Student(SId,Sname,Sage,Ssex)
学生表
Course(CId,Cname,TId)
课程表
SC(SId,CId,score) 成绩表
Teacher(TId,Tname) 教师表
问题:
1、查询“
select a.SId from (select sId,score
from SC where CId='001') a,(select sId,score
from SC where CId='002') b
where a.score>b.score and a.sId=b.sId;
2、查询平均成绩大于60分的同学的学号和平均成绩;
select SId,avg(score)
from sc
group by SId
having avg(score) >60;
3、查询所有同学的学号、姓名、选课数、总成绩;
select Student.SId,Student.Sname,count(SC.CId),sum(score)
from Student left Outer join SC on Student.SId=SC.SId
group by
Student.SId,Sname
4、查询姓“李”的老师的个数;
select count(distinct(Tname))
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;
select Student.SId,Student.Sname
from Student
where SId not in (select distinct( SC.SId) from SC,Course,Teacher where SC.CId=Course.CId and Teacher.TId=Course.TId and
Teacher.Tname='叶平');
6、查询学过“
select Student.SId,Student.Sname from Student,SC where Student.SId=SC.SId and SC.CId='001'and exists( Select * from SC as SC_2 where SC_2.SId=SC.SId and SC_2.CId='002');
7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
select SId,Sname
from Student
where SId in (select SId from SC ,Course ,Teacher where SC.CId=Course.CId and Teacher.TId=Course.TId and Teacher.Tname='叶平' group by SId having count(SC.CId)=(select count(CId) from Course,Teacher where Teacher.TId=Course.TId and
Tname='叶平'));
8、查询课程编号“
Select SId,Sname from (select
Student.SId,Student.Sname,score ,(select score from SC SC_2 where SC_2.SId=Student.SId and
SC_2.CId='002') score2
from Student,SC where Student.SId=SC.SId and CId='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;
select SId,Sname
from Student
where SId not in (select Student.SId from Student,SC where S.SId=SC.SId and score>60);
10、查询没有学全所有课的同学的学号、姓名;
select Student.SId,Student.Sname
from Student,SC
where Student.SId=SC.SId group by Student.SId,Student.Sname having count(CId) <(select count(CId) from Course);
11、查询至少有一门课与学号为“
select SId,Sname from Student,SC where Student.SId=SC.SId and CId
in select CId from SC where SId='1001';
12、查询至少学过学号为“
select distinct SC.SId,Sname
from Student,SC
where Student.SId=SC.SId and CId in (select CId from SC where SId='001');
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;
update SC set score=(select avg(SC_2.score)
from SC SC_2
where SC_2.CId=SC.CId ) from
Course,Teacher where
Course.CId=SC.CId and Course.TId=Teacher.TId and Teacher.Tname='叶平');
14、查询和“
select SId from SC where CId in (select CId from SC where SId='1002')
group by SId
having count(*)=(select count(*) from SC where SId='1002');
15、删除学习“叶平”老师课的SC表记录;
Delect SC
from course ,Teacher
where Course.CId=SC.CId and
Course.TId= Teacher.TId 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“
号课的平均成绩;
Insert SC select SId,'002',(Select avg(score)
from SC where CId='002') from Student where SId not in (Select SId from SC where CId='002');
17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示: 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
SELECT SId as 学生ID
,(SELECT score FROM SC WHERE SC.SId=t.SId AND CId='004') AS 数据库
,(SELECT score FROM SC WHERE SC.SId=t.SId AND CId='001') AS 企业管理
,(SELECT score FROM SC WHERE SC.SId=t.SId AND CId='006') AS 英语
,COUNT(*) AS 有效课程数, AVG(t.score)
AS 平均成绩
FROM SC AS t
GROUP BY SId
ORDER BY avg(t.score)
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
SELECT L.CId As 课程ID,L.score AS 最高分,R.score AS 最低分
FROM SC L ,SC AS R
WHERE L.CId = R.CId and
L.score = (SELECT MAX(IL.score)
FROM SC AS IL,Student AS IM
WHERE L.CId = IL.CId and IM.SId=IL.SId
GROUP BY IL.CId)
AND
R.Score = (SELECT MIN(IR.score)
FROM SC AS IR
WHERE R.CId = IR.CId
GROUP BY IR.CId
