SQL server练习

/*
--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
*/
if exists (select * from sysdatabases where name='sct')
	drop database sct;
go

create database sct;
go
use sct;
go
/*
create database mydatabase
on 
(name=mydatabase_data,filename='c:\mydatabase.mdf',size=5MB,filegrowth=15%)
log on
(name=mydatabase_log,filename='c:\mydatabase.ldf',size=3MB,filegrowth=15%)
*/
use sct;
/*
select name from sysobjects where type = 'U';

由于系统表sysobjects保存的都是数据库对象,其中type表示各种对象的类型，具体包括:
U = 用户表
S = 系统表
C = CHECK 约束
D = 默认值或 DEFAULT 约束
F = FOREIGN KEY 约束
L = 日志
FN = 标量函数
IF = 内嵌表函数
P = 存储过程
PK = PRIMARY KEY 约束（类型是 K）
RF = 复制筛选存储过程
TF = 表函数
TR = 触发器
UQ = UNIQUE 约束（类型是 K）
V = 视图
X = 扩展存储过程及相关的对象信息。 
*/


if object_id('student') is not null
	drop table student;
go
create table student(
	s# varchar(10),
	sname nvarchar(10),
	sage datetime,
	ssex nvarchar(10)
);
insert into student(s#,sname,sage,ssex) values ('01',N'赵雷','1990-01-01',N'男');
insert into student(s#,sname,sage,ssex) 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');


if object_id('course') is not null
	drop table course;
go
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');


if object_id('teacher') is not null
	drop table teacher;
go
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');


if object_id('sc') is not null
	drop table sc;
go
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

create proc show_course
as
select * from course
go

create proc show_sc
as
begin
select * from sc
end
go

create proc show_student
as
begin
select * from student
end
go

create proc show_teacher
as
begin
select * from teacher
end
go

exec show_course
exec show_sc
exec show_student
exec show_teacher

--1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
--1.1、查询同时存在"01"课程和"02"课程的情况
select a.*, b.score [课程'01'的分数], c.score [课程'02'的分数]
from student a , sc b , sc c
where a.s# = b.s# and a.s# = c.s# 
and b.c# = '01' and c.c# = '02' and b.score > c.score;

--1.2、查询同时存在"01"课程和"02"课程的情况和存在"01"课程但可能不存在"02"课程的情况(不存在时显示为null)(以下存在相同内容时不再解释)
select a.*, b.score [课程'01'的分数], c.score [课程'02'的分数]
from student a
left join sc b on a.s# = b.s# and b.c# = '01'
left join sc c on a.s# = c.s# and c.c# = '02'
where b.score > isnull(c.score, 0)

select a.*, b.score
from student a left join sc b on a.s# = b.s# and b.c# = '01'

select a.*, c.score
from student a left join sc c on a.s# = c.s# and c.c# = '02'

--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 [01课程成绩],c.score [02课程成绩]
from student as a left join sc as b on a.s# = b.s# and b.c# = '01'
left join sc as c on a.s# = c.s# and c.c# = '02'
where  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 [01课程成绩],c.score [02课程成绩]
from student as a left join sc as b on a.s# = b.s# and b.c# = '01'
left join sc as 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 * from sc

select sc.s# [学号], student.sname [姓名], avg(sc.score) as [平均成绩]
from sc join student on sc.s# = student.s#
group by sc.s#,student.sname
having avg(sc.score) > 60 

select sc.s# [学号], student.sname [姓名], cast(avg(sc.score) as decimal(10,1)) as [平均成绩]
from sc join student on sc.s# = student.s#
group by sc.s#,student.sname
having cast(avg(sc.score) as decimal(10,1)) > 60

select a.S# , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S# = b.S#
group by a.S# , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60 
order by a.S#

--4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
--4.1、查询在sc表存在成绩的学生信息的SQL语句。
select student.s#, student.sname, cast(avg(sc.score) as decimal(10,1)) [平均成绩]
from student join sc on student.s# = sc.s#
group by student.s#, student.sname
having avg(sc.score) < 60

--4.2、查询在sc表中不存在成绩的学生信息的SQL语句。
select *
from student left join sc on student.s# = sc.s#
where sc.score is null

select a.S# , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0) avg_score
from Student a left join sc b
on a.S# = b.S#
group by a.S# , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 
order by a.S#

