假如已有學生表和課程表的話，再加一個選課表就可以了。選課表與學生表，課程表都是1對N的關系，這樣學生表和課程表就可以是N對N的關系了。
學生表-----studentid-----選課表-----courseid-----課程表
Create table StudentCourse (
id int Indentity(1,1) not null,
StudentId varchar(10) not null,
CourseId varchar(10) not null,
Score int
)
然后依次用insert to 插入數據
至于樓主的那些要求都是數據上的問題，自行控制吧

CREATE TABLE sc(
sno VARCHAR(12) NOT null,//學生編號
cno VARCHAR(4) NOT NULL,//課程編號
cname varchar(10),//數據類型要跟你的學生表和課程表都一樣 課程名
PRIMARY KEY(sno,cno),//設置聯合主鍵
FOREIGN KEY (sno) REFERENCES student(sno),//設置外鍵
FOREIGN KEY (cno) REFERENCES course(cno)//設置外鍵
);

你是要做一個查詢呢，還是要做什么，看你寫的象是要做一個程序。

問題一：select sno, sname from student
where sno in(select sno from sc
where cno =(select cno from couse
where cname="計算機原理"))
問題二：select cname from couse
where cno in(select cno from sc
where sno=(select sno from student
where sname ="周星馳"))
問題三：select sno, sname from student
where sno in(select sno from sc
where count(cno)=5

第一個
select s.sno, s.sname
from student s, sc t
where s.sno = t.sno
and t.cno = (
select c.cno
from couse c
where c.cname = '計算機原理'
);
第二個：
select c.cname
from student s, couse c, sc t
where c.cno = t.cno and t.sno = s.sno and s.sname = '周星馳';
第三個：
select s.sno, s.sname
from student s
where s.sno in (
select t.sno
from sc t
group by t.sno
having count(t.cno) = 5
);
我已經在本地數據庫建表測試過了，如果有什么問題，可以再聯系我。

你那個couse表名字錯誤了吧，應該是course才對吧？答案如下

問題一：查詢選了”計算機原理“的學生學號和姓名？
SELECT student.sno, student.sname
FROM student, sc, couse
WHERE student.sno = sc.sno
AND couse.cname = '計算機原理'
AND couse.cno = sc.cno

問題二：查詢”周星馳“同學選修的課程名字？
SELECT cname
FROM couse
WHERE cno IN (SELECT sc.cno FROM sc, student
WHERE sc.sno = student.sno
AND student.sname = '周星馳')

問題三：選秀了5門課程的學生學號和姓名？
SELECT sno, sname
FROM student
WHERE sno IN (SELECT sno FROM sc
GROUP BY sno
HAVING COUNT(cno) = 5)

加油！取得答案后好好研究下自己掌握才是最重要的。

這些問題都不難的，在企業管理器里面稍微動手寫一下，相信花點時間不難出來結果

問題一：

select sno,sname from student where sno in
(
select sno from sc where cno =
(
select cno from course where cname='計算機原理'
)
)

問題二：

select cname from course where cno in
(
select cno from sc where sno =
(
select sno from student where sname = '周星弛'
)
)

問題三：

select sno,sname from student where sno in
(
select sno from sc group by sno having count(*) = 5
)

用子查詢
select 姓名
from 學生,選課,課程
where 學生.學號=選課.學號 and 選課.課程號=課程.課程號 and 課程.課程名='數據庫'
and 學生.學號 in
(select 學生.學號
from 學生,選課,課程
where 學生.學號=選課.學號 and 選課.課程號=課程.課程號 and 課程.課程名='操作系統';
}

您好，從您的SQL語法中猜測，你是使用T-SQL在SQL SERVER中建立數據庫。
您的前兩個SQL中把primarykey 寫成primary key中間加個空格。
第三個語句，把identify換成大寫，即可。我已經在 sqlserver2000中測試通過了語句。呵呵。
希望能給你提供點信息。祝你好運。