);
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
SELECT t.CId AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩
,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
FROM SC T,Course
where t.CId=course.CId
GROUP BY
t.CId
ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=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(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004)
SELECT SUM(CASE WHEN CId ='001' THEN score ELSE 0 END)/SUM(CASE CId WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分
,100 * SUM(CASE WHEN CId = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN CId = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数
,SUM(CASE WHEN CId
= '002' THEN score ELSE 0 END)/SUM(CASE CId WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分
,100 * SUM(CASE WHEN CId = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN CId = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数
,SUM(CASE WHEN CId
= '003' THEN score ELSE 0 END)/SUM(CASE CId WHEN '003' THEN 1 ELSE 0 END) AS UML平均分
,100 * SUM(CASE WHEN CId = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN CId = '003' THEN 1 ELSE 0 END) AS UML及格百分数
,SUM(CASE WHEN CId
= '004' THEN score ELSE 0 END)/SUM(CASE CId WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分
,100 * SUM(CASE WHEN CId = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN CId = '004' THEN 1 ELSE 0 END) 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
SELECT max(Z.TId)
AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.CId AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C
,Teacher AS Z
where T.CId=C.CId and C.TId=Z.TId
GROUP BY
C.CId
ORDER BY AVG(Score) 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.SId 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(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC LEFT JOIN SC AS T1
ON SC.SId = T1.SId AND
T1.CId = '001'
LEFT JOIN SC AS T2
ON SC.SId = T2.SId AND
T2.CId = '002'
LEFT JOIN SC AS T3
ON SC.SId = T3.SId AND
T3.CId = '003'
LEFT JOIN SC AS T4
ON SC.SId = T4.SId AND
T4.CId = '004'
WHERE student.SId=SC.SId and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.SId =
T1.SId AND T1.CId = 'k1'
LEFT JOIN sc AS T2
ON sc.SId = T2.SId AND
T2.CId = 'k2'
LEFT JOIN sc AS T3
ON sc.SId = T3.SId AND
T3.CId = 'k3'
LEFT JOIN sc AS T4
ON sc.SId = T4.SId AND
T4.CId = 'k4'
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);
23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
SELECT SC.CId as 课程ID, Cname as 课程名称
,SUM(CASE WHEN
score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]
,SUM(CASE WHEN
score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]
,SUM(CASE WHEN
score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]
,SUM(CASE WHEN
score < 60 THEN 1 ELSE 0 END) AS [60 -]
FROM SC,Course
where SC.CId=Course.CId
GROUP BY
SC.CId,Cname;
24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT SId,AVG(score) AS 平均成绩
FROM SC
GROUP BY SId
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
SId as 学生学号,平均成绩
FROM (SELECT SId,AVG(score) 平均成绩
FROM SC
GROUP BY SId
) AS T2
ORDER BY 平均成绩 desc;
25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT t1.SId as 学生ID,t1.CId as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.CId= CId
ORDER BY score DESC
)
ORDER BY
t1.CId;
26、查询每门课程被选修的学生数
select cId,count(SId) from sc group by CId;
27、查询出只选修了一门课程的全部学生的学号和姓名
select SC.SId,Student.Sname,count(CId) AS 选课数
from SC ,Student
where SC.SId=Student.SId group by SC.SId
,Student.Sname having count(CId)=1;
28、查询男生、女生人数
Select count(Ssex)
as 男生人数 from Student group by Ssex having Ssex='男';
Select count(Ssex)