--5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
--5.1、查询所有有成绩的SQL。
select a.s#, a.sname, count(b.c#) [选课总数],sum(b.score)[课程总成绩]
from student a inner join sc b on 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#, a.sname, count(b.c#) [选课总数],sum(b.score)[课程总成绩]
from student a left join sc b on a.s# = b.s#
group by a.s#,a.sname
order by a.s#

--6、查询"李"姓老师的数量 
--方法1
select count(Tname) ["李"姓老师的数量] from Teacher where Tname like N'李%'
--方法2
select count(Tname) ["李"姓老师的数量] from Teacher where left(Tname,1) = N'李'

--7、查询学过"张三"老师授课的同学的信息 
select student.*
from teacher join course on teacher.t# = course.t#
join sc on sc.c# = course.c#
join student on sc.s# = student.s#
where 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 * from student
except
select student.*
from teacher join course on teacher.t# = course.t#
join sc on sc.c# = course.c#
join student on sc.s# = student.s#
where tname = N'张三'
order by Student.S#

select m.*
from Student m 
where S# not in 
(
	select distinct SC.S# from SC , Course , Teacher 
	where SC.C# = Course.C# and Course.T# = Teacher.T# and Teacher.Tname = N'张三'
) 
order by m.S#

--9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
--方法1
select Student.* from Student , SC 
where Student.S# = SC.S# and SC.C# = '01' 
and exists 
(Select 1 from SC SC_2 where SC_2.S# = SC.S# and SC_2.C# = '02') 
order by Student.S#

--方法2
select Student.* from Student , SC 
where Student.S# = SC.S# and SC.C# = '02' 
and exists 
(Select 1 from SC SC_2 where SC_2.S# = SC.S# and SC_2.C# = '01') 
order by Student.S#

--方法3
select m.* from Student m where S# in
(
  select S# from
  (
    select distinct S# from SC where C# = '01'
    union all
    select distinct S# from SC where C# = '02'
  ) t group by S# having count(1) = 2 
)
order by m.S#

--方法4
select * from student
where student.s# in
(
	select distinct a.s#
	from sc a join sc b
	on ((a.c# = '01' and b.c# = '02') or (a.c# = '02' and b.c# = '01')) and a.s# = b.s#
	group by a.s#
)
order by 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 join sc on student.s# = sc.s# and sc.c# = '01'
where not exists
(select * from sc a where student.s# = a.s# and a.c# = '02')

--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 
select * from course;
select count(*) from course
select * from sc;

select student.*
from student join
(
	select s#
	from
	(
		select s#,count(s#) as c
		from sc
		group by s#
	)t where t.c < (select count(*) from course)
)ts on student.s# = ts.s#;

--11.1、
select Student.*
from Student , SC 
where Student.S# = SC.S# 
group by Student.S# , Student.Sname , Student.Sage , Student.Ssex 
having count(C#) < (select count(C#) from Course) 

--11.2
select Student.*
from Student left join SC 
on Student.S# = SC.S# 
group by Student.S# , Student.Sname , Student.Sage , Student.Ssex 
having count(C#) < (select count(C#) from Course) 

--11.3查询学全所有课程的同学的信息
--1
select student.*
from student join
(
	select s#
	from
	(
		select s#,count(s#) as c
		from sc
		group by s#
	)t where t.c = (select count(*) from course)
)ts on student.s# = ts.s#;

--2
select student.*
from student join
(
	select s#
	from
	(
		select s#,count(s#) as c
		from sc
		group by s#
	)t where t.c in (select count(*) from course)
)ts on student.s# = ts.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
select student.*
from student join 
(
	select distinct a.s#
	from sc as a
	where a.s# <> '01' and a.c# in
	(
		select distinct c#
		from sc
		where s# = '01'
	)
) as b on student.s# = b.s#
order by student.s#

select distinct Student.* 
from Student , SC 
where Student.S# = SC.S# 
	and SC.C# in 
		(select C# from SC where S# = '01') 
	and Student.S# <> '01'
			
--13、查询和"01"号的同学学习的课程完全相同的其他同学的信息
select Student.* 
from Student 
where S# in
(
	select distinct SC.S# 
	from SC 
	where S# <> '01' 
		and SC.C# in (select distinct C# from SC where S# = '01') 
	group by SC.S# 
	having count(1) = (select count(1) from SC where S#='01')
) 

