数据表介绍

--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号

--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名

--4.成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数

学生表 Student

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));insert into Student values('01' , '赵雷' , '1990-01-01' , '男');insert into Student values('02' , '钱电' , '1990-12-21' , '男');insert into Student values('03' , '孙风' , '1990-12-20' , '男');insert into Student values('04' , '李云' , '1990-12-06' , '男');insert into Student values('05' , '周梅' , '1991-12-01' , '女');insert into Student values('06' , '吴兰' , '1992-01-01' , '女');insert into Student values('07' , '郑竹' , '1989-01-01' , '女');insert into Student values('09' , '张三' , '2017-12-20' , '女');insert into Student values('10' , '李四' , '2017-12-25' , '女');insert into Student values('11' , '李四' , '2012-06-06' , '女');insert into Student values('12' , '赵六' , '2013-06-13' , '女');insert into Student values('13' , '孙七' , '2014-06-01' , '女');

科目表 Course

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));insert into Course values('01' , '语文' , '02');insert into Course values('02' , '数学' , '01');insert into Course values('03' , '英语' , '03');

教师表 Teacher

create table Teacher(TId varchar(10),Tname varchar(10));insert into Teacher values('01' , '张三');insert into Teacher values('02' , '李四');insert into Teacher values('03' , '王五');

成绩表 SC

create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));insert into SC values('01' , '01' , 80);insert into SC values('01' , '02' , 90);insert into SC values('01' , '03' , 99);insert into SC values('02' , '01' , 70);insert into SC values('02' , '02' , 60);insert into SC values('02' , '03' , 80);insert into SC values('03' , '01' , 80);insert into SC values('03' , '02' , 80);insert into SC values('03' , '03' , 80);insert into SC values('04' , '01' , 50);insert into SC values('04' , '02' , 30);insert into SC values('04' , '03' , 20);insert into SC values('05' , '01' , 76);insert into SC values('05' , '02' , 87);insert into SC values('06' , '01' , 31);insert into SC values('06' , '03' , 34);insert into SC values('07' , '02' , 89);insert into SC values('07' , '03' , 98);

---

练习题目

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
 

select * from Student RIGHT JOIN (select t1.SId, class1, class2 from(select SId, score as class1 from sc where sc.CId = '01')as t1,(select SId, score as class2 from sc where sc.CId = '02')as t2where t1.SId = t2.SId AND t1.class1 > t2.class2)ron Student.SId = r.SId;select * from  (select t1.SId, class1, class2from(SELECT SId, score as class1 FROM sc WHERE sc.CId = '01') AS t1,(SELECT SId, score as class2 FROM sc WHERE sc.CId = '02') AS t2where t1.SId = t2.SId and t1.class1 > t2.class2) rLEFT JOIN StudentON Student.SId = r.SId;1.1 查询同时存在" 01 "课程和" 02 "课程的情况select * from(select * from sc where sc.CId = '01') as t1,(select * from sc where sc.CId = '02') as t2where t1.SId = t2.SId;

1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )
 

select * from(select * from sc where sc.CId = '01') as t1left join(select * from sc where sc.CId = '02') as t2on t1.SId = t2.SId;select * from(select * from sc where sc.CId = '02') as t2right join(select * from sc where sc.CId = '01') as t1on t1.SId = t2.SId;

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select * from scwhere sc.SId not in (select SId from scwhere sc.CId = '01')AND sc.CId= '02';

1. 查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩
   这里只用根据学生ID把成绩分组，对分组中的score求平均值，最后在选取结果中AVG大于60的即可. 

select student.SId,sname,ss from student,(select SId, AVG(score) as ss from sc  GROUP BY SIdHAVING AVG(score)> 60)rwhere student.sid = r.sid;select Student.SId, Student.Sname, r.ss from Student right join(select SId, AVG(score) AS ss from scGROUP BY SIdHAVING AVG(score)> 60)r on Student.SId = r.SId;select s.SId,ss,Sname from(select SId, AVG(score) as ss from sc  GROUP BY SIdHAVING AVG(score)> 60)r left join(select Student.SId, Student.Sname fromStudent)s on s.SId = r.SId;

1. 查询在 SC 表存在成绩的学生信息

select DISTINCT student.*from student,scwhere student.SId=sc.SId

4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和
联合查询不会显示没选课的学生：

select student.sid, student.sname,r.coursenumber,r.scoresumfrom student,(select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from scgroup by sc.sid)rwhere student.sid = r.sid;

