-- 一对一 -- 身份证表 CREATE TABLE idcard ( idcard_num CHAR(18) PRIMARY KEY, name VARCHAR2(20), address VARCHAR2(100), minzu VARCHAR2(20) ); -- 学生表 CREATE TABLE stu( sno VARCHAR2(10) PRIMARY KEY, name VARCHAR2(20), idcard_num CHAR(18) ); INSERT INTO idcard(idcard_num, name, address, minzu) VALUES ('123456789012345678','小芳','珠海南方','汉族'); INSERT INTO idcard(idcard_num, name, address, minzu) VALUES ('123456789087654321','小军','珠海东方','傣族'); commit select * from idcard insert into stu(sno, name, idcard_num) values ('001','小芳','123456789012345678'); insert into stu(sno, name, idcard_num) values ('002','小军','123456789087654321'); commit select * from idcard; select * from stu; update stu set name='大芳' where sno='001' commit select sno,t1.name,t1.idcard_num,address from stu t1,idcard t2 where t1.idcard_num = t2.idcard_num select * from stu -- 故意修改身份证号码,让它关联不上 update stu set idcard_num='123456789012345677' where idcard_num='123456789012345678' commit; -- 本质上是一个内关联 -- 关联不上的数据,不显示 select sno,t1.name,t1.idcard_num,address from stu t1,idcard t2 where t1.idcard_num = t2.idcard_num ---------- -- 左关联left join -- (以左为主,左边关联不上的数据,继续显示) select sno,t1.name,t1.idcard_num,address from stu t1,idcard t2 where t1.idcard_num = t2.idcard_num(+) --传统的写法 select * from stu t1 left join idcard t2 on t1.idcard_num = t2.idcard_num