标签：问题 null convert avg isnull str 最好 shape art
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
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创建表注意：1.课程从001开始
2.
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.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“001”并且也学过编号“002”课程的同学的学号、姓名；
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“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；
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“1001”的同学所学相同的同学的学号和姓名；
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“001”同学所有一门课的其他同学学号和姓名；
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“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 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“003”课程的同学学号、2、
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,CourseC,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;
标签：问题 null convert avg isnull str 最好 shape art
原文地址：http://www.cnblogs.com/story1/p/7941384.html
