MySQL创建三张关系表实操介绍

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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;

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create table tbl_sub (

id int not null primary key auto_increment,

subject varchar(45) not null

)engine=innodb default charset=utf8;

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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)

);

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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);

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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;

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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;

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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;