Oracle 建表(一对多)代码及相关约束示例
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建表(一对多)代码及相关约束
create table t_class(
c_id number(3) primary key,
c_name varchar2(20) not null
);
create table t_stu(
s_id number(5) primary key,
s_name varchar2(8) not null,
sex char(2) default '男',
birthday date,
school_age number(2) check(school_age>0),
school_score number(5,2),
c_id number(3),
id_card char(18) unique,
foreign key(c_id) references t_class(c_id)
);
附：测试数据：
insert into t_class values(1,'订单班');
insert into t_class values(2,'开发班');
insert into t_class values(3,'美工班');
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (1, 'a', '男', to_date('01-01-1980', 'dd-mm-yyyy'), 24, 90, 1, null);
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (2, 'b', '男', to_date('11-05-1981', 'dd-mm-yyyy'), 23, 80, 1, null);
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (3, 'c', '女', to_date('19-09-1982', 'dd-mm-yyyy'), 22, 50, 1, null);
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (4, 'd', '女', to_date('28-01-1984', 'dd-mm-yyyy'), 21, 80, 2, null);
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (5, 'e', '男', to_date('28-02-1984', 'dd-mm-yyyy'), 21, 90, 2, null);
insert into T_STU (S_ID, S_NAME, SEX, BIRTHDAY, SCHOOL_AGE, SCHOOL_SCORE, C_ID, ID_CARD)
values (6, 'f', '男', to_date('16-10-1986', 'dd-mm-yyyy'), 21, 50, 2, null);
commit;
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