Question

哪位高手帮忙解决几个关于数据库的问题（sql server）

有下列关系：
S (SNO,SNAME) 学生关系。SNO 为学号，SNAME 为姓名
C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号，CNAME 为课程名，CTEACHER 为任课教师
SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩
1. 找出没有选修过“李明”老师讲授课程的所有学生姓名
--实现代码:

2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩
--实现代码:

3. 列出既学过“1”号课程，又学过“2”号课程的所有学生姓名
　　--实现代码:
　
4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号
　　--实现代码:
　
5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩

Answer 1

Student 学生表

Course 课程表

SC 成绩表

Teacher 教师表

问题：

1、查询“001”课程比“002”课程成绩高的所有学生的学号；

selecta.S# froma， b

where a.scoreb.score and a.s#=b.s#;

2、查询平均成绩大于60分的同学的学号和平均成绩；

selectS#，avg

from sc

group by S# having avg 60;

3、查询所有同学的学号、姓名、选课数、总成绩；

selectStudent.S#，Student.Sname，count，sum

from Student left Outer join SC on Student.S#=SC.S#

group by Student.S#，Sname

4、查询姓“李”的老师的个数；

selectcount)

from Teacher

where Tname like ‘李%‘;

5、查询没学过“叶平”老师课的同学的学号、姓名；

selectStudent.S#，Student.Sname

from Student

where S# not infrom SC，Course，Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname=‘叶平‘);

6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

selectStudent.S#，Student.Sname from Student，SC where Student.S#=SC.S# and SC.C#=‘001‘and exists;

7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；

selectS#，Sname

from Student

where S# in = from Course，Teacher where Teacher.T#=Course.T# and Tname=‘叶平‘));

8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；

selectS#，Sname fromscore2

from Student，SC where Student.S#=SC.S# and C#=‘001‘) S_2 where score2 score;

9、查询所有课程成绩小于60分的同学的学号、姓名；

selectS#，Sname

from Student

where S# not in ;

10、查询没有学全所有课的同学的学号、姓名；

selectStudent.S#，Student.Sname

from Student，SC

where Student.S#=SC.S# group by Student.S#，Student.Sname having count from Course);

11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名；

selectS#，Sname from Student，SC where Student.S#=SC.S# and C# inselectC# from SC where S#=‘1001‘;

12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；

selectdistinct SC.S#，Sname

from Student，SC

where Student.S#=SC.S# and C# in ;

13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩；

updateSC set score=

from SC SC_2

where SC_2.C#=SC.C# ) from Course，Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname=‘叶平‘);

14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名；

selectS# from SC where C# in

group by S# having count= from SC where S#=‘1002‘);

15、删除学习“叶平”老师课的SC表记录；

Delect SC

from course ，Teacher

where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname=‘叶平‘;

16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、

号课的平均成绩；

Insert SCselectS#，‘002‘，

from SC where C#=‘002‘) from Student where S# not in ;

17、按平均成绩从高到低显示所有学生的“”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID，，，企业管理，英语，有效课程数，有效平均分

selectS# as 学生ID

， AS 数据库

， AS 企业管理

， AS 英语

，COUNT AS 有效课程数， AVG AS 平均成绩

FROM SC AS t

GROUP BY S#

ORDER BY avg

18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分

selectL.C# As 课程ID，L.score AS 最高分，R.score AS 最低分

FROM SC L ，SC AS R

WHERE L.C# = R.C# and

L.score =

FROM SC AS IL，Student AS IM

WHERE L.C# = IL.C# and IM.S#=IL.S#

GROUP BY IL.C#)

AND

R.Score =

FROM SC AS IR

WHERE R.C# = IR.C#

GROUP BY IR.C#

);

