三個班的課程表

3. 用EXCEL作的課程表，有了總表（含班級、星期、節次），怎麼使之自動生成一個給各個班級和老師的小課程表

通過Word的郵件合並功能實現。作法如下：
1、打開Word，創建主文檔。設計課程表的頁面為主文檔，將Excel課程總表上的欄位名輸入。
2、打開「工具」欄，選「郵件合並」，依次點擊→主文檔→創建→套用信函→活動窗口。
3、選取數據源，打開Excel的課程總表。
4、在郵件合並幫助器的「獲取數據」按鈕下點擊「打開數據源」，選取文件。
5、插入合並域。逐一指定插入的對應位置，在郵件合並幫助器的左上角點擊插入合並域。再點擊工具欄的「郵件合並」按鈕即大功告成，可以逐記錄瀏覽或按指定記錄瀏覽。
這種方法可實現每條記錄(每個班級)的單頁輸出，效果很佳，既可全部列印，也可指定班級分別列印。
如果要按老師列印，重復上述步驟，按照老師代課班級選取欄位及合並域（Excel的單元格內容）。

4. SQL查詢問題：我有三張表 A表是一個班的學員表，B班是這個班的課程表，C是這個班的課程評價表。

先用B表算出應評價課程總數，然後減去C表中的學生已評價數（group by xh）就是結果了吧。我現在沒有資料庫沒發幫你測試最終語句。

5. 大學不是每一個人的課表都不一樣嗎為什麼我們還是一個班一張課程表

大學課程基本上是一樣的，只有一些選修課根據自己愛好來選擇，所以每個人的課表不一樣。在大學剛開始的時候還沒有選修，那麼大家都一樣

6. 3個老師教四個班級，怎麼排課程表

A老師－－1班
B老師－－2班
C老師－－3班
－－4班－－預習
然後輪流，每個上午4節課，每個班都有一節是預習或復習課，或作業課。

7. 手頭上有好幾個班級的課程表，如何整理出他們之間共同的空閑時間

我覺得你可以這樣列：
某人星期一第三節課空，則記為1.3；同理，星期二第五節空，則記為2.5
這樣一來就可以比一下有相同數字的就行了

8. 五個老師四個班級,語數英三門功課怎麼排課程表

因為要求語文來、英語不自相鄰，則用插空法；先排數學、政治和化學3個科目，有A33=6種排法，排好後有4個空位，再將語文、英語插入到4個空位中，有C42=6種情況，由分步計數原理，可得共有6×6=36種情況；故答案為36．

9. 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