--14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select student.*
from student 
where s# in (
	select distinct s# from student
	except
	select s# from sc 
	where c# in (
		select c#
		from teacher join course on teacher.t# = course.t#
		where tname = N'张三'
		)
)

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#

select t# from teacher where tname = '张三'
select c# from course where t# = '01'
select * from sc;

--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 * from sc where score < 60

select s#,count(c#)
from sc
where score < 60
group by s#
having count(c#) > 1

select student.s#,student.sname,avg(sc.score) as [平均成绩]
from student join sc on student.s# = sc.s#
where student.s# in
(
	select s#
	from sc
	where score < 60
	group by s#
	having count(c#) > 1
)
group by student.s#,student.sname

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.score
from student, sc
where student.s# = sc.s# and student.s# in
	(
		select distinct s#
		from sc
		where sc.c# = '01' and sc.score < 60
	) and sc.c# = '01'
order by sc.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
--17.1 SQL 2000 静态
select student.s#,student.sname,
	max(case cname when N'语文' then sc.score else null end) as [语文],
	max(case cname when N'数学' then sc.score else null end) as [数学],
	max(case cname when N'英语' then sc.score else null end) as [英语],
	cast(avg(sc.score) as decimal (18,2)) as 平均分
from student left join sc on student.s# = sc.s#
			 left join course on sc.c# = course.c#
group by student.s#, student.sname
order by student.s#

--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)

--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.*,sc.*
from course join sc on course.c# = sc.c#

select course.c#,count(sc.s#)
from course join sc on course.c# = sc.c#
group by course.c#

select course.c#,course.cname,
	max(sc.score) as [最高分],
	min(sc.score) as [最低分],
	cast(avg(sc.score) as decimal(10,2)) as [平均分],
	cast(sum(case when sc.score >= 60 then 1 else 0 end)*1.0/(select count(1) from sc b where b.c# = course.c#) as decimal(10,2)) [及格率>=60],
	cast(sum(case when 70 <= sc.score and sc.score < 80 then 1 else 0 end)*1.0/(select count(1) from sc b where b.c# = course.c#) as decimal(10,2)) [中等率70-80],
	cast(sum(case when 80 <= sc.score and sc.score < 90 then 1 else 0 end)*1.0/(select count(1) from sc b where b.c# = course.c#) as decimal(10,2)) [优良率80-90],
	cast(sum(case when 90 <= sc.score  then 1 else 0 end)*1.0/(select count(1) from sc b where b.c# = course.c#) as decimal(10,2)) [优秀率>=90]
from course join sc on course.c# = sc.c#
group by course.c#,course.cname

--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 * from sc order by c#

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) + 1 
from sc t 
order by t.c# , px

--19.2 sql 2005用rank,DENSE_RANK完成
--Score重复时保留名次空缺(rank完成)
select t.* , px = rank() over(partition by c# order by score desc) 
from sc t 
order by t.C# , px 

--Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by c# order by score desc) 
from sc t 
order by t.C# , px 

--20、查询学生的总成绩并进行排名
--20.1 查询学生的总成绩
select * from sc order by sc.s#

select sc.s#,sum(sc.score) total_score,rank() over(order by sum(sc.score)) final_rank
from sc
group by sc.s#
order by final_rank

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 t2.s#, t2.sname,t2.score, count(t1.score)+1 rank
from
(
	select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t1 
right join
(
	select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t2
on t1.score > t2.score
group by t2.s#,t2.sname,t2.score
order by rank

select t2.s#, t2.sname,t2.score, count(distinct t1.score)+1 rank
from
(
	select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t1 
right join
(
	select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t2
on t1.score > t2.score
group by t2.s#,t2.sname,t2.score
order by rank

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 student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname

select t1.s#,t1.sname,t1.score,rank() over(order by t1.score desc) as [排名]
from
(select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t1

select t1.s#,t1.sname,t1.score,dense_rank() over(order by t1.score desc) as [排名]
from
(select student.s# as s#,student.sname as sname,isnull(sum(sc.score),0) as score
	from student left join sc on student.s# = sc.s#
	group by student.s#,student.sname
) as t1

