三个班的课程表

3. 用EXCEL作的课程表，有了总表（含班级、星期、节次），怎么使之自动生成一个给各个班级和老师的小课程表

通过Word的邮件合并功能实现。作法如下：
1、打开Word，创建主文档。设计课程表的页面为主文档，将Excel课程总表上的字段名输入。
2、打开“工具”栏，选“邮件合并”，依次点击→主文档→创建→套用信函→活动窗口。
3、选取数据源，打开Excel的课程总表。
4、在邮件合并帮助器的“获取数据”按钮下点击“打开数据源”，选取文件。
5、插入合并域。逐一指定插入的对应位置，在邮件合并帮助器的左上角点击插入合并域。再点击工具栏的“邮件合并”按钮即大功告成，可以逐记录浏览或按指定记录浏览。
这种方法可实现每条记录(每个班级)的单页输出，效果很佳，既可全部打印，也可指定班级分别打印。
如果要按老师打印，重复上述步骤，按照老师代课班级选取字段及合并域（Excel的单元格内容）。

4. SQL查询问题：我有三张表 A表是一个班的学员表，B班是这个班的课程表，C是这个班的课程评价表。

先用B表算出应评价课程总数，然后减去C表中的学生已评价数（group by xh）就是结果了吧。我现在没有数据库没发帮你测试最终语句。

5. 大学不是每一个人的课表都不一样吗为什么我们还是一个班一张课程表

大学课程基本上是一样的，只有一些选修课根据自己爱好来选择，所以每个人的课表不一样。在大学刚开始的时候还没有选修，那么大家都一样

6. 3个老师教四个班级，怎么排课程表

A老师－－1班
B老师－－2班
C老师－－3班
－－4班－－预习
然后轮流，每个上午4节课，每个班都有一节是预习或复习课，或作业课。

7. 手头上有好几个班级的课程表，如何整理出他们之间共同的空闲时间

我觉得你可以这样列：
某人星期一第三节课空，则记为1.3；同理，星期二第五节空，则记为2.5
这样一来就可以比一下有相同数字的就行了

8. 五个老师四个班级,语数英三门功课怎么排课程表

因为要求语文来、英语不自相邻，则用插空法；先排数学、政治和化学3个科目，有A33=6种排法，排好后有4个空位，再将语文、英语插入到4个空位中，有C42=6种情况，由分步计数原理，可得共有6×6=36种情况；故答案为36．

9. 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