废话不多说,来看例子
一、建表,导入测试数据
create table temp1 ( aid VARCHAR2(5) not null, car VARCHAR2(10) not null ); create table temp2 ( bid VARCHAR2(5) not null, username VARCHAR2(10) not null ); create table temp3 ( cid VARCHAR2(5) not null, dogname VARCHAR2(10) not null ); insert into temp1(aid,car) values('001','benz'); insert into temp1(aid,car) values('001','BMW'); insert into temp1(aid,car) values('001','ford'); insert into temp1(aid,car) values('001','jeep'); insert into temp1(aid,car) values('002','jeep'); insert into temp1(aid,car) values('003','hongqi'); insert into temp2(bid,username) values('001','mayun'); insert into temp3(cid,dogname) values('001','lily'); insert into temp3(cid,dogname) values('001','lucy'); insert into temp3(cid,dogname) values('002','xiaohua');
查一下数据长什么样:
select * from temp1; select * from temp2; select * from temp3;
| temp1 | temp2 | temp3 |
![]() |
![]() |
![]() |
二、左连接测试
–1.左连接,把左边的全部查出来,右边有的则匹配,没有则为null
select * from temp1 t1 left join temp2 t2 on t1.aid=t2.bid ;

select * from temp2 t2 left join temp1 t1 on t2.bid=t1.aid ;

–2.若是三张表,通过两个left join来连接,则把前面两张表先left join之后当作一张表,然后再与第三张表left join,同理,多张表的left join 以此类推
select * from temp1 t1 left join temp2 t2 on t1.aid=t2.bid left join temp3 t3 on t2.bid=t3.cid ;

select * from temp3 t3 left join temp1 t1 on t3.cid=t1.aid left join temp2 t2 on t3.cid=t2.bid;

–3.right join 与left join相对应,会将右边的数据全部查出来(例子略)
— 一年多以后回过头来,发现第三张表的数据没有造好,也不想更正了,将就看吧,见谅
发布者:全栈程序员-站长,转载请注明出处:https://javaforall.net/202608.html原文链接:https://javaforall.net
