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;
以上是MySQL如何建立三張關係表的詳細內容。更多資訊請關注PHP中文網其他相關文章!