ else null 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 ,
cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C
group by m.C , m.Cname , (
case when n.score >= 85 then ‘85-100‘
when n.score >= 70 and n.score < 85 then ‘70-85‘
when n.score >= 60 and n.score < 70 then ‘60-70‘
else ‘0-60‘
end)
order by m.C , m.Cname , 分数段
--纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
case when n.score >= 85 then ‘85-100‘
when n.score >= 70 and n.score < 85 then ‘70-85‘
when n.score >= 60 and n.score < 70 then ‘60-70‘
else ‘0-60‘
end) ,
count(1) 数量 ,
cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C
group by all m.C , m.Cname , (
case when n.score >= 85 then ‘85-100‘
when n.score >= 70 and n.score < 85 then ‘70-85‘
when n.score >= 60 and n.score < 70 then ‘60-70‘
else ‘0-60‘
end)
order by m.C , m.Cname , 分数段
--24、查询学生平均成绩及其名次
select Student.S,Student.Sname, cast(AVG(score)as decimal(18,2))as avg_score, RANK() over(order by cast(AVG(score)as decimal(18,2)) desc ) ranking from SC,Student where Student.S=sc.S group by Student.S,Student.Sname
--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 SC ,Student where Student.S=sc.S group by Student.S,Student.Sname having count(sc.C)=2
--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
alter table student alter column s varchar(10) not null--修改student表中的S字段为not null,这样才能修改为主键
格式：
ALTER TABLE table_name
ALTER COLUMN column_name column_type NOT NULL;
alter table student add constraint pk_s primary key(s)--增加student表中主键为s
alter table sc add constraint fk_s foreign key(s) references student(s)on delete cascade on update cascade
--增加sc表中的S为student的外键，当student中的某个S被删除后，则对应的sc表中的分数也会被删除。
创建索引语法：
CREATE [索引类型] INDEX 索引名称
ON 表名(列名)
WITH FILLFACTOR = 填充因子值0~100
GO
create index index_s on student(s)--创建student s为索引
总结：
1.什么是索引：数据库中的索引是某个表中一列或多列值的集合和相应的指向表中物理标识这些值的数据页的逻辑指针清单。
2.分类：
唯一索引(UNIQUE)：不允许两行具有相同的索引值（创建了唯一约束，系统将自动创建唯一索引）
主键索引：主键索引要求主键中的每个值是唯一的，（创建主键自动创建主键索引）
聚集索引(CLUSTERED)：表中各行的物理顺序与键值的逻辑（索引）顺序相同，表中只能包含一个聚集索引，主键列默认为聚集索引
非聚集索引(NONCLUSTERED)：表中各行的物理顺序与键值的逻辑（索引）顺序不匹配，表中可以有249个非聚集索引
3.创建索引的标准：用于频繁搜索的列；用于对数据进行排序的列
注意：如果表中仅有几行，或列中只包含几个不同的值，不推荐创建索引，因为SQL Server 在小型表中用索引搜索数据所花的时间比逐行搜索更长。
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原文地址：http://www.cnblogs.com/roytan/p/5255004.html