11

1. select * from SC
2. select Sname，Sage from Student where Sdept = '計算機'
3. select Sno,Cno,Grade from SC where Grade >= 70 and Grade <= 80
4. select Sname,Sage from Student where Sage between 18 and 20 and Ssex = '男'
5. select top 1 Grade from SC where Cno = 'C01'
6. select max(Sage),min(Sage) from Student
7. select Sdept,sum(Sno) from Student group by Sdept
8. select course.Cname,sum(sc.Sno),max(Grade) from SC
join studet on Student.Sno = SC.Sno
join Course on Course.Cno = SC.Cno
group by course.cname,max(grade)

9. select sum(Cno),avg(Grade) from SC
join Course on Course.Cno = SC.Cno
join Student on Student.Sno= SC.Sno
order by SC.Sno

10. select Stuent.Sno，Stuent.Sname，sum(Grade) A from SC
join Student on Student.Sno = SC.Sno
group by sc.Sno,student.Sname
having A > 200

11. select Student.Sname,Student.Sdept from Student
join Course on Course.Cno = SC.Cno
join SC on SC.Sno = Student.Sno
where SC.Cno = 'C02'

12. select Student.sname,course.cno,sc.grade from sc
join student on student.sno = sc.sno
join course on course.cno = sc.cno
where sc.grade >= 80
order by sc.grade desc

13. select cno,cname from
(
select course.cno,course.cname,sun(sno) from student
join course on course.cno = sc.cno
join sc on sc.sno = student.sno
group by cno,cname
having sun(sno) > 0
)

14. ① select student.sname,student.sdept from
(
select student.sname,student.sdept,course.cname from student
join sc on sc.sno = student.sno
join course on course.cno = sc.cno
where course.cname = 'C01'
)

② select student.sno,student.sname from
(
select student.sno,student.sname,student.sdept,sc.grade from sc
join student on student.sno = sc,sno
where student.sdept = '信息' and sc.grade >= 80
)

③ select top 1 student.sname from
(
select student.sname,student.sdept,sum(sc.grade) from sc
join student on student.sno = sc.sno
where student.sdept = '計算機'
group by student.sname,student.sdept
order by
)

15. delete from sc where grade < 50
16. update sc set grade += 5 from sc
join course on course.cno = sc.cn
join student on student.sno = sc.sno
where student.sno =
(
select student.sno from sc
join course on course.cno = sc.cno
where course.cname = 'c01'
)

17. update sc set grade += 10 from sc
join student on student.sno = sc.sno
where student.sno =
(
select student.sno from student
join sc on sc.sno = student.sno
join course on course.cno = sc.cno
where student.sdept = '計算機' and course.cname = '計算機文化基礎'
)

18. create view [A] as
select student.sno,student.sname,student.sdept,course.cno,course.cname,sc.grade from sc
join student on student.sno = sc.sno
join course on course.cno = sc.cno

19. create view [A] as
select student.sno,avg(sc.grade) from sc
join student on student.sno = sc.sno
group by student.sno

20. create view [A] as
select student.sno,sum(sc.grade) from sc
join student on student.sno = sc.sno
group by student.sno

21. create index A on student(sname)
22. 不會

select * from course
select Sname,Sage from Student where sdept = '計算機系'
select * from sc where 70 <= grade and grade =>80

select Sname,Sage from Student where sdept = '計算機系' and Ssex='男'
。。。。

自己看手冊把，都很簡單的，題太多，不好回答，，

先做題，后看答案哦！ 相信乖乖做完的你肯定會有不小的收獲的~
SQL代碼如下：
CREATE TABLE student
(Sno varchar(20) NOT NULL,
Sname varchar(20) NOT NULL,
sex varchar(20) NOT NULL,
age INT NOT NULL,
dept varchar(20) NOT NULL,
PRIMARY KEY (Sno)
);

CREATE TABLE course
(Cno varchar(20) NOT NULL,
Cname varchar(20) NOT NULL,
hours VARCHAR(20) NOT NULL,
PRIMARY KEY (Cno)
);

CREATE TABLE SC
(Sno varchar(20) NOT NULL,
Cno varchar(20) NOT NULL,
grade INT ,
PRIMARY KEY (Sno,Cno)
);

