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 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';

常用SQL50句的更多相关文章

  1. 经典SQL50句

    50个常用的sql语句 Student(S#,Sname,Sage,Ssex) 学生表 Course(C#,Cname,T#) 课程表 SC(S#,C#,score) 成绩表 Teacher(T#,T ...

  2. JavaScript常用語句

    1.document.write(""); 输出语句2.JS中的注释为//3.传统的HTML文档顺序是:    document->html->(head,body)4 ...

  3. it工程师常用英文自我介绍常用用语

      Good morning ! It is really my honor to have this opportunity for an interview, I hope i can make ...

  4. HTML 4.01+5基礎知識

    HTML 4.01+5 1.Html結構:html>head+body 2.Html快捷鍵:!加Tab(在sublime中) 3.雙標籤: ①常用標籤 h1.h2.h3.h4.h5.h6 p.c ...

  5. 【托业】【新托业TOEIC新题型真题】学习笔记3-题库二->P5-6

    --------------------------------------单词-------------------------------------- oppose vt. 反对:对抗,抗争 v ...

  6. 技术分享,学术报告presentation 常用的承接句

    前言 现在即使是搞技术,做科研的,也需要在不同的场合,用ppt来做分享,做汇报,做总结. 如果国际会议,研讨会,或者在外企,国外工作,英文的presentation就更加必不可少.英语的提升需要大家从 ...

  7. Presentation 常用的承接句——技术分享、学术报告串联全场不尴尬

    前言 现在即使是搞技术,做科研的,也需要在不同的场合,用ppt来做分享,做汇报,做总结. 如果国际会议,研讨会,或者在外企,国外工作,英文的presentation就更加必不可少.英语的提升需要大家从 ...

  8. 【原】实时渲染中常用的几种Rendering Path

    [原]实时渲染中常用的几种Rendering Path 本文转载请注明出处 —— polobymulberry-博客园 本文为我的图形学大作业的论文部分,介绍了一些Rendering Path,比较简 ...

  9. Unity3D中常用的数据结构总结与分析

    来到周末,小匹夫终于有精力和时间来更新下博客了.前段时间小匹夫读过一份代码,对其中各种数据结构灵活的使用赞不绝口,同时也大大激发了小匹夫对各种数据结构进行梳理和总结的欲望.正好最近也拜读了若干大神的文 ...

随机推荐

  1. SegNet网络的Pytorch实现

    1.文章原文地址 SegNet: A Deep Convolutional Encoder-Decoder Architecture for Image Segmentation 2.文章摘要 语义分 ...

  2. python学习之面向对象

    目录 __main__,__name__ __module__,__class__ __init__ __del__ __repr__,__str__ __mro__ __call__ __new__ ...

  3. redis 与 序列化

    概念 序列化:把对象转化为可传输的字节序列过程称为序列化. 反序列化:把字节序列还原为对象的过程称为反序列化. 为什么需要序列化 序列化最终的目的是为了对象可以跨平台存储,和进行网络传输.而我们进行跨 ...

  4. machine learning(13) -- solving the problem of overfitting:regularization

    solving the problem of overfitting:regularization 发生的在linear regression上面的overfitting问题 发生在logistic ...

  5. Go 语言解释器 Yaegi

    Yaegi 是一个优雅的 Go 语言解释器,可以执行 Go 脚本和插件. 特性 完整支持 Go 语言规范 用 Go 编写,只使用标准库 简单的解释器 API: New(), Eval(), Use() ...

  6. LeetCode 317. Shortest Distance from All Buildings

    原题链接在这里:https://leetcode.com/problems/shortest-distance-from-all-buildings/ 题目: You want to build a ...

  7. 题解 [51nod1225]余数之和

    题解 [51nod1225]余数之和 题面 解析 首先可以发现,\(a\)%\(b\)\(=a-b*\lfloor a/b \rfloor\). 而对于一段连续的\(b\)来说\(\lfloor a/ ...

  8. 改变某个对象的CSS样式时,不要使用JS直接添加样式,

    重绘: 使用js改变网页的背景颜色 浏览器会把整个网页的颜色重新在画一遍,导致性能降低 回流: 只要改变某个DOM对象的宽或者高,浏览器就会重新再计算网页结构,重新生成一次,导致性能严重降低. CSS ...

  9. php面向对象之数据隐藏

    什么是数据隐藏? 看到这个有的人会觉得挺不理解的.在前面的文章中,介绍类的时候,我们说定义变量用的关键词是public,但是不止这一个,还有public.private.protected.stati ...

  10. AcWing P379 捉迷藏 题解

    Analysis 这道题因为我们要给能到达的两个点都连上,又由于n<=200,所以我们可以用n³的传递闭包来建边,再用匈牙利算法来求二分图最大点独立集. #include<iostream ...