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今天小編給大家分享一下MySQL如何創(chuàng)建三張關系表的相關知識點,內容詳細,邏輯清晰,相信大部分人都還太了解這方面的知識,所以分享這篇文章給大家參考一下,希望大家閱讀完這篇文章后有所收獲,下面我們一起來了解一下吧。
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,"語文"); insert into tbl_sub values (0,"數(shù)學"); insert into tbl_sub values (0,"英語"); 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如何創(chuàng)建三張關系表”這篇文章的所有內容,感謝各位的閱讀!相信大家閱讀完這篇文章都有很大的收獲,小編每天都會為大家更新不同的知識,如果還想學習更多的知識,請關注億速云行業(yè)資訊頻道。
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