--21、查询不同老师所教不同课程平均分从高到低显示 
select * from teacher

select * from teacher left join course on teacher.t# = course.t#

select teacher.t#,teacher.tname,course.c#,course.cname, cast(avg(sc.score) as decimal(10,2)) avg_score
from teacher left join course on teacher.t# = course.t#
             left join sc on course.c# = sc.c#
group by teacher.t#,teacher.tname,course.c#,course.cname
order by avg_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
--22.1 sql 2000用子查询完成
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 b.*, px = (select count(distinct a.score)  from sc a where a.c# = b.c# and a.score > b.score) + 1
from sc b
)m
where m.px between 2 and 3
order by m.c#,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 sc.*, rank() over(partition by sc.c# order by sc.score desc) as px
from sc

select *
from
(
	select sc.*, rank() over(partition by sc.c# order by sc.score desc) as px
	from sc
)m
where m.px between 2 and 3
order by c#

--Score重复时合并名次(DENSE_RANK完成)
select *
from
(
	select sc.*, dense_rank() over(partition by sc.c# order by sc.score desc) as px
	from sc
)m
where m.px between 2 and 3
order by c#

--23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 
--23.1 统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]
--横向显示
select *
from sc join course on sc.c# = course.c#
order by sc.c#;

select sc.c#,course.cname,
	count(sc.s#) as 总人数,
	sum(case when (sc.score between 85 and 100) then 1 else 0 end) as [85-100],
	cast(((sum(case when (sc.score between 85 and 100) then 1 else 0 end)*1.0)/count(sc.s#)) as decimal(10,2)) as '[85-100]%',
	sum(case when (sc.score between 70 and 85) then 1 else 0 end) as [85-70],
	sum(case when (sc.score between 60 and 70) then 1 else 0 end) as [70-60],
	sum(case when (sc.score between 0 and 60) then 1 else 0 end) as [60-0]
from sc join course on sc.c# = course.c#
group by sc.c#,course.cname

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] ,n.score [分数],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)  
from Course m , sc n
where m.C# = n.C# 


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), all子句的使用
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 sc.c#, course.cname,
	cast(sum(case when sc.score >= 85 and sc.score <= 100 then 1 else 0 end)*1.0/count(sc.s#) as decimal(10,2)) as '100-85百分比',
	cast(sum(case when sc.score >= 70 and sc.score <  85 then 1 else 0 end)*1.0/count(sc.s#)  as decimal(10,2)) as '85-70百分比',
	cast(sum(case when sc.score >= 60 and sc.score <  70 then 1 else 0 end)*1.0/count(sc.s#) as decimal(10,2))as '70-60百分比',
	cast(sum(case when sc.score >=  0 and sc.score <  60 then 1 else 0 end)*1.0/count(sc.s#) as decimal(10,2)) as '<60百分比'
from course join sc on course.c# = sc.c#
group by sc.c#,course.cname

--纵向显示1(显示存在的分数段)
select sc.c#,course.cname,
	case
		when sc.score >= 85 and sc.score <= 100 then '85-100'
		when sc.score >= 70 and sc.score <  85 then '70-85'
		when sc.score >= 60 and sc.score <  70 then '60-70'
		when sc.score >=  0 and sc.score <  60 then '0-60'
	end as 分数段
from course join sc on course.c# = sc.c#
group by sc.c#,course.cname,case
	when sc.score >= 85 and sc.score <= 100 then '85-100'
	when sc.score >= 70 and sc.score <  85 then '70-85'
	when sc.score >= 60 and sc.score <  70 then '60-70'
	when sc.score >=  0 and sc.score <  60 then '0-60'
	end

select a.c#,course.cname,
	case
		when a.score >= 85 and a.score <= 100 then '85-100'
		when a.score >= 70 and a.score <  85 then '70-85'
		when a.score >= 60 and a.score <  70 then '60-70'
		when a.score >=  0 and a.score <  60 then '0-60'
	end as 分数段, cast(count(a.s#)*1.0/(select count(*) from sc b where b.c# = a.c#)as decimal(10,2)) as 百分比
from course join sc as a on course.c# = a.c#
group by a.c#,course.cname,case
	when a.score >= 85 and a.score <= 100 then '85-100'
	when a.score >= 70 and a.score <  85 then '70-85'
	when a.score >= 60 and a.score <  70 then '60-70'
	when a.score >=  0 and a.score <  60 then '0-60'
	end

