課程表的欄位類型
❶ 學生—課程"資料庫中包含學生表,課程表,學生選課表3個表,分別是:
您好，您這樣：
--1
select Sname,Sage from Student where Sage<(select Sage from Student where Sno='某一學生') and Sdept='數學系'
--2
select Sname from Student where Sno in (select Sno from SC)
--3
select Sname from Student where Sno in (select Sno from SC group by Sno having count(*)=(select count(*) from Course ))
❷ 在資料庫中存儲學生一周的課程表，資料庫如何設計
您好，可以這樣設計
資料庫需要設計一天的8節課 8個欄位 分別存 8節課 主鍵設置成星期 然後還有班級欄位，班級欄位可以做關聯欄位;
舉例
----------------------
星期(主鍵) 1節 2節 3節 4節 5節 6節 7節 8節 班級
星期三 物理 數學 體育 語文 英語 計算機 物理 生物 一班
星期四 物理 數學 體育 語文 英語 計算機 物理 生物 三班
根據前台選擇後台對應查詢，比如選擇一班，會根據今天的星期做判斷主鍵值，然後在根據查詢條件班級。就會把一般今天的所有8節課程列出來。
❸ 大學資料庫題 ，建立一個課程表C,有課程號Cno,課程名稱Cn,學分xf,課時ks,課程類型
create table t_C (cno int primary key auto increment,
cn varchar(20),
xf int(11),
ks int(11),
kl varchar(30));
cno int primary key auto是設置cno為主鍵並且自增
❹ 給定資料庫中,包括「學生表」,「課程表」,「選課表」等三個數據表,表結構如下： 學生表：學號（C，6），姓
用LINQ（語言集成查詢）完成你說的這些任務很方便，具體的細節問題還得你自己做。下面舉出兩個例子供你參考，這兩個例子用數組代替資料庫，對資料庫也可用同樣的方法。
例1，用bool值作關鍵字區分兩組的結果，注意group子句中的表達式產生的結果。
public class Student
{
public string Name { get; set; }
public int ID { get; set; }
public List<int> Scores;
}
public static List<Student> GetStudents()
{
//實例化數據源，注意各成員內部成績序列的初始化
List<Student> students = new List<Student>
{
new Student {Name="張勇", ID=1, Scores= new List<int> {97, 72, 81, 60}},
new Student {Name="王磊", ID=2 Scores= new List<int> {75, 84, 91, 39}},
new Student {Name="孫敏", ID=3, Scores= new List<int> {88, 94, 65, 85}},
new Student {Name="劉曉", ID=4, Scores= new List<int> {97, 89, 85, 82}},
new Student {Name="揚帆", ID=5, Scores= new List<int> {35, 72, 91, 70}}
};
return students;
}
static void Main()
{
List<Student> students = GetStudents();
//用true或false分組，查詢變數是IEnumerable<IGrouping<bool, Student>>類型
var booleanGroupQuery = from student in students
group student by student.Scores.Average() >= 80;
foreach (var studentGroup in booleanGroupQuery)
{
Console.WriteLine(studentGroup.Key == true ? "平均分高於80" : "平均分低於80");
foreach (var student in studentGroup)
Console.WriteLine("{0,4} {1} {2}", student.ID, student.Name, student.Scores.Average());
}
}
輸出：
平均分低於80
1 張勇 77.5
2 王磊 72.25
5 揚帆 67
平均分高於80
3 孫敏 83
4 劉曉 88.25
例2，查詢人名先排序再分組，可以用姓作為分組關鍵字。
string[] Name = {"張明","劉新","王宏","劉洋","張媛","張寶","王金貴","劉東","王凱","劉芳"};
var sortedGroups = from name in Name
orderby name
group name by name[0] into newGroup
select newGroup;
foreach (var nameGroup in sortedGroups)
{
Console.WriteLine(nameGroup.Key);
foreach (var name in nameGroup)
Console.WriteLine(" "+name);
}
輸出：
劉
劉東 劉芳 劉新 劉洋
王
王宏 王金貴 王凱
張
張寶 張明 張媛
輸出結果按姓分類，組內排序。
《C#編程指南》（清華大學出版社2011年1月出版，可網購）第5章和第17章專門討論LINQ查詢更新資料庫。
❺ 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
❻ 1．在學生資料庫中有三張數據表，即學生表、課程表和選課表，三張表的表數據如下，寫出創建學生表、課程
create table student(S_no char(9), S_name char(20), S_sex char(2), S_age int, S_department char(20))
create table course (C_no char(10), C_name char(20), C_teacher char(20) )
create table choice(S_no char(9), C_no char(10), S_score int)
insert into student(S_no, S_name, S_sex, S_age, S_department)
values('001', '李志強', '男', 20, '計算機系' )
insert into student(S_no, S_name, S_sex, S_age, S_department)
values('002', '張 亮', '男', 21 '建築工程系' )
insert into student(S_no, S_name, S_sex, S_age, S_department)
values('003', '李 平', '女', 19 '計算機系' )
insert into course(C_no, C_name, C_teacher )
values('C001', 'C語言', '王雅新' )
insert into course(C_no, C_name, C_teacher )
values('C002', '數據結構', '和海蓮' )
insert into course(C_no, C_name, C_teacher )
values('C003', 'SQL Server2000', '陳 紅' )
insert into choice(S_no, C_no, S_score)
values('001', 'C001', 83)
insert into choice(S_no, C_no, S_score)
values('001', 'C002', 80 )
insert into choice(S_no, C_no, S_score)
values('002', 'C001', 75)
insert into choice(S_no, C_no, S_score)
values('002', 'C004', 85)
insert into choice(S_no, C_no, S_score)
values('003', 'C002', 88)
insert into choice(S_no, C_no, S_score)
values('003', 'C003', 86)
❼ 資料庫中包含學生表、課程表、學生選課表3個表,分別是什麼
1、SELECT
課程號,
課程名,
課程學分
FROM
課程
WHERE
課程號
IN(SELECT
課程號
FROM
選課
GROUP
BY
課程號
HAVING
COUNT(學生號)
BETWEEN
2
AND
4
)
2、SELECT
MAX(成績)
-
MIN(成績)
AS
分數之差
FROM
選課
倘若要每門課程相差
SELECT
課程,
MAX(成績)
-
MIN(成績)
AS
分數之差
FROM
選課
GROUP
BY
課程
❽ 資料庫中建立學生表課程表選課表及其聯系
以 access 為例，建立數據表如下：
1、學生表
❾ 假設當前有課程表,課程表的結構包括(課程名,學時,學分),刪除課程表中的學時欄位
在命令窗口中輸入
alter table 課程 drop column 學時
如果提示找不到表,可以先輸入set defa to 表所在路徑
如set defa to d:\練習1
或者用菜單,選擇"文件|打開",文件類型選"表",把課程表打開,然後選擇菜單"顯示|表設計器",選中"學時",然後單擊"刪除"按鈕.
❿ 在資料庫teacher_class下，寫出創建教師表、課程表、授課表的SQL語句，各表結構如下所示：
（1）alter table tc add 『修課類別』 nchar（1）；
（2）alter table tc modify 『修課類別』 nchar（2）;
（3）alter table tc drop column 『修課類別』；