选修课程表

❸ 高中科课程表

高一上：语文数学英语都上必修1、2，物理化学政治历史地理上必修1，生物会上一半必修1
高一版下：语文数学英语权都上必修3、4，物理化学政治历史地理上必修2，生物会上另一半必修1
高二上;语文数学英语都上必修5,语文上一本选修大多是 中国古代诗歌散文欣赏 数学上选修2-1英语选修6
理科;物理上3-1化学上选修4化学反应原理，生物上必修2
高二下：语文继续上中国古代诗歌散文欣赏 先秦诸子选读 数学上2-2和2-3 英语选修7、8
物理上选修3-2和选修3-4（选考题）3-5（选考题）化学上选修五有机化学，选修三化学反应原理，生物上必修三
高三上，英语可能上选修9，物理把没上完的选修上完，其他的开始总复习

❹ 跪求大神帮帮忙！！！有一个【学生选修课】数据库，数据库中包括三个表，学生表，课程表，成绩表

//自己把中文替换成英文字段名 我用的sql server数据库
select 学号、姓名、性别、年龄、所在系 from 学生 order by 年龄 desc, 学号 asc
select 学号，姓名 from 学生 where 姓名 in( select 姓名 from 学生 group by 姓名 having count(*)>1 )
update 成绩 set 成绩=0 where CNO=1
如果/不是除法的话 只是字符串 （除法暂时有问题）下班了 明天再来写
select C.课程号,D.课程名,D.成绩 from COURSE C,(
select CONVERT(VARCHAR(50),A.排名)+'/'+CONVERT(VARCHAR(50),B.ZS) as '排名/人数',a.学号,a.课程名,a.成绩 FROM (
select ROW_NUMBER() over(partition BY 课程名 order by 成绩 desc) as 排名 ,* from GRADE where 学号='1') A,
(SELECT COUNT(*) as zs,课程名
FROM GRADE group by 课程名 ) B where a.课程名=b.课程名) D WHERE C.课程名=D.课程名

❺ 中医药大学的课表（必修基础课和选修课）以及教材的问题

你好，孙广仁的中国中医药出版社的书好些，是国家十一五规划教材。学习中回医推荐依次学习中医基础答理论，中医诊断学，中药学，方剂学，中医内科学等等，后面的诊断学，中药学，方剂学，内科学和中医基础理论是一个版面的，都是绿色的。在优酷网还有对应的讲课专辑，配起来看效果更好。可以看看《高等教育中医专业自学考试指导丛书》
你先把这几本搞定，你就入门了，后面学什么你自己应该也知道了。

❻ 西安外事学院每年选修课课表在哪查

登录学校的教务在线主页就可以查到每年的选修课课表，必须登录你的用户名。

❼ SQL语句的一道题 三个基本表：学生表(Student)、课程表（Course）、学生选课表（SC）

1. select * from SC
2. select Sname，Sage from Student where Sdept = '计算机'
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 = '计算机'
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 = '计算机' and course.cname = '计算机文化基础'
)

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. 不会

❽ 超级课程表怎么加入选修课

选修课需要手动添加的哦

❾ 在SQL中创建学生表课程表还有成绩表怎么 （1）查询有多少同学选修了课程。（2） 查询有多少同学没有选课

您的问题描述不是很清楚。我的理解是如果成绩表里有学生ID和课程ID字段，那么可以用学生表左连接成绩表，新结果集右表有值的就是答案1，没有的就是答案2。