是否可以在相交表上实现参照完整性，以便其父代在外键列上具有相同的值？

Question

示例Oracle SQL代码：

create table t1 (
        t1a integer,
        t1b varchar2(30)
);
alter table t1 add constraint t1_pk primary key (t1a);
insert into t1 values (1, 'A');
insert into t1 values (2, 'B');

create table t11 (
        t11a integer,
        t1a  integer,
        t11b varchar2(30)
);
alter table t11 add constraint t11_pk primary key (t11a);
alter table t11 add constraint t11_t1_fk foreign key (t1a) references t1(t1a);
insert into t11 values (10, 1, 'A1');
insert into t11 values (11, 2, 'B1');

create table t12 (
        t12a integer,
        t1a  integer,
        t12b varchar2(30)
);
alter table t12 add constraint t12_pk primary key (t12a);
alter table t12 add constraint t12_t1_fk foreign key (t1a) references t1(t1a);
insert into t12 values (20, 1, 'A2');
insert into t12 values (21, 2, 'B2');

create table t1112 (
        t11a integer,
        t12a integer
);
alter table t1112 add constraint t1112_pk primary key (t11a, t12a);
alter table t1112 add constraint t1112_t11_fk foreign key (t11a) references t11(t11a);
alter table t1112 add constraint t1112_t12_fk foreign key (t12a) references t12(t12a);

create or replace trigger t1112_trg before insert or update on t1112 for each row
declare
        t11a_v  integer;
        t12a_v  integer;
begin
        select t11.t1a into t11a_v from t11 where t11.t11a = :new.t11a;
        select t12.t1a into t12a_v from t12 where t12.t12a = :new.t12a;
        if (t11a_v != t12a_v) then
                raise_application_error(-20000, 'Mismatch');
        end if;
end;
/

顶级表：t1。t11和t12具有对t1的外键引用。t1112是t1和t2之间的交集表。我要确保写入t1112的任何行都具有t11和t12的组合，以便它们都指向t1中的同一行。

在示例中，我尝试使用触发器来实现它。有没有办法通过参照完整性约束来做到这一点？

Answer 1

使用实例化视图：

CREATE MATERIALIZED VIEW t11_t12_mv
   BUILD IMMEDIATE
   AS SELECT t11a, t12a
      FROM   t11 INNER JOIN t12 ON t11.t1a = t12.t1a;

ALTER TABLE t11_t12_mv ADD CONSTRAINT t11a_t12a__pk PRIMARY KEY ( t11a, t12a );

create table t1112 (
  t11a integer,
  t12a integer,
  CONSTRAINT t1112_pk PRIMARY KEY (t11a, t12a),
  CONSTRAINT t1112_fk FOREIGN KEY (t11a, t12a) REFERENCES t11_t12_mv ( t11a, t12a )
);

然后：

INSERT INTO t1112 VALUES ( 10, 20 );
INSERT INTO t1112 VALUES ( 11, 21 );

Works。但是：

INSERT INTO t1112 VALUES ( 10, 21 );

引发异常：

ORA-02291: integrity constraint (USER_4_F249D.T1112_FK) violated - parent key not found 

sqlfiddle here

Answer 2

最简单的方法是在表t11和t12上创建其他唯一约束：

ALTER TABLE t11 ADD CONSTRAINT t11_un UNIQUE ( t11a, t1a );
ALTER TABLE t12 ADD CONSTRAINT t12_un UNIQUE ( t12a, t1a );

将列t1a添加到表T1112

ALTER TABLE t1112 ADD (t1a integer);

在t1112上拖放并重新创建参照约束：

ALTER TABLE T1112 DROP CONSTRAINT T1112_T11_FK;
ALTER TABLE T1112 ADD CONSTRAINT T1112_T11_FK FOREIGN KEY (t11a, t1a ) REFERENCES T11 (t11a, t1a);
ALTER TABLE T1112 DROP CONSTRAINT T1112_T12_FK;
ALTER TABLE T1112 ADD CONSTRAINT T1112_T12_FK FOREIGN KEY (t12a, t1a ) REFERENCES T11 (t12a, t1a);

这样，他们现在使用两个唯一约束。连接表中的共享列t1a将确保t11和t12中的键重新连接到t1。