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一.
Student(S#,Sname,Sage,Ssex) 学生表 Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 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； select a.S# from (select s#,score from SC where C#='001') a,(select s#,score
from SC where C#='002') b
where a.score>b.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； select S#,avg(score) from sc group by S# having avg(score) >60; 3、查询所有同学的学号、姓名、选课数、总成绩； select Student.S#,Student.Sname,count(SC.C#),sum(score)
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； 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.S#,Student.Sname
from Student where S# not in (select distinct( SC.S#) from 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； select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='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 S#,Sname from Student where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) 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； Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2
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； select S#,Sname from Student where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60); 10、查询没有学全所有课的同学的学号、姓名； select Student.S#,Student.Sname
from Student,SC where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) 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； select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# 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； select distinct SC.S#,Sname
from Student,SC where Student.S#=SC.S# and C# in (select C# from SC where S#='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.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； select S# from SC where C# in (select C# from SC where S#='1002')
group by S# having count(*)=(select 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 into SC select S#,'002',(Select avg(score)
from SC where C#='002') from Student where S# not in (Select S# from SC where C#='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 S# as 学生ID ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='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 S# 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.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 = (SELECT MAX(IL.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 = (SELECT MIN(IR.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 SELECT t.C# 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.C#=course.C# GROUP BY t.C# 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 C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分 ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数 ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '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.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# 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.C#=C.C# and C.T#=Z.T#
GROUP BY C.C# 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.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(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.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(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.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(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.C# 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.C#=Course.C# GROUP BY SC.C#,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 S#,AVG(score) AS 平均成绩 FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次, S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩 FROM SC
GROUP BY S#
) 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.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1
WHERE score IN (SELECT TOP 3 score FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
26、查询每门课程被选修的学生数 select c#,count(S#) from sc group by C#;
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.S#,Student.Sname,count(C#) AS 选课数 from SC ,Student where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=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 C#,Avg(score) from SC group by C# order by Avg(score),C# 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.S# ,avg(score)
from Student,SC where Student.S#=SC.S# group by SC.S#,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.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60;
35、查询所有学生的选课情况； SELECT SC.S#,SC.C#,Sname,Cname
FROM SC,Student,Course where SC.S#=Student.S# and SC.C#=Course.C# ;
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.S#,student.Sname,SC.C#,SC.score
FROM student,Sc WHERE SC.score>=70 AND SC.S#=student.S#;
37、查询不及格的课程，并按课程号从大到小排列 select c# from sc where scor e <60 order by C# ;
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.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='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.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# );
41、查询各个课程及相应的选修人数 select count(*) from sc group by C#;
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.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
43、查询每门功成绩最好的前两名 SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数 FROM SC t1
WHERE score IN (SELECT TOP 2 score FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
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 C# as 课程号,count(*) as 人数 from sc group by C#
order by count(*) desc,c#
45、检索至少选修两门课程的学生学号 select S# from sc group by s#
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 C#,Cname
