数据库中有四张表：Teacher(教师)表、Student(学生)表、Course(课 程)表和StuCur(选课)表

1. 数据库中有四张表：Teacher(教师)表、Student(学生)表、Course(课 程)表和StuCur(选课)表

1) select s.Name from Student s join StuCur c on s.id=c.sid
join Course cu on cu.id=c.cid where cu.Name='自然'
2) select c.Name from  Course c join StuCur sc on c.id=sc.CID
join Student s on sc.sid=s.id where s.EntranceTime between '1999-01-01' and '2012-12-31'

2. 有四张表：Student 学生表 ，Course 课程表 ，SC成绩表 ，Teacher 教师表，求SQL语句

把表结构发来，我给你写

3. 有四张表 student(s#,sname,sage,sex)学生表，Course(c#,cname,t#)课程表，SC（s#,c#,score）成绩表，

全部通过测试-----
第一题
select a.sno,count(a.sno) as 选课数,Sum(c.score) as 总成绩 
from student as a,course as b,SC as c
where a.sno=c.sno and b.cno=c.cno
group by a.sno
第二题
select sno,sname from student
where sno in(
select distinct(d.sno)
from student as d,sc as e
where
d.sno=e.sno and e.sno'001' and
 e.cno in
(
select b.cno
from student as a,course as b,SC as c
where a.sno=c.sno and b.cno=c.cno and a.sno='001'
))
第三题
update sc
set score=(select e.cavgScore from(
select cno as classno,avg(score) as cavgScore
from sc
where cno in(
select cno
from course as a,teacher as b
where tname='叶平')
group by cno) as e where classno=cno)
第四题
delete from sc
where sc.cno in
(
   select a.cno
   from course as a,teacher as b
   where a.tno=b.tno and b.tname='叶平'
)

4. 有学生头选课系统,数据库中有四个表,分别为: Student(S#, Sname, Sage, Ssex)

1
select a.S#, avg(b.Score) Student a,SC b where a.S#=b.S# group by a.S#
having avg(b.Score) > 60
 
2
select a.S#, avg(b.Score) Student a,SC b where a.S#=b.S#  and S# IN 
(
select S# from SC where Score  2 
)
group by a.S#
 
3
select S#,Sname from where S# not in 
(select a.S# from Student a,Course b,SC c,Teacher d 
where 
a.S#=c.S# and 
b.C#=c.C# and
b.T#=d.T# and 
d.Tname='叶平')

5. Student 学生表 ，Course 课程表 ，SC成绩表 ，Teacher 教师表，sql操作运用

建表语句 

CREATE TABLE student
(
s#    INT,
sname nvarchar(32),
sage  INT,
ssex  nvarchar(8)
) 

CREATE TABLE course
(
c#    INT,
cname nvarchar(32),
t#    INT
) 

CREATE TABLE sc
(
s#    INT,
c#    INT,
score INT
) 

CREATE TABLE teacher
(
t#    INT,
tname nvarchar(16)
) 


插入测试数据语句 

insert into Student select 1,N'刘一',18,N'男' union all
select 2,N'钱二',19,N'女' union all
select 3,N'张三',17,N'男' union all
select 4,N'李四',18,N'女' union all
select 5,N'王五',17,N'男' union all
select 6,N'赵六',19,N'女' 

insert into Teacher select 1,N'叶平' union all
select 2,N'贺高' union all
select 3,N'杨艳' union all
select 4,N'周磊'

insert into Course select 1,N'语文',1 union all
select 2,N'数学',2 union all
select 3,N'英语',3 union all
select 4,N'物理',4

insert into SC 
select 1,1,56 union all 
select 1,2,78 union all 
select 1,3,67 union all 
select 1,4,58 union all 
select 2,1,79 union all 
select 2,2,81 union all 
select 2,3,92 union all 
select 2,4,68 union all 
select 3,1,91 union all 
select 3,2,47 union all 
select 3,3,88 union all 
select 3,4,56 union all 
select 4,2,88 union all 
select 4,3,90 union all 
select 4,4,93 union all 
select 5,1,46 union all 
select 5,3,78 union all 
select 5,4,53 union all 
select 6,1,35 union all 
select 6,2,68 union all 
select 6,4,71


