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','小明','珠海','汉族');
INSERT INTO idcard(idcard_num,name,address,minzu)
VALUES('123456789012345688','小方','珠海','彝族');
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;