create database  school;
use school;#学生表
create table `Student`
(`Sno` varchar(20) not null COMMENT '人名',`Sname` varchar(20) not null COMMENT '姓名',`Ssex` varchar(20) not null COMMENT '性别',`Sbirthday` datetime COMMENT '出生日期',`Class` varchar(20) COMMENT '班级'
);
#课程表
create table `Course`(`Cno` varchar(20) not null COMMENT '课程号',`Cname` varchar(20) not null COMMENT '课程名称',`Tno` varchar(20) not null COMMENT '教工编号'
);
#成绩表
Create table `Score` (`Sno` varchar(20) not null COMMENT '学号',`Cno` varchar(20) not NULL comment '课程号',`Degree` DECIMAL(4,1) null COMMENT '成绩'
);
#教师表
create table `Teacher` (`Tno` varchar(20) not null COMMENT '教工编号',`Tname` varchar(20) not null COMMENT '教工姓名',`Tsex` varchar(20) not null COMMENT '教工性别',`Tbirthday` datetime null COMMENT '教工出生日期',`Prof` varchar(20) null COMMENT '职称',`Depart` varchar(20) null COMMENT '教工所在部门'
);insert into `Student` value
(108,'曾华','男','1977-09-01',95033),
(105,'匡明','男','1975-10-02',95031),
(107,'王丽','女','1976-01-23',95033),
(101,'李军','男','1976-02-20',95033),
(109,'王芳','女','1975-02-10',95031),
(103,'陆军','男','1974-06-03',95031);insert into `Course` value
('3-105','计算机导论',825),
('3-245','操作系统',804),
('6-166','数字电路',856),
('9-888','高等数学',831);insert into `Score` value
(103,'3-245',86),
(105,'3-245',75),
(109,'3-245',68),
(103,'3-105',92),
(105,'3-105',88),
(109,'3-105',76),
(101,'2-105',64),
(107,'3-105',91),
(108,'3-105',78),
(101,'6-166',85),
(107,'6-166',79),
(108,'6-166',81);insert into `Teacher` value
(804,'李城','男','1958-12-02','副教授','计算机系'),
(856,'张旭','男','1969-03-12','讲师','电子工程系'),
(825,'王萍','女','1972-05-05','助教','计算机系'),
(831,'刘冰','女','1977-08-14','助教','电子工程系');

select Sname, Ssex, Class from student;

select distinct depart from teacher;

select * from student;

select Sno, Cno, degree from score where degree >= 60 and degree < 80;

select Sno, Cno, degree from score where degree = 85 or degree = 86 or degree = 88;#或另一种写法select * from score where degree in (85, 86, 88)

select * from student where Class = '95031' or Ssex = '女';

select * from student order by Class desc;

select * from score order by Cno asc, degree desc;
# asc, ascending order(升序排列）, desc, descending order(降序排列）, 默认按升序排列

select count(Class) as '95031班学生数'  from student where Class = '95031';

select * from score where degree = (select max(degree) from score);
# 直接写 select * from score where degree = max(degree)不行，会报错

select Cno, avg(degree) as '课程平均分' from score group by Cno ; 

select cno, avg(degree) from score where cno like '3%' group by cno having count(sno) > 4;
#参考网站是5名学生，则查询不到，应为4名学生
#like '3%' 以3开头，或like '%3' 以3结尾, 或包含'%12%'\
#HAVING增加的原因是WHERE关键字无法与合计函数一起使用

SELECT SNO FROM SCORE WHERE DEGREE BETWEEN 70 AND 90;

select student.Sname, score.Cno, score.Degree from student, score where student.Sno = score.Sno;#另一种inner join 写法select student.Sname, score.Cno, score.Degree from student inner join score on student.Sno = score.Sno

select score.Sno, course.Cname, score.Degree from  score, course where course.Cno = score.Cno;#或另一种inner join写法select score.Sno, course.Cname, score.Degree from  score INNER JOIN course on course.Cno = score.Cno;

select student.Sname, course.Cname, score.Degree from student, course, score WHERE student.Sno = score.Sno and score.Cno = course.Cno;

select avg(degree) as '95033班平均分' from score where sno in (select sno from student where Class = '95033');

create table `grade`(`low` int   COMMENT '人名',`upp` int not null COMMENT '课长', `weight` varchar(20) not null COMMENT '等级' ); insert into `grade` values(90,100,'A'); insert into `grade` values(80,89,'B'); insert into `grade` values(70,79,'C'); insert into `grade` values(60,69,'D'); insert into `grade` values(0,59,'E');#网络上的grede，错误, rank关键字不能用，sql存在rank函数,经测试更改为weight可以

#select Sno,Cno,Degree,rank from grade join Score on Score.Degree between low and upp;
#select Sno,Cno,Degree,rank from Score, grade where Degree between low and upp;以上为网络上的答案，在Navicat 11.1.13测试错误，实际也是因为rank关键字的原因

select Sno, Cno, weight from Score, grade where (score.Degree > grade.low and score.Degree < grade.upp);或者是select Sno, Cno, weight from Score, grade where Degree between low and upp;或者是select Sno, Cno, weight from Score inner join grade where Degree between low and upp

select sno from score where score.cno = '3-105' and score.degree>(select Degree from Score where Cno = '3-105' and Sno = '109');#上面是我的答案，题目要求是查询到所有同学的记录，个人是把所有的同学找出来了，下面是网络上的答案，把满足条件的同学的特征又补全了。select * from Student,Score where Score.Cno = '3-105' and Student.Sno = Score.Snoand Score.Degree>(select Degree from Score where Cno = '3-105' and Sno = '109');

select * from Score a where Degree<(select MAX(Degree)from  Score b
where a.Cno = b.Cno) and Sno in(select Sno from Score group by  Sno having  count(*) > 1);

select * from Student,Score where Student.Sno = Score.Sno and Score.Degree>
(select Degree from Score where Cno='3-105' and Sno = '109');#此为网络上答案，个人感觉不对，第一要求只查询课程号3-105的课程，实际查询少限制条件，第二信息不全select * from student, course, score where score.sno = student.sno and score.cno = course.cno and score.cno = '3-105' and score.degree > (select degree from score where cno = '3-105' and sno = '109')

select Sno, Sname, Sbirthday from student where YEAR(Sbirthday) = (select year(Sbirthday) from student where sno = '107');#用到了year()函数

select * from student, course, score where student.sno =score.sno and score.cno = course.cno and course.tno = (select tno from teacher where tname = '张旭');#以上为我的答案，将学生的信息补充的较为完整，以下是网络答案，两个where in 跟score.cno = course.cno student.sno = score.sno 等价，也是另一种写法select Sno,Cno,Degree from Score where Cno in(select Cno from Course where Tno in(select Tno from Teacher where Tname = '张旭'));

select DISTINCT tname from teacher where tno in (select tno from course where cno in (select cno from score group by cno having count(sno) > 5));#之前的联等的写法行不通，因为没想到group by 与having count和where之间的连接方式

select * from student where Class='95033' or Class='95031'

select cno from score where degree >= 85;

select * from teacher, student, course, score where score. sno = student.sno and score.cno = course.cno and teacher.tno =course.tno and teacher.depart = '计算机系';#此处有歧义的是成绩表到底包含哪些内容，通常学生姓名、性别、学号包含，所以这里我把信息特征补的较为齐全select sno,Cno ,Degree from Score where Cno in (select Cno from Course where Tno in (select tno from Teacher where Depart='计算机系'))#以上为网络答案

select Tname,Prof from Teacher a where Prof not in(select Prof from Teacher b where a.Depart!=b.Depart);#此处感觉略奇怪，不同职称的应该有遗漏

select Cno,Sno,Degree from Score a where (select Degree from Score b where Cno='3-105' and
b.Sno = a.Sno)>=(select Degree from Score c where Cno='3-245' and c.Sno = a.Sno)order by
Degree desc ;

select cno, sno, degree from score a where (select degree from score b where cno = '3-105' and a.sno = b.sno) > (select degree from score c where cno = '3-245' and a.sno = c.sno);

select distinct sname as name, ssex as sex, sbirthday as birthday from student union select distinct tname as name, tsex as sex, tbirthday as birthday from teacher （ORDER BY birthday） ;#这里order by birthday 是自己加的，按生日升序排列，实际不用

select distinct sname as name, ssex as sex, sbirthday as birthday from student where ssex = '女' union select distinct tname as name, tsex as sex, tbirthday as birthday from teacher where tsex = '女';

SELECT sno, cno, degree from score a  where a.degree < (select avg(Degree) from score b where a.sno = b.sno); 

select tname, depart from teacher where tname in (select distinct tname from teacher, course, score, student where course.tno = teacher.tno and course.cno = score.cno and score.sno =student.sno);

select tname, depart from teacher where tname not in (select distinct tname from teacher, course, score, student where course.tno = teacher.tno and course.cno = score.cno and score.sno =student.sno);

select Class from student where Ssex='男' group by Class having count(*)>1;

select * from Student where Sname not like ('王%');

select Sname,YEAR(GETDATE())-year(Sbirthday) from student;#getdate函数不可用

select MAX(Sbirthday) as 最大,MIN(Sbirthday) as 最小 from student;

select * from Student order by Class desc ,Sbirthday asc;#注意生日数字越大，年龄越小

select Tname,Cname from Teacher,Course where teacher.Tsex='男' and Teacher.Tno = Course.Tno;

select sno, cno, degree from score where degree = (select max(Degree) from score);

select sname from student where ssex = (select ssex from student where sname = '李军') and sname not in ('李军');

select Sname from Student where Ssex=(select Ssex from Student where Sname='李军')and Sname not in ('李军')and Class=(select Class from Student where Sname='李军');

select Sno,Degree from Score where Sno in(select Sno from Student where Ssex='男')and Cno in (select Cno from Course where Cname='计算机导论')；

id    学生姓名    课程名称    课程成绩
1    张三         Linux       85
2    张三         MySQL       92
3    张三         Java        87
4    李四         Linux       96
5    李四         MySQL       100
7    王五         Linux       91
8    王五         MySQL       83
9    王五         Java        98

学生姓名    Linux    MySQL    Java
张三          85      92      87
李四          96      89      100
王五          91      83      98

#创建数据表CREATE TABLE tb_lemon_grade (id INT(10) NOT NULL AUTO_INCREMENT PRIMARY KEY,student_name VARCHAR(20) DEFAULT NULL,course VARCHAR(20) DEFAULT NULL,score FLOAT DEFAULT '0');#更改显示形式SELECT  student_name,
MAX(IF(COURSE = 'Linux',SCORE,0)) 'Linux',
MAX(IF(COURSE = 'MySQL',SCORE,0)) 'MySQL',
MAX(IF(COURSE = 'Java',SCORE,0)) 'Java'
FROM lemon_grade
group by student_name;

year   month amount
1991   1     1.1
1991   2     1.2
1991   3     1.3
1991   4     1.4
1992   1     2.1
1992   2     2.2
1992   3     2.3
1992   4     2.4year m1   m2   m3   m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4 select year,
(select amount from   aaa m where month=1   and m.year=aaa.year) as m1,
(select amount from   aaa m where month=2   and m.year=aaa.year) as m2,
(select amount from   aaa m where month=3   and m.year=aaa.year) as m3,
(select amount from   aaa m where month=4   and m.year=aaa.year) as m4
from aaa   group by year

 姓名  类别          花费李  看电影         30李  吃饭           100李  旅游           500王  吃饭           500王  看电影         100王  买衣服         700姓名     TOP1类别        TOP1花费         TOP2类别       TOP2花费            TOP3类别         TOP3花费
李       旅游            500               吃饭            100               看电影            30
王       买衣服          700               吃饭            500               看电影            100

#decode(条件,值1,翻译值1,值2,翻译值2,...值n,翻译值n,缺省值)

create proc StuProc
@sname varchar(100)
as
begin
select S#,Sname,Sage,Ssex from student where sname=@sname
end
go