INSERT INTO student VALUES ('9512101','李勇','男',19,'計算機系');
INSERT INTO student VALUES ('9512102','劉晨','男',20,'計算機系');
INSERT INTO student VALUES ('9512103','王敏','女',20,'計算機系');
INSERT INTO student VALUES ('9521101','張立','男',22,'信息系');
INSERT INTO student VALUES ('9521102','吳賓','女',21,'信息系');
INSERT INTO student VALUES ('9521103','張海','男',20,'信息系');
INSERT INTO student VALUES ('9531101','錢小力','女',18,'數學系');
INSERT INTO student VALUES ('9531102','王大力','男',19,'數學系');

INSERT INTO course VALUES ('C01','計算機文化學','70');
INSERT INTO course VALUES ('C02','VB','90');
INSERT INTO course VALUES ('C03','計算機網絡','80');
INSERT INTO course VALUES ('C04','數據庫基礎','108');
INSERT INTO course VALUES ('C05','高等數學','180');
INSERT INTO course VALUES ('C06','數據結構','72');

INSERT INTO SC VALUES ('9512101','C01',90);
INSERT INTO SC VALUES ('9512101','C02',86);
INSERT INTO SC VALUES ('9512101','C06',NULL);
INSERT INTO SC VALUES ('9512102','C02',78);
INSERT INTO SC VALUES ('9512102','C04',66);
INSERT INTO SC VALUES ('9521102','C01',82);
INSERT INTO SC VALUES ('9521102','C02',75);
INSERT INTO SC VALUES ('9521102','C04',92);
INSERT INTO SC VALUES ('9521102','C05',50);
INSERT INTO SC VALUES ('9521103','C02',68);
INSERT INTO SC VALUES ('9521103','C06',NULL);
INSERT INTO SC VALUES ('9531101','C01',80);
INSERT INTO SC VALUES ('9531101','C05',95);
INSERT INTO SC VALUES ('9531102','C05',85);

-- 1.分別查詢學生表和學生修課表中的全部數據
SELECT * FROM student
SELECT * FROM course

-- 2.查詢成績在70-80分之間的學生的學號、課程號和成績
SELECT *
FROM SC
WHERE grade > 70 and grade < 80

-- 3.查詢C01號課程成績最高的分數
SELECT max(grade)
FROM SC
WHERE Cno = 'C01'
GROUP BY Cno

-- 4.查詢學生都選修了哪些課程，要求列出課程號
SELECT Cno
FROM SC
GROUP BY Cno

-- 5.查詢C02號課程的所有學生的平均成績、最高成績和最低成績。
SELECT AVG(grade),MAX(grade),MIN(grade)
FROM SC
WHERE Cno = 'C02'
group BY Cno

-- 6.統計每個系的學生人數
SELECT dept,count(*) as `人數`
FROM student
GROUP BY dept

-- 7.統計每門課程的修課人數和考試最高分
SELECT Cno, count(*),max(grade)
FROM SC
group BY Cno

-- 8.統計每個學生的選課門數，并按選課門數的遞增順序顯示結果
SELECT Sno, count(*)
FROM SC
GROUP BY Sno
ORDER BY count(*)

-- 9.統計選修課的學生總數和考試的平均成績
SELECT Cno, count(*), avg(grade)
FROM SC
GROUP BY Cno

-- 10.查詢選課門數超過2門的學生的平均成績和選課門數
SELECT Sno, count(*),avg(grade)
FROM SC
GROUP BY Sno
having count(*) > 2

-- 11.列出總成績超過200分的學生，要求列出學號、總成績
SELECT Sno,sum(grade)
FROM SC
GROUP BY Sno
HAVING sum(grade) > 200

-- 12.查詢選修了C02號課程的學生的姓名和所在系。
SELECT student.Sname,student.dept,SC.Cno
FROM student join SC ON student.Sno = SC.Sno -- 選取內容在兩張表中，第一種方法將兩張表聯結起來
WHERE Cno = 'C02'

SELECT Sname,dept
FROM student
WHERE Sno in (SELECT Sno
FROM SC
WHERE Cno = 'C02') -- 第二種方法，采用子查詢法。

-- 13.查詢成績80分以上的學生的姓名、課程號和成績，并按成績的降序排列結果。
SELECT S.Sname,SC.Cno,SC.grade
FROM student AS S JOIN SC ON S.Sno = SC.Sno
WHERE SC.grade > 80
ORDER BY SC.grade DESC