--纵向显示2(显示存在的分数段，不存在的分数段用0显示)
/*
//这种方式不行
select a.c#,course.cname,
	case
		when a.score >= 85 and a.score <= 100 then '85-100'
		when a.score >= 70 and a.score <  85 then '70-85'
		when a.score >= 60 and a.score <  70 then '60-70'
		when a.score >=  0 and a.score <  60 then '0-60'
	end as 分数段, cast(count(a.s#)*1.0/(select count(*) from sc b where b.c# = a.c#)as decimal(10,2)) as 百分比
from course , sc as a 
where course.c# = a.c#
group by all a.c#,course.cname,case
	when a.score >= 85 and a.score <= 100 then '85-100'
	when a.score >= 70 and a.score <  85 then '70-85'
	when a.score >= 60 and a.score <  70 then '60-70'
	when a.score >=  0 and a.score <  60 then '0-60'
	end
*/

select course.c#,course.cname,
	case
		when a.score >= 85 and a.score <= 100 then '85-100'
		when a.score >= 70 and a.score <  85 then '70-85'
		when a.score >= 60 and a.score <  70 then '60-70'
		else '0-60'
	end as 分数段, cast(count(a.s#)*1.0/(select count(*) from sc b where b.c# = course.c#)as decimal(10,2)) as 百分比
from course , sc as a
where course.c# = a.c#
group by all course.c#,course.cname, case
	when a.score >= 85 and a.score <= 100 then '85-100'
	when a.score >= 70 and a.score <  85 then '70-85'
	when a.score >= 60 and a.score <  70 then '60-70'
	else '0-60'
	end
order by course.c#,course.cname

--24、查询学生平均成绩及其名次 
--24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，
--分平均成绩重复时保留名次空缺和不保留名次空缺两种。
select sc.s#, cast(avg(sc.score) as decimal(10,2)) as 'avgscore'
from sc
group by sc.s#
order by avg(sc.score)

----平均成绩重复时保留名次空缺
select t.s#,t.avgscore,
	px = (select count(*) 
		  from (select sc.s#, cast(avg(sc.score) as decimal(10,2)) as 'avgscore'
				from sc
		  group by sc.s#) a 
		 where a.avgscore >= t.avgscore)
from (
select sc.s#, cast(avg(sc.score) as decimal(10,2)) as 'avgscore'
from sc
group by sc.s#
)t
order by px

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

----平均成绩重复时不保留名次空缺
select t.s#,t.avgscore,
	px = (select count(distinct a.avgscore) 
		  from (select sc.s#, cast(avg(sc.score) as decimal(10,2)) as 'avgscore'
				from sc
		  group by sc.s#) a 
		 where a.avgscore >= t.avgscore
		)
from (
select sc.s#, cast(avg(sc.score) as decimal(10,2)) as 'avgscore'
from sc
group by sc.s#
)t
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 student.s# as s#, student.sname as sname,sc.c# as c#, sc.score as score
from student left join sc on student.s# = sc.s#

select student.s# as s#, student.sname as sname,avg(sc.score) avgscore
from student left join sc on student.s# = sc.s#
group by student.s#,student.sname

select student.s# as s#, student.sname as sname,isnull(avg(sc.score),0) avgscore,px = rank() over(order by avg(sc.score) desc)
from student left join sc on student.s# = sc.s#
group by student.s#,student.sname

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 student.s# as s#, student.sname as sname,isnull(avg(sc.score),0) avgscore,px = dense_rank() over(order by avg(sc.score) desc)
from student left join sc on student.s# = sc.s#
group by student.s#,student.sname

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 course.c#,cname,score,px = rank() over(partition by course.c# order by sc.score desc)
from course left join sc on course.c#= sc.c#

select *
from
(
select course.c#,cname,score,px = rank() over(partition by course.c# order by sc.score desc)
from course left join sc on course.c#= sc.c#
)t
where t.px in (1,2,3)

--25.2 分数重复时不保留名次空缺，合并名次
select *
from
(
select course.c#,cname,score,px = dense_rank() over(partition by course.c# order by sc.score desc)
from course left join sc on course.c#= sc.c#
)t
where t.px in (1,2,3)

--sql 2000用子查询实现
select course.c#,course.cname,score
from course left join sc on course.c#= sc.c#

select course.c#,course.cname,bsc.score, 
px = (select count(distinct a.score) from sc a where a.c#= bsc.c# and a.score >= bsc.score)
from course left join sc as bsc on course.c#= bsc.c#
order by c#,bsc.score

select *
from(
select course.c#,course.cname,bsc.score as score, 
px = (select count(distinct a.score) from sc a where a.c#= bsc.c# and a.score >= bsc.score)
from course left join sc as bsc on course.c#= bsc.c#
)t where t.px in (1,2,3)