问题

问题： 1、查询“001”课程比“002”课程成绩高的所有学生的学号；  select a.S# from (select s#,score from SC where C#='001') a,(select s#,score  from SC where C#='002') b  where a.score>b.score and a.s#=b.s#; 2、查询平均成绩大于60分的同学的学号和平均成绩；    select S#,avg(score)    from sc    group by S# having avg(score) >60; 3、查询所有同学的学号、姓名、选课数、总成绩；  select Student.S#,Student.Sname,count(SC.C#),sum(score)  from Student left Outer join SC on Student.S#=SC.S#  group by Student.S#,Sname 4、查询姓“李”的老师的个数；  select count(distinct(Tname))  from Teacher  where Tname like '李%'; 5、查询没学过“叶平”老师课的同学的学号、姓名；    select Student.S#,Student.Sname    from Student      where S# not in (select distinct( SC.S#) from SC,Course,Teacher where  SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平'); 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；  select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002'); 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；  select S#,Sname  from Student  where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher  where Teacher.T#=Course.T# and Tname='叶平')); 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；  Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2  from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 60); 10、查询没有学全所有课的同学的学号、姓名；    select Student.S#,Student.Sname    from Student,SC    where Student.S#=SC.S# group by  Student.S#,Student.Sname having count(C#) <(select count(C#) from Course); 11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名；    select distinct S#,Sname from Student,SC where Student.S#=SC.S# and SC.C# in (select C# from SC where S#='1001'); 12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；    select distinct SC.S#,Sname    from Student,SC    where Student.S#=SC.S# and C# in (select C# from SC where S#='001'); 13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩；    update SC set score=(select avg(SC_2.score)    from SC SC_2    where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名；    select S# from SC where C# in (select C# from SC where S#='1002')    group by S# having count(*)=(select count(*) from SC where S#='1002'); 15、删除学习“叶平”老师课的SC表记录；    Delect SC    from course ,Teacher      where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平'; 16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、    号课的平均成绩；    Insert SC select S#,'002',(Select avg(score)    from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002'); 17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分    SELECT S# as 学生ID        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语        ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩    FROM SC AS t    GROUP BY S#    ORDER BY avg(t.score)   18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分    SELECT L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分    FROM SC L ,SC AS R    WHERE L.C# = R.C# and        L.score = (SELECT MAX(IL.score)                      FROM SC AS IL,Student AS IM                      WHERE L.C# = IL.C# and IM.S#=IL.S#                      GROUP BY IL.C#)        AND        R.Score = (SELECT MIN(IR.score)                      FROM SC AS IR                      WHERE R.C# = IR.C#                  GROUP BY IR.C#                    ); 
自己写的:select c# ,max(score)as 最高分 ,min(score) as 最低分 from dbo.sc  group by c# 19、按各科平均成绩从低到高和及格率的百分数从高到低顺序    SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩        ,100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数    FROM SC T,Course    where t.C#=course.C#    GROUP BY t.C#    ORDER BY 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004）    SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分        ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数        ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分        ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数        ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分        ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数        ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分        ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数  FROM SC


21、查询不同老师所教不同课程平均分从高到低显示
SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C ,Teacher AS Z
where T.C#=C.C# and C.T#=Z.T#
GROUP BY C.C#
ORDER BY AVG(Score) DESC 
22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004）
[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
SELECT  DISTINCT top 3
SC.S# As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC  LEFT JOIN SC AS T1
ON SC.S# = T1.S# AND T1.C# = '001'
LEFT JOIN SC AS T2
ON SC.S# = T2.S# AND T2.C# = '002'
LEFT JOIN SC AS T3
ON SC.S# = T3.S# AND T3.C# = '003'
LEFT JOIN SC AS T4
ON SC.S# = T4.S# AND T4.C# = '004'
WHERE student.S#=SC.S# and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.S# = T1.S# AND T1.C# = 'k1'
LEFT JOIN sc AS T2
ON sc.S# = T2.S# AND T2.C# = 'k2'
LEFT JOIN sc AS T3
ON sc.S# = T3.S# AND T3.C# = 'k3'
LEFT JOIN sc AS T4
ON sc.S# = T4.S# AND T4.C# = 'k4'
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 