19、按各科平均成绩从低到高和及格率的百分数从高到低顺序

selectt.C# AS 课程号，maxAS 课程名，isnull，0) AS 平均成绩

，100 * SUM=60 THEN 1 ELSE 0 END)/COUNT AS 及格百分数

FROM SC T，Course

where t.C#=course.C#

GROUP BY t.C#

ORDER BY 100 * SUM=60 THEN 1 ELSE 0 END)/COUNT DESC

20、查询如下课程平均成绩和及格率的百分数: 企业管理（001），马克思（002），OOUML （003），数据库（004）

selectSUM/SUM AS 企业管理平均分

，100 * SUM/SUM AS 企业管理及格百分数

，SUM/SUM AS 马克思平均分

，100 * SUM/SUM AS 马克思及格百分数

，SUM/SUM AS UML平均分

，100 * SUM/SUM AS UML及格百分数

，SUM/SUM AS 数据库平均分

，100 * SUM/SUM AS 数据库及格百分数

FROM SC

21、查询不同老师所教不同课程平均分从高到低显示

selectmax AS 教师ID，MAX AS 教师姓名，C.C# AS 课程ID，MAX AS 课程名称，AVG AS 平均成绩

FROM SC AS T，Course AS C ，Teacher AS Z

where T.C#=C.C# and C.T#=Z.T#

GROUP BY C.C#

ORDER BY AVG DESC

22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004）

[学生ID]，[学生姓名]，企业管理，马克思，UML，数据库，平均成绩

SELECT DISTINCT top 3

SC.S# As 学生学号，

Student.Sname AS 学生姓名 ，

T1.score AS 企业管理，

T2.score AS 马克思，

T3.score AS UML，

T4.score AS 数据库，

ISNULL + ISNULL + ISNULL + ISNULL as 总分

FROM Student，SC LEFT JOIN SC AS T1

ON SC.S# = T1.S# AND T1.C# = ‘001‘

LEFT JOIN SC AS T2

ON SC.S# = T2.S# AND T2.C# = ‘002‘

LEFT JOIN SC AS T3

ON SC.S# = T3.S# AND T3.C# = ‘003‘

LEFT JOIN SC AS T4

ON SC.S# = T4.S# AND T4.C# = ‘004‘

WHERE student.S#=SC.S# and

ISNULL + ISNULL + ISNULL + ISNULL

NOT IN

+ ISNULL + ISNULL + ISNULL

FROM sc

LEFT JOIN sc AS T1

ON sc.S# = T1.S# AND T1.C# = ‘k1‘

LEFT JOIN sc AS T2

ON sc.S# = T2.S# AND T2.C# = ‘k2‘

LEFT JOIN sc AS T3

ON sc.S# = T3.S# AND T3.C# = ‘k3‘

LEFT JOIN sc AS T4

ON sc.S# = T4.S# AND T4.C# = ‘k4‘

ORDER BY ISNULL + ISNULL + ISNULL + ISNULL DESC);

Answer 2

学生吧，给你贴一部分代码，后面的自己动一点点脑筋就出来了

1. 找出没有选修过“李明”老师讲授课程的所有学生姓名
--实现代码:
SELECT S.SNMAE
FROM S
INNER JOIN SC ON S.SNO = SC.SNO
INNER JOIN C ON C.CNO = SC.CNO
WHERE CTEACHER <> '李明'

2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩
--实现代码:
SELECT S.SNAME, B.SCGRADE
FROM S
INNER JOIN (SELECT SC.SNO, COUNT(*) AS CNT, AVG(SCGRADE) AS SCGRADE
FROM SC
INNER JOIN C ON SC.CNO = C.CNO
GROUP BY SC.SNO)B ON S.CNO = B.SNO
WHERE EXISTS(SELECT 1 FROM SC WHERE SCGRADE < 60 AND S.SNO = SC.SNO GROUP BY SC HAVING COUNT(*) >= 2)

3. 列出既学过“1”号课程，又学过“2”号课程的所有学生姓名
　　--实现代码:
SELECT S.SNAME
FROM S
WHERE EXISTS(SELECT 1
FROM SC
INNER JOIN C ON SC.CNO = C.CNO
WHERE S.SNO = SC.SNO AND C.CNAME IN ('1', '2')
GROUP BY SC.SNO
HAVING COUNT(*) = 2)