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 

--26、查询每门课程被选修的学生数 
select course.c#,cname, isnull(选修人数,0) 选修人数
from course left join
(
select c#,count(s#) [选修人数]
from sc
group by c#
)t on course.c# = t.c#

--27、查询出只有两门课程的全部学生的学号和姓名
select s#
from sc 
group by s#
having count(c#) = 2

select *
from student
where student.s# in(
select s#
from sc 
group by s#
having count(c#) = 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 ssex,count(s#) as 人数
from student
group by ssex

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 student.sname like N'%风%'

select charindex(N'风',N'风中')
select charindex(N'风',N'中风中')
select charindex(N'风',N'中')

select student.*
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 
exec show_student

select a.sname,count(a.s#)
from student a,student b
where a.sname = b.sname and a.s# != b.s#
group by a.sname

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) 
exec show_student

select *
from student
where year(student.sage) = 1990

select * from Student where datepart(yy,sage) = 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 *
from course left join sc on course.c# = sc.c#

select course.c#,course.cname,cast(avg(sc.score) as decimal(10,2)) [平均成绩]
from course left join sc on course.c# = sc.c#
group by course.c#,course.cname
order by 平均成绩 desc,course.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 student.s#,student.sname,cast(avg(sc.score) as decimal(10,2)) [平均成绩]
from student join sc on student.s# = sc.s#
group by student.s#,student.sname
having avg(sc.score) >= 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 student.sname,t.score from student,(
select sc.s#,sc.score
from course join sc on course.c# = sc.c#
where course.cname = N'数学' and sc.score < 60) as t
where student.s# = t.s#

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.score
from student join sc on student.s# = sc.s#
			 left join course on sc.c# = course.c#

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 s#,c#,score
from sc
where sc.s# not in (select s# from sc where sc.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# 

select student.sname,course.cname,score
from student join
(
select s#,c#,score
from sc
where sc.s# not in (select s# from sc where sc.score < 70)
)t on student.s# = t.s# 
join course on t.c# = course.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 distinct student.s#,student.sname
from sc join student on sc.s# = student.s#
where sc.c# = '01' and sc.score >= 80

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 sc.c#,course.cname,count(sc.s#) as 人数
from course left join sc on course.c# = sc.c#
group by sc.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.*,sc.score
from course join sc on course.c# = sc.c#
			join teacher on course.t# = teacher.t# and teacher.tname = N'张三'
			join student on student.s# = sc.s#
order by sc.score desc

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.*,sc.score
from course join sc on course.c# = sc.c#
			join teacher on course.t# = teacher.t# and teacher.tname = N'张三'
			join student on student.s# = sc.s#
where sc.score = (
select max(sc.score) as maxscore
from course join sc on course.c# = sc.c#
			join teacher on course.t# = teacher.t# and teacher.tname = N'张三'
			join student on student.s# = sc.s#
group by sc.c#)

--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 *
from sc a join sc b on a.c# = b.c# and a.score = b.score and a.s# != b.s#

select a.s#,a.c#,a.score
from sc a join sc b on a.c# = b.c# and a.score = b.score and a.s# != b.s#
order by a.c#

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
from
(select sc.*,px = rank() over( partition by sc.c# order by sc.score desc)
from sc)t
where t.px between 1 and 2

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 c#, count(s#) as 选修人数
from sc
group by c#
having count(s#) >= 5
order by count(s#) desc,c#

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 s#
from sc
group by s#
having count(c#) >= 2

--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 sc join student on sc.s# = student.s#
group by student.s#,student.sname,student.sage,student.ssex
having count(c#) = (select count(*) 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#)
)

select * from student
select * from course
select * from student , course 

--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 student.*,year(getdate()) - year(sage) [年龄] from student
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 (month(getdate()) < month(sage)) or (month(getdate()) = month(sage)) and day(getdate()) < day(sage) then year(getdate())-year(sage)-1 else year(getdate())-year(sage)   
				
		  end
from student

--CONVERT(data_type(length),data_to_be_converted,style)
--data_type(length) 规定目标数据类型（带有可选的长度）。data_to_be_converted 含有需要转换的值。style 规定日期/时间的输出格式。
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