-- 14.查詢計算機系男生修了‘數據庫基礎’的學生姓名、性別、成績
SELECT S.Sname,S.sex,SC.grade
FROM student AS S JOIN SC ON S.Sno = sc.Sno
JOIN course ON sc.Cno = course.Cno
WHERE S.dept = '計算機系' AND S.sex = '男' AND course.Cname = '數據庫基礎'

SELECT Sname,sex,SC.Grade
FROM student
inner join SC ON Cno IN(SELECT Cno FROM course WHERE Cname='數據庫基礎') -- 顯示成績的條件,僅聯結了兩張表，On 后面也可以跟不同的聯結條件。
AND student.Sno=SC.Sno -- 顯示成績的學生的學號
WHERE dept='計算機系' AND sex='男'

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

-- 15.查詢哪些學生的年齡相同，要求列出年齡相同的學生的姓名和年齡。
SELECT S1.Sname,S1.sex
FROM student AS S1, student AS S2
WHERE S1.age = S2.age AND S1.Sname <> S2.Sname
GROUP BY S1.Sname
ORDER BY S1.age

-- 16.查詢哪些課程沒有人選，要求列出課程號和課程名。
SELECT Cno,Cname
FROM course
WHERE Cno NOT in (SELECT Cno
FROM sc
GROUP BY Cno)

-- 17.查詢有考試成績的所有學生的姓名、修課名稱及考試成績。
-- 要求將查詢結果放在一張新的永久表(假設新表名為new-sc)中。
CREATE TABLE new_sc AS -- 將查詢結果放在新表中的方法。
SELECT S.Sname,course.Cname,sc.grade
FROM student AS S inner JOIN sc ON S.Sno = sc.Sno
INNER JOIN course ON course.Cno = sc.Cno
WHERE grade IS NOT NULL

-- 18.分別查詢信息系和計算機系的學生姓名、性別、修課名稱、修課成績
SELECT S.Sname,S.sex,course.Cname,sc.grade
FROM student AS S inner JOIN sc ON S.Sno = sc.Sno
INNER JOIN course ON course.Cno = sc.Cno
WHERE S.dept = '信息系'
UNION
SELECT S.Sname,S.sex,course.Cname,sc.grade
FROM student AS S inner JOIN sc ON S.Sno = sc.Sno
INNER JOIN course ON course.Cno = sc.Cno
WHERE S.dept = '計算機系'

-- 19.用子查詢實現如下查詢：
-- （1）查詢選修了C01號課程的學生的姓名和所在系
SELECT Sname,dept
FROM student
WHERE Sno in (SELECT Sno
FROM sc
WHERE Cno = 'C01')

-- （2）查詢數學系成績80分以上的學生的學號和姓名。
SELECT Sno,Sname
FROM student
WHERE dept = '數學系' AND Sno in (SELECT Sno FROM sc WHERE grade > 80)

-- （3）查詢計算機系學生所選的課程名
SELECT Cname
FROM course
WHERE Cno in (SELECT Cno
FROM sc
where Sno in (SELECT Sno
FROM student
WHERE dept = '計算機系'))

-- 20.將計算機系成績高于80分的學生的修課情況插入到另一張表中，分兩種情況實現:
-- （1）在插入的過程中建表
SELECT Sno,Sname
FROM(
SELECT Sno,Sname
FROM student
WHERE dept = '數學系' AND Sno in (SELECT Sno FROM sc WHERE grade > 80)) AS newtable

-- （2）先建一個新表，然后再插入數據
CREATE TABLE newtable2 AS
SELECT Sno,Sname
FROM student
WHERE dept = '數學系' AND Sno in (SELECT Sno FROM sc WHERE grade > 80)

-- 21.刪除修課成績小于50分的學生的修課記錄
DELETE FROM sc
WHERE grade < 50 or grade IS NULL

-- 22.將所有選修了C01課程的學生的成績加10分。
UPDATE sc
SET grade = grade + 10
WHERE Cno = 'C01'

老師留的作業嗎？為你好，自己做吧。