as 女生人数 from Student group by Ssex having Ssex='女';
29、查询姓“张”的学生名单
SELECT Sname FROM Student WHERE Sname like '张%';
30、查询同名同性学生名单,并统计同名人数
select Sname,count(*) from Student group by Sname having count(*)>1;;
31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
from student
where CONVERT(char(11),DATEPART(year,Sage))='1981';
32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
Select CId,Avg(score) from SC group by CId
order by Avg(score),CId DESC ;
33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
select Sname,SC.SId ,avg(score)
from Student,SC
where Student.SId=SC.SId group by SC.SId,Sname having avg(score)>85;
34、查询课程名称为“数据库”,且分数低于60的学生姓名和分数
Select Sname,isnull(score,0)
from Student,SC,Course
where SC.SId=Student.SId and
SC.CId=Course.CId and Course.Cname='数据库'and score <60;
35、查询所有学生的选课情况;
SELECT SC.SId,SC.CId,Sname,Cname
FROM SC,Student,Course
where SC.SId=Student.SId and
SC.CId=Course.CId ;
36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
SELECT distinct student.SId,student.Sname,SC.CId,SC.score
FROM student,Sc
WHERE SC.score>=70 AND
SC.SId=student.SId;
37、查询不及格的课程,并按课程号从大到小排列
select cId from sc where scor e <60 order by CId ;
38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
select SC.SId,Student.Sname
from SC,Student where SC.SId=Student.SId and Score>80 and CId='003';
39、求选了课程的学生人数
select count(*) from sc;
40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
select Student.Sname,score
from Student,SC,Course C,Teacher
where Student.SId=SC.SId and SC.CId=C.CId and C.TId=Teacher.TId and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where CId=C.CId );
41、查询各个课程及相应的选修人数
select count(*) from sc group by CId;
42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
select distinct A.SId,B.score from SC A ,SC B where A.Score=B.Score and A.CId <>B.CId ;
43、查询每门功成绩最好的前两名
SELECT t1.SId as 学生ID,t1.CId as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.CId= CId
ORDER BY score DESC
)
ORDER BY
t1.CId;
44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数降序排列,若人数相同,按课程号升序排列
select CId as 课程号,count(*) as 人数
from sc
group by CId
order by count(*) desc,cId
45、检索至少选修两门课程的学生学号
select SId
from sc
group by sId
having count(*) > = 2
46、查询全部学生都选修的课程的课程号和课程名
select CId,Cname
from Course
where CId in (select cId from sc group by cId)
47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
select Sname from Student where SId not in (select SId from Course,Teacher,SC where Course.TId=Teacher.TId and SC.CId=course.CId and Tname='叶平');
48、查询两门以上不及格课程的同学的学号及其平均成绩
select SId,avg(isnull(score,0)) from SC where SId in (select SId from SC where score <60 group by SId having count(*)>2)group by SId;
49、检索“
select SId from SC where CId='004'and score <60 order by score desc;
50、删除“
delete from Sc where SId='001'and CId='001';
问题描述:
本题用到下面三个关系表:
CARD 借书卡。 CNO 卡号,NAME 姓名,CLASS 班级
BOOKS 图书。 BNO 书号,BNAME 书名,AUTHOR 作者,PRICE 单价,QUANTITY 库存册数
BORROW 借书记录。 CNO 借书卡号,BNO 书号,RDATE 还书日期
备注:限定每人每种书只能借一本;库存册数随借书、还书而改变。
要求实现如下15个处理:
1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束。
2. 找出借书超过5本的读者,输出借书卡号及所借图书册数。
3. 查询借阅了"水浒"一书的读者,输出姓名及班级。
4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期。
5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者。
6. 查询现有图书中价格最高的图书,输出书名及作者。
7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出。
8. 将"C01"班同学所借图书的还期都延长一周。
9. 从BOOKS表中删除当前无人借阅的图书记录。
10.如果经常按书名查询图书信息,请建立合适的索引。
11.在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表)。
12.建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名)。
13.查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出。