from Course where C# in (select c# from sc group by c#)
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 S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# 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 S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#; 49、检索“004”课程分数小于60，按分数降序排列的同学学号 select S# from SC where C#='004'and score <60 order by score desc;
50、删除“002”同学的“001”课程的成绩 delete from Sc where S#='001'and C#='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 (S#,SN,SD,SA) S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄 C (C#,CN ) C#,CN 分别代表课程编号、课程名称 SC ( S#,C#,G ) S#,C#,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’C2’的学员姓名和所属单位 3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位 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 [S#] IN( SELECT [S#] FROM C,SC WHERE C.[C#]=SC.[C#] 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’C2’的学员姓名和所属单位 --实现代码： SELECT S.SN,S.SD FROM S,SC WHERE S.[S#]=SC.[S#] AND SC.[C#]='C2' 3. 使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位 --实现代码： SELECT SN,SD FROM S WHERE [S#] NOT IN( SELECT [S#] FROM SC
WHERE [C#]='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 [S#] IN( SELECT [S#] FROM SC
RIGHT JOIN C ON SC.[C#]=C.[C#]
GROUP BY [S#] HAVING COUNT(*)=COUNT(DISTINCT [S#]))
5. 查询选修了课程的学员人数 --实现代码： SELECT 学员人数=COUNT(DISTINCT [S#]) 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 [S#] IN( SELECT [S#] FROM SC
GROUP BY [S#] HAVING COUNT(DISTINCT [C#])>5) if not object_id('cj')is null drop table cj go create table cj(stuName nvarchar(10),KCM nvarchar(10),cj numeric(5,2)) insert into cj select '张三','语文',98 union select '李四','语文',89
union select '王五','语文',67
union select '周攻','语文',56
union select '张三','数学',89 union select '李四','数学',78
union select '王五','数学',90
union select '周攻','数学',87
方法一： select stuname from (select stuName,kcm,(select count(*) from cj where stuname!=a.stuname and kcm=a.kcm and cj>a.cj) cnt from cj a) x
group by stuname having max(cnt)<=1 go 方法二： SELECT stuname FROM cj1 a where cj IN(SELECT TOP 2 cj FROM cj1 WHERE kcm=a.kcm ORDER BY cj desc)
GROUP BY stuname HAVING(count(1)>1) 方法三： select distinct stuname from cj a where not exists(select kcm from cj b where a.stuname=stuname
and (select count(*) from cj where kcm=b.kcm and stuname!=a.stuname and cj>b.cj)>1)
S (S#,SN,SD,SA)
S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄 C (C#,CN ) C#,CN 分别代表课程编号、课程名称 SC ( S#,C#,G )
S#,C#,G 分别代表学号、所选修的课程编号、学习成绩 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 [S#] IN( Select [S#] FROM C,SC Where C.[C#]=SC.[C#] 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.[S#]=SC.[S#] AND SC.[C#]='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 [S#] NOT IN( Select [S#] FROM SC Where [C#]='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 [S#] IN( Select [S#] FROM SC RIGHT JOIN C ON SC.[C#]=C.[C#] GROUP BY [S#]
HAVING COUNT(*)=COUNT([S#])) 5. 查询选修了课程的学员人数 --实现代码: Select 学员人数=COUNT(DISTINCT
[S#]) 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 [S#] IN( Select [S#] FROM SC GROUP BY [S#] HAVING COUNT(DISTINCT
[C#])>5) 题目2: 问题描述: 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: Select SNAME FROM S
Where NOT EXISTS( Select * FROM SC,C Where SC.CNO=C.CNO AND CNAME='李明' AND SC.SNO=S.SNO) 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.SNO,S.SNAME,AVG_SCGRADE=AVG(SC.SCGRADE) FROM S,SC,( Select SNO FROM SC Where
SCGRADE<60 GROUP BY SNO HAVING COUNT(DISTINCT CNO)>=2 )A Where
S.SNO=A.SNO AND SC.SNO=A.SNO GROUP BY S.SNO,S.SNAME 3. 列出既学过“ --实现代码: Select
S.SNO,S.SNAME FROM S,( Select SC.SNO FROM SC,C Where SC.CNO=C.CNO AND C.CNAME
IN('1','2') GROUP BY SNO HAVING COUNT(DISTINCT CNO)=2 )SC Where S.SNO=SC.SNO 4. 列出“ --实现代码: Select
S.SNO,S.SNAME FROM S,( Select SC1.SNO FROM SC SC1,C C1,SC SC2,C C2 Where
SC1.CNO=C1.CNO AND C1.NAME='1' AND SC2.CNO=C2.CNO AND C2.NAME='2' AND
SC1.SCGRADE>SC2.SCGRADE )SC Where S.SNO=SC.SNO 5. 列出“ --实现代码: Select
S.SNO,S.SNAME,SC.[1号课成绩],SC.[2号课成绩] FROM S,( Select SC1.SNO,[1号课成绩]=SC1.SCGRADE,[2号课成绩]=SC2.SCGRADE FROM
SC SC1,C C1,SC SC2,C C2 Where SC1.CNO=C1.CNO AND C1.NAME='1' AND SC2.CNO=C2.CNO
AND C2.NAME='2' AND SC1.SCGRADE>SC2.SCGRADE )SC Where S.SNO=SC.SNO 求其中同一个号码的两次通话之间间隔大于10秒的通话记录ID 例如：6，7，8，9，10条记录均符合 ID 主叫号码 被叫号码 通话起始时间
通话结束时间 通话时长 1 98290000 0215466546656 2007-02-01
09:49:53.000 2007-02-01 09:50:16.000 23 2 98290000 021546654666 2007-02-01 09:50:29.000 2007-02-01
09:50:41.000 12 3 98290000 021546654666 2007-02-01 09:50:58.000 2007-02-01
09:51:12.000 14 4 68290900 0755133329866 2007-02-01
10:04:31.000 2007-02-01 10:07:13.000 162 5 78290000 0755255708638 2007-02-01
10:48:26.000 2007-02-01 10:49:23.000 57 6 78290000 0755821119109 2007-02-01
10:49:39.000 2007-02-01 10:52:55.000 196 7 78290000 035730928370 2007-02-01 11:30:45.000 2007-02-01
11:31:58.000 73 8 78290000 0871138889904 2007-02-01
11:33:47.000 2007-02-01 11:35:00.000 73 9 68290000 035730928379 2007-02-01 11:52:20.000 2007-02-01
11:54:56.000 156 10 68290000
0298521811199 2007-02-01 12:44:45.000 2007-02-01 12:45:04.000 19 答案： SELECT DISTINCT a.*
FROM dbo.hc a left join dbo.hc b ON a.主叫号码=b.主叫号码 WHERE
a.id<>b.id AND (DATEDIFF(second,a.通话起始时间,b.通话结束时间)>10 AND DATEDIFF(second,b.通话起始时间,a.通话结束时间)>10) Sql Server关于按周统计的问题 统计Sql Server里一个销售明细表里某个时间段的销售额，而且要按周进行比较，以下是该语句的写法： select sum(销售金额), datename(week, 销售日期-1) from sales
where 销售日期 betwee begindate and enddate group by datename(week, 销售日期-1) 注意：这里之所以要把销售日期-1是因为sql server默认的一周的第一天是星期天，而我们习惯的统计是以星期一到星期天计算的，所以减一。 |
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