如要显示没选课的学生(显示为NULL)，需要使用join:

select s.sid, s.sname,r.coursenumber,r.scoresumfrom ((select student.sid,student.snamefrom student)sleft join(selectsc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumberfrom scgroup by sc.sid)ron s.sid = r.sid);

4.2 查有成绩的学生信息
这一题涉及到in和exists的用法，在这种小表中，两种方法的效率都差不多，但是请参考SQL查询中in和exists的区别分析
当表2的记录数量非常大的时候，选用exists比in要高效很多.
EXISTS用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值True或False.
结论：IN()适合B表比A表数据小的情况
结论：EXISTS()适合B表比A表数据大的情况

select * from studentwhere exists (select sc.sid from sc where student.sid = sc.sid);select * from studentwhere student.sid in (select sc.sid from sc);

1. 查询「李」姓老师的数量

select count(*)from teacherwhere tname like '李%';

1. 查询学过「张三」老师授课的同学的信息
   多表联合查询

select student.* from student,teacher,course,scwherestudent.sid = sc.sidand course.cid=sc.cidand course.tid = teacher.tidand tname = '张三';

1. 查询没有学全所有课程的同学的信息
   因为有学生什么课都没有选，反向思考，先查询选了所有课的学生，再选择这些人之外的学生.

select * from studentwhere student.sid not in (select sc.sid from scgroup by sc.sidhaving count(sc.cid)= (select count(cid) from course));

1. 查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息
    

select * from studentwhere student.sid in (select sc.sid from scwhere sc.cid in(select sc.cid from scwhere sc.sid = '01'));

10.查询没学过"张三"老师讲授的任一门课程的学生姓名
仍然还是嵌套，三层嵌套， 或者多表联合查询

select * from studentwhere student.sid not in(select sc.sid from sc where sc.cid in(select course.cid from course where course.tid in(select teacher.tid from teacher where tname = "张三")));select * from studentwhere student.sid not in(select sc.sid from sc,course,teacherwheresc.cid = course.cidand course.tid = teacher.tidand teacher.tname= "张三");

11.查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
 

select student.SId, student.Sname,b.avgfrom student RIGHT JOIN(select sid, AVG(score) as avg from scwhere sid in (select sid from scwhere score<60GROUP BY sidHAVING count(score)>1)GROUP BY sid) b on student.sid=b.sid;

1. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息
   双表联合查询，在查询最后可以设置排序方式，语法为ORDER BY ***** DESC\ASC;

select student.*, sc.score from student, scwhere student.sid = sc.sidand sc.score < 60and cid = "01"ORDER BY sc.score DESC;

1. 按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select *  from scleft join (select sid,avg(score) as avscore from scgroup by sid)ron sc.sid = r.sidorder by avscore desc;

1. 查询各科成绩最高分、最低分和平均分：

以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

selectsc.CId ,max(sc.score)as 最高分,min(sc.score)as 最低分,AVG(sc.score)as 平均分,count(*)as 选修人数,sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率from scGROUP BY sc.CIdORDER BY count(*)DESC, sc.CId ASC

1. 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
    

select a.cid, a.sid, a.score, count(b.score)+1 as rankfrom sc as aleft join sc as bon a.score<b.score and a.cid = b.cidgroup by a.cid, a.sid,a.scoreorder by a.cid, rank ASC;

1. 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺
   这里主要学习一下使用变量。在SQL里面变量用@来标识。

set @crank=0;select q.sid, total, @crank := @crank +1 as rank from(select sc.sid, sum(sc.score) as total from scgroup by sc.sidorder by total desc)q;

1. 统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比
    

select course.cname, course.cid,sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) as "[100-85]",sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) as "[85-70]",sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) as "[70-60]",sum(case when sc.score<=60 and sc.score>0 then 1 else 0 end) as "[60-0]"from sc left join courseon sc.cid = course.cidgroup by sc.cid;

1. 查询各科成绩前三名的记录
    

select * from scwhere (select count(*) from sc as awhere sc.cid = a.cid and sc.score<a.score)< 3order by cid asc, sc.score desc;

19.查询每门课程被选修的学生数

select cid, count(sid) from scgroup by cid;

20.查询出只选修两门课程的学生学号和姓名
嵌套查询

select student.sid, student.sname from studentwhere student.sid in(select sc.sid from scgroup by sc.sidhaving count(sc.cid)=2);

联合查询

select student.SId,student.Snamefrom sc,studentwhere student.SId=sc.SId  GROUP BY sc.SIdHAVING count(*)=2；