14.假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句。
15.对CARD表做如下修改:
a. 将NAME最大列宽增加到10个字符(假定原为6个字符)。
b. 为该表增加1列NAME(系名),可变长,最大20个字符。
1. 写出建立BORROW表的SQL语句,要求定义主码完整性约束和引用完整性约束
--实现代码:
CREATE TABLE BORROW(
CNO int FOREIGN KEY REFERENCES CARD(CNO),
BNO int FOREIGN KEY REFERENCES BOOKS(BNO),
RDATE datetime,
PRIMARY KEY(CNO,BNO))
2. 找出借书超过5本的读者,输出借书卡号及所借图书册数
--实现代码:
SELECT CNO,借图书册数=COUNT(*)
FROM BORROW
GROUP BY CNO
HAVING COUNT(*)>5
3. 查询借阅了"水浒"一书的读者,输出姓名及班级
--实现代码:
SELECT * FROM CARD c
WHERE EXISTS(
SELECT * FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO
AND b.BNAME=N'水浒'
AND a.CNO=c.CNO)
4. 查询过期未还图书,输出借阅者(卡号)、书号及还书日期
--实现代码:
SELECT * FROM BORROW
WHERE RDATE<GETDATE()
5. 查询书名包括"网络"关键词的图书,输出书号、书名、作者
--实现代码:
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE BNAME LIKE N'%网络%'
6. 查询现有图书中价格最高的图书,输出书名及作者
--实现代码:
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE PRICE=(
SELECT MAX(PRICE)
FROM BOOKS)
7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者,输出其借书卡号,并按卡号降序排序输出
--实现代码:
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO AND b.BNAME=N'计算方法'
AND NOT EXISTS(
SELECT * FROM BORROW aa,BOOKS bb
WHERE aa.BNO=bb.BNO
AND bb.BNAME=N'计算方法习题集'
AND aa.CNO=a.CNO)
ORDER BY a.CNO DESC
8. 将"C01"班同学所借图书的还期都延长一周
--实现代码:
UPDATE b SET RDATE=DATEADD(Day,7,b.RDATE)
FROM CARD a,BORROW b
WHERE a.CNO=b.CNO
AND a.CLASS=N'C01'
9. 从BOOKS表中删除当前无人借阅的图书记录
--实现代码:
DELETE A FROM BOOKS a
WHERE NOT EXISTS(
SELECT * FROM BORROW
WHERE BNO=a.BNO)
10. 如果经常按书名查询图书信息,请建立合适的索引
--实现代码:
CREATE CLUSTERED INDEX IDX_BOOKS_BNAME ON BOOKS(BNAME)
11. 在BORROW表上建立一个触发器,完成如下功能:如果读者借阅的书名是"数据库技术及应用",就将该读者的借阅记录保存在BORROW_SAVE表中(注ORROW_SAVE表结构同BORROW表)
--实现代码:
CREATE TRIGGER TR_SAVE ON BORROW
FOR INSERT,UPDATE
AS
IF @@ROWCOUNT>0
INSERT BORROW_SAVE SELECT i.*
FROM INSERTED i,BOOKS b
WHERE i.BNO=b.BNO
AND b.BNAME=N'数据库技术及应用'
12. 建立一个视图,显示"力01"班学生的借书信息(只要求显示姓名和书名)
--实现代码:
CREATE VIEW V_VIEW
AS
SELECT a.NAME,b.BNAME
FROM BORROW ab,CARD
a,BOOKS b
WHERE ab.CNO=a.CNO
AND ab.BNO=b.BNO
AND a.CLASS=N'力01'
13. 查询当前同时借有"计算方法"和"组合数学"两本书的读者,输出其借书卡号,并按卡号升序排序输出
--实现代码:
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO
AND b.BNAME IN(N'计算方法',N'组合数学')
GROUP BY a.CNO
HAVING COUNT(*)=2
ORDER BY a.CNO DESC
14. 假定在建BOOKS表时没有定义主码,写出为BOOKS表追加定义主码的语句
--实现代码:
ALTER TABLE BOOKS ADD PRIMARY KEY(BNO)
15.1 将NAME最大列宽增加到10个字符(假定原为6个字符)
--实现代码:
ALTER TABLE CARD ALTER COLUMN NAME varchar(10)
15.2 为该表增加1列NAME(系名),可变长,最大20个字符
--实现代码:
ALTER TABLE CARD ADD 系名 varchar(20)
问题描述:
为管理岗位业务培训信息,建立3个表:
S (SId,SN,SD,SA) SId,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄
C (CId,CN ) CId,CN 分别代表课程编号、课程名称
SC ( SId,CId,G ) SId,CId,G 分别代表学号、所选修的课程编号、学习成绩
要求实现如下5个处理:
1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
2. 使用标准SQL嵌套语句查询选修课程编号为’C
3. 使用标准SQL嵌套语句查询不选修课程编号为’C
4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
5. 查询选修了课程的学员人数
6. 查询选修课程超过5门的学员学号和所属单位
1. 使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
--实现代码:
SELECT SN,SD FROM S
WHERE [SId] IN(
SELECT [SId] FROM C,SC
WHERE C.[CId]=SC.[CId]
AND CN=N'税收基础')
2. 使用标准SQL嵌套语句查询选修课程编号为’C
--实现代码:
SELECT S.SN,S.SD FROM S,SC
WHERE S.[SId]=SC.[SId]
AND SC.[CId]='C2'
3. 使用标准SQL嵌套语句查询不选修课程编号为’C
--实现代码:
SELECT SN,SD FROM S
WHERE [SId] NOT IN(
SELECT [SId] FROM SC
WHERE [CId]='C5')
4. 使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [SId] IN(
SELECT [SId] FROM SC
RIGHT JOIN C ON SC.[CId]=C.[CId]
GROUP BY [SId]
HAVING COUNT(*)=COUNT(DISTINCT [SId]))
5. 查询选修了课程的学员人数
--实现代码:
SELECT 学员人数=COUNT(DISTINCT [SId]) FROM SC
6. 查询选修课程超过5门的学员学号和所属单位
--实现代码:
SELECT SN,SD FROM S
WHERE [SId] IN(
SELECT [SId] FROM SC
GROUP BY [SId]
HAVING COUNT(DISTINCT [CId])>5)