您可以使用受控冗余和复合FK约束来做到这一点:
CREATE TABLE offr (
offr_id INT NOT NULL,
coy_id INT NOT NULL,
PRIMARY KEY (offr_id),
FOREIGN KEY (coy_id) REFERENCES ins_coy (coy_id),
UNIQUE KEY (offr_id, coy_id)
);
我添加了一个复合唯一键(offr_id,coy_id)来支持subscribe
表上的复合FK约束。
CREATE TABLE provide (
coy_id INT NOT NