今天小编给大家分享一下MySQL如何创建三张关系表的相关知识点，内容详细，逻辑清晰，相信大部分人都还太了解这方面的知识，所以分享这篇文章给大家参考一下，希望大家阅读完这篇文章后有所收获，下面我们一起来了解一下吧。
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create table tbl_stu (
id int not null primary key auto_increment,
name varchar(45) not null
)engine=innodb default charset=utf8;
create table tbl_sub (
id int not null primary key auto_increment,
subject varchar(45) not null
)engine=innodb default charset=utf8;
create table tbl_scores(
id int not null primary key auto_increment,
stu_id int,
sub_id int
score decimal(5,2),
constraint sco_stu foreign key(stu_id) references tbl_stu(id),
constraint sco_sub foreign key(sub_id) references tbl_sub(id)
);
insert into tbl_stu values (0,"小王");
insert into tbl_stu values (0,"小宋");
insert into tbl_stu values (0,"小李");
insert into tbl_sub values (0,"语文");
insert into tbl_sub values (0,"数学");
insert into tbl_sub values (0,"英语");
insert into tbl_scores values (0,1,1,90);
insert into tbl_scores values (0,1,2,70);
insert into tbl_scores values (0,1,3,82);
insert into tbl_scores values (0,2,1,95);
insert into tbl_scores values (0,2,2,70);
insert into tbl_scores values (0,2,3,84);
insert into tbl_scores values (0,3,1,85);
insert into tbl_scores values (0,3,2,86);select s3.name,s2.subject,s1.score from tbl_scores as s1
inner join tbl_sub as s2 on s1.sub_id = s2.id
inner join tbl_stu as s3 on s1.sub_id = s3.id;
select s3.name,avg(s1.score) from tbl_scores as s1
inner join tbl_stu as s3 on s1.sub_id = s3.id
group by s3.name;
select s3.name,sum(s1.score) as s from tbl_scores as s1
inner join tbl_stu as s3 on s1.stu_id = s3.id
group by s3.name order by s desc;
以上就是“MySQL如何创建三张关系表”这篇文章的所有内容，感谢各位的阅读！相信大家阅读完这篇文章都有很大的收获，小编每天都会为大家更新不同的知识，如果还想学习更多的知识，请关注亿速云行业资讯频道。
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