23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
SELECT SC.C# as 课程ID, Cname as 课程名称
,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]
,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]
,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]
,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]
FROM SC,Course
where SC.C#=Course.C#
GROUP BY SC.C#,Cname; 

24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT S#,AVG(score) AS 平均成绩
FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩
FROM SC
GROUP BY S#
) AS T2
ORDER BY 平均成绩 desc;

原文地址：http://www.cnblogs.com/qixuejia/p/3637735.html

6. 求用java语言编写一个学生成绩管理系统，使用sql数据库（包含5张表:Student,Course,Class,Teacher,Score)

你可以现在数据库里面建立一张视图，视图里面的SQL语句就是查询该数据库中的表明，这就是访问系统表的内容，只提供单独一个列，然后展示给客户，当客户选定某张表的时候，就将该表进行数据备份，可以将表中数据导出到excel，实现方式就是查询该表全部内容，导入excel中，进行保存。
导出数据的时候，最后给个客户提示，备份进行中，备份完成后，将文件路径展示给用户，以便访问

7. 基于student表,course表，score 表，利用sql 语句查询1900年以前出生的学生

create table student
(
s_no varchar(20),
s_name varchar(20),
s_birthday date
)
insert into student (s_no ,s_name,s_birthday  ) values ('001','a','1989-02-01');
insert into student (s_no ,s_name,s_birthday  ) values ('002','b','1900-03-04');
insert into student (s_no ,s_name,s_birthday  ) values ('003','c','1985-11-21');
insert into student (s_no ,s_name,s_birthday  ) values ('004','d','1995-12-14');
insert into student (s_no ,s_name,s_birthday  ) values ('005','e','1994-04-01');

create table course
(
c_no varchar(20),
c_name varchar(20)
)
insert into course (c_no ,c_name ) values ('1','计算机');
insert into course (c_no ,c_name ) values ('2','会计');


create table score 
(
s_no varchar(20),
c_no varchar(20),
sc decimal(18,2) ,
)
insert into score (s_no ,c_no ,sc ) values ('001','1',90);
insert into score (s_no ,c_no ,sc ) values ('001','2',80);
insert into score (s_no ,c_no ,sc ) values ('002','1',88);
insert into score (s_no ,c_no ,sc ) values ('003','1',100);
insert into score (s_no ,c_no ,sc ) values ('003','2',98);
insert into score (s_no ,c_no ,sc ) values ('004','1',96);
insert into score (s_no ,c_no ,sc ) values ('005','2',95);




select c.*,DATEDIFF (year,s_birthday,getdate()) as age from student c inner join 
(select a.*,b.s_no from course a inner join score b  on a.c_no=b.c_no where a.c_name ='会计') d
on c.s_no=d.s_no

8. 设在学生数据库中有三张表，表结构如下所示： Student(学生表): Student表的主键：sno(学号) Course(课程

(1) create table S_C
      (sno char(10) not null
      ,cno char(8)   not null
      ,score int        null
      ,constraint PK_SC primary key (sno,cno)
      )
(2)insert into Student (sno,sname,ssex)
    values('1010','李小丽','女')
(3)create index IND_CName on Course (cname)
(4)update student set sage=23 where sno='1005'
(5)delete from course where cname='管理信息系统'
(6)select cno,cname,ctime
    from course
    where teacher='李元'
    order by cno ASC
(7)select sno,sum(score) as score
    from S_C
    group by sno
(8)create view V_Student
    as
    select *
    from student
    where ssex='男' and sage>=18 and sage<=24
(9)select sno,sname
    from student
    where sno in (select sno from S_C where cno='001')
(10)select A.sno,A.sname
      from student A
      left join S_C B on A.sno=B.sno
      left join Course C on B.cno=C.cno
     where C.cname='关系数据库'
(11)select sno,sname from
        (select A.sno,A.sname,count(1) as count_
       from student A
       left join S_C B on A.sno=B.sno
       group by A.sno,A.sname) A
       where count_>3
(12)select C.sname
        from
        (select * from S_C where cno='002') A
       inner join
      (select * from S_C where cno='004' on) B on A.sno=B.sno
        left join student C on A.sno=C.sno