21.查询男生、女生人数

select ssex, count(*) from studentgroup by ssex;

22.查询名字中含有「风」字的学生信息

select *from studentwhere student.Sname like '%风%'

23.查询同名学生名单，并统计同名人数
找到同名的名字并统计个数

select sname, count(*) from studentgroup by snamehaving count(*)>1;

嵌套查询列出同名的全部学生的信息

select * from studentwhere sname in (select sname from studentgroup by snamehaving count(*)>1);

24.查询 1990 年出生的学生名单

select *from studentwhere YEAR(student.Sage)=1990;

25.查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select sc.cid, course.cname, AVG(SC.SCORE) as average from sc, coursewhere sc.cid = course.cidgroup by sc.cidorder by average desc,cid asc;

26.查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩
 

select student.sid, student.sname, AVG(sc.score) as aver from student, scwhere student.sid = sc.sidgroup by sc.sidhaving aver > 85;

27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

select student.sname, sc.score from student, sc, coursewhere student.sid = sc.sidand course.cid = sc.cidand course.cname = "数学"and sc.score < 60;

28.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

select student.sname, cid, score from studentleft join scon student.sid = sc.sid;

29.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 student.sname, course.cname,sc.score from student,course,scwhere sc.score>70and student.sid = sc.sidand sc.cid = course.cid;

30.查询存在不及格的课程
可以用group by 来取唯一，也可以用distinct

select cid from scwhere score< 60group by cid;select DISTINCT sc.CIdfrom scwhere sc.score <60;

31.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

select student.sid,student.snamefrom student,scwhere cid="01"and score>=80and student.sid = sc.sid;

32.求每门课程的学生人数

select sc.CId,count(*) as 学生人数from scGROUP BY sc.CId;

33.成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩
用having max()理论上也是对的，但是下面那种按分数排序然后取limit 1的更直观可靠

select student.*, sc.score, sc.cid from student, teacher, course,scwhere teacher.tid = course.tidand sc.sid = student.sidand sc.cid = course.cidand teacher.tname = "张三"having max(sc.score);select student.*, sc.score, sc.cid from student, teacher, course,scwhere teacher.tid = course.tidand sc.sid = student.sidand sc.cid = course.cidand teacher.tname = "张三"order by score desclimit 1;

34.成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩
为了验证这一题，先修改原始数据

UPDATE sc SET score=90where sid = "07"and cid ="02";

这样张三老师教的02号课就有两个学生同时获得90的最高分了。
这道题的思路继续上一题，我们已经查询到了符合限定条件的最高分了，这个时候只用比较这张表，找到全部score等于这个最高分的记录就可，看起来有点繁复。

select student.*, sc.score, sc.cid from student, teacher, course,scwhere teacher.tid = course.tidand sc.sid = student.sidand sc.cid = course.cidand teacher.tname = "张三"and sc.score = (select Max(sc.score)from sc,student, teacher, coursewhere teacher.tid = course.tidand sc.sid = student.sidand sc.cid = course.cidand teacher.tname = "张三");

35.查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
同上，在这里用了inner join后会有概念是重复的记录：“01 课与 03课”=“03 课与 01 课”，所以这里取唯一可以直接用group by

select  a.cid, a.sid,  a.score from sc as ainner joinsc as bon a.sid = b.sidand a.cid != b.cidand a.score = b.scoregroup by cid, sid;

36.查询每门功成绩最好的前两名
 

select a.sid,a.cid,a.score from sc as aleft join sc as bon a.cid = b.cid and a.score<b.scoregroup by a.cid, a.sidhaving count(b.cid)<2order by a.cid;

37.统计每门课程的学生选修人数（超过 5 人的课程才统计）

select sc.cid, count(sid) as cc from scgroup by cidhaving cc >5;

38.检索至少选修两门课程的学生学号

select sid, count(cid) as cc from scgroup by sidhaving cc>=2;

39.查询选修了全部课程的学生信息

select student.*from sc ,studentwhere sc.SId=student.SIdGROUP BY sc.SIdHAVING count(*) = (select DISTINCT count(*) from course )

41. 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

select student.SId as 学生编号,student.Sname  as  学生姓名,TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄from student

42.查询本周过生日的学生

select *from studentwhere WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());

43.查询下周过生日的学生

select *from studentwhere WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;

44.查询本月过生日的学生

select *from studentwhere MONTH(student.Sage)=MONTH(CURDATE());

45.查询下月过生日的学生

select *from studentwhere MONTH(student.Sage)=MONTH(CURDATE())+1;