設有學生成績表sc

① 設有以下3個基本表：學生S，成績SC，課程G

1
insert into s(SNO,SNAME,SAGE) values('990750','錢江',18)
2
update SC set GRADE=null from SC
inner join G on SC.CNO=G.CNO
where CNAME='數學' and GRADE<60
3
select SNO,CNO from SC where GRADE is null
4
select SNO from SC
inner join G on SC.CNO=G.CNO
where TNAME='王老專師屬'

② 設有一個教學資料庫ecation，包括student、course和sc三個表

create table sc(
sno char(100),
cno char(100),
grade number
);
alter table sc add constraint pk_1 primary key(sno,cno)

alter table sc add constraint sno_1 foreign key(sno) references student(sno)
alter table sc add constraint cno_1 foreign key(cno) references course(cno)
select a.*,b.*,c.grade from student a, course b ,sc c
where a.sno=c.sno and b.cno=c.cno and a.name like'張%'
select a.*,b.*,c.grade from student a, course b ,sc c
where a.sno=c.sno and b.cno=c.cno and a.xi_name in('IS','EB')
select sex,sum(age)/count(*) from student a
group by sex
select a.Stu_no,a.name,b.course_name,c.grade from student a, course b ,sc c
where a.sno=c.sno and b.cno=c.cno
select a.stu_no,c.grade from student a, course b ,sc c
where a.sno=c.sno and b.cno=c.cno and b.cno='1'
order by c.grade desc
alter table student add(inyear char(4))
update sc set grade =60 where grade between 55 and 59
delete sc where sno='200910001'
insert student into
(sno,
name,
sex
)
values
('20110001',
'張小濤',
'男'
)

③ 有四張表：Student 學生表 ，Course 課程表 ，SC成績表 ，Teacher 教師表，求SQL語句

select sc.* from sc,course where sc.課程號=course.課程號 and course.課程名='3-105' and sc.成績 between 60 and 80;
select sc.* from sc where sc.成績 in(85,56,88);
3. 這個有疑問
我的理解是，學生參加了多門考試，所有成回績都在70-90之間的。答取這部分人的學號。
select 學號 from (select 學號，max(成績) 最高分，min(成績) 最低分 from sc group by 學號) a where a.最高分 < 90 and a.最低分 >70

④ 設有學生選課表SC(學號，課程號，成績),用SQL命令檢索

C1、C5在表中是兩條記錄哦

⑤ 設有表示學生選課的三張表,學生S(學號,姓名,性別,年齡,身份證號),課程C(課號,課名),選課SC(學號,課號,成績

以下圖為例

學號 課號 成績
X1 K1 90
X1 K2 90
X1 K3 90

X2 K1 90
X2 K2 90
X2 K3 90

X3 K1 90
X3 K3 90

則會發現 學號+課號 具有唯一性（鍵或碼的充要條件）回，單獨的學號或者課答號也不具有唯一性，所以不能單獨作為鍵或碼。 同時注意，圖中成績都為90分的話，那麼成績在該表討論 鍵或碼 的問題中是沒有意義的。如果成績改成排名的話，並且強制排名只能具有唯一性的話，排名就可以單獨成為 鍵或碼。

判斷鍵或碼的時候主要是看是否具有唯一性。

所以 C

⑥ SQL語句執行問題：有一張表sc如下：查詢每位學生的最高成績。

對於每個學號，不存在比這個學號的分數低的數據全部查出來。
反之就是得到每個學號最高的分數了。

另外，你也可以

select sno,max(grade) from sc group by sno

⑦ 設有學生課程表SC（學號，課程名，成績）用SQL檢索同時選修課程號為"C1"和"C5"的學生的學號的正確命令是

你那個是永遠為假的，一個課程號怎麼可能同時為2個值呢，可以這樣寫

SELECT 學號內
FROM SC
WHERE 課程容號 IN ('C1', 'C5')
GROUP BY 學號
HAVING COUNT(DISTINCT 課程號) = 2

或者
SELECT 學號
FROM SC t1
WHERE 課程號 = 'C1'
AND EXISTS (SELECT 1 FROM SC WHERE 學號 = t1.學號 AND 課程號 = 'C5')

⑧ 假設在教務管理系統資料庫中有學生Student、課程關系Course、選修關系SC、試用關系代數表示下列查詢。

從學生，當然1。選擇姓名，成績在那裡（student.學生數=得分。學生證）和（版course.課程名稱=得分。權課程名稱）和（score.學期= '02-03/1'）和（ coourse。課程名=數據結構）
2。無法查詢，因為沒有老師的名字欄位。

從學生，當然，得分3。選擇名稱，其中（student.學生數=得分。學生證）和（course.課程名稱=得分。課程名稱），並得分。成績> =（選擇

4。說每門課檢查出那些誰得分最高等級的名稱

5，從學生刪除，成績在那裡的學生。學號=得分。學號和成績。得分= 0
6。更新設置得分得分。得分=得分。分數*（1 +0.05），其中（course.課程名=得分。課程名稱），當然，課程號= 02和（score.分數< AVG（score.分數））

⑨ 有四張表：Student 學生表 ，Course 課程表 ，SC成績表 ，Teacher 教師表，求SQL語句

把表結構發來，我給你寫

⑩ SQL語句的一道題 三個基本表：學生表(Student)、課程表（Course）、學生選課表（SC）

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