課程表綜合成績怎麼算

⑴ 已知三個關系模式(學生表,課程表,成績表)用SQL語句實現以下內容

CREATE TABLE 選課表

(學號 CHAR(8) REFERENCES 學生表(學號),
課程編號 CHAR(6) REFERENCES 課程表(課程編號),
成績 smallint)
GO
INSERT INTO 學生表 VALUES('20100101','李丹','女','1993-6-6','管理系')

GO
SELECT 學號,姓名,所在系

FROM 學生表
GO
SELECT 學生表.學號,姓名
FROM 學生表 JOIN 選課表 ON 學生表.學號=選課表.學號
WHERE 課程編號='A01-02'
GO
SELECT *
FROM 學生表
WHERE 姓名 LIKE '[趙李張]%'
GO
SELECT 課程名稱,COUNT(*) AS 選修總人數

INTO 修課統計
FROM 課程表 JOIN 選課表 ON 課程表.課程編號=選課表.課程編號
GROUP BY 課程名稱

⑵ 求sql語句 有一張課程表（course）和一張成績表(score)

select sc_name from [成績表內容] where sc_number like '14%' and sc_sex = 0;

select a,co_name , b.sc_name b. sc_score from [課程表] a , [成績表] b where a.co_id =sc_courseid and b.sc_score < 60

select sc_name , sum(sc_score) / count(sc_score) range from [成績表] group by sc_name

order by range

⑶ 如何查詢成績以及查看課表

查看雲南省昭通市鎮雄縣塘房鎮塘中吉茂騫初二考試成績

⑷ 根據學生表，課程表和成績表，查詢學生的學號，姓名，課程號，課程名和分數。（多表查詢

直接將三表關聯查詢就可以了 select b.姓名,c.課程,a.成績 from 成績專表 a,學生表屬 b,課程表 c where a.學生學號 = b.學號 and a.課程編號 = c.編號 and a.成績 < 60

⑸ SQL查詢，學生表，課程表，成績表的一個查詢要求如下怎麼寫呢

selects.sid,s.sname,sc.scorefromstudent,course,scwherestudent.sid=sc.sidandcourse.cid=sc.cidandcourse.cname='影視設計';

注意一下，你自己寫的sid和sname中間沒有逗號，到時候一定執行報錯

⑹ 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

⑺ 用SQL對學生表、課程表、成績表進行多表查詢。

CREATE TABLE 選課表

(學號 CHAR(8) REFERENCES 學生表內(學號),
課程編號 CHAR(6) REFERENCES 課程表(課程編號),
成績 smallint)
GO
INSERT INTO 學生表 VALUES('20100101','李丹','女','1993-6-6','管理系容')

GO
SELECT 學號,姓名,所在系

FROM 學生表
GO
SELECT 學生表.學號,姓名
FROM 學生表 JOIN 選課表 ON 學生表.學號=選課表.學號
WHERE 課程編號='A01-02'
GO
SELECT *
FROM 學生表
WHERE 姓名 LIKE '[趙李張]%'
GO
SELECT 課程名稱,COUNT(*) AS 選修總人數

INTO 修課統計
FROM 課程表 JOIN 選課表 ON 課程表.課程編號=選課表.課程編號
GROUP BY 課程名稱

⑻ 在學生表，課程表，成績表和老師表四個表中，查詢課程為'資料庫'的，且成績在70到80之間的學生信息

要有表結構才可以寫sql

⑼ 超級課程表查的成績是綜合成績還是原始成績

綜合成績，你的學校教務系統顯示分數是多少就是多少

⑽ 寫一個sql;比如有課程表，學員表，分數表；查詢出每個課程的前三名同學；考慮優化問題；大神答出追加100分

fra_id是分抄數吧
select * from 分數表襲 a ( join 課程表 b on a.obj_id=b.obj_id）join 學員表 c on a.s_id=c.s_id
where a.s_id in (select top 3 fra_id from 分數表 group by obj_id,s_id) group by b.obj_id,c.s_id

不知道對不