一个项目涉及到的50个Sql语句(验证中)

Posted 2020-06-21

--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