How to create three relational tables in MySQL

1.Create student table

create table tbl_stu (

id int not null primary key auto_increment,

name varchar(45) not null

)engine=innodb default charset=utf8;

2.Create subject table

create table tbl_sub (

id int not null primary key auto_increment,

subject varchar(45) not null

)engine=innodb default charset=utf8;

3.Create score table

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)
);

4. Insert data

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);

5. Query all scores

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;

6. Query the average score of students

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;

7.Total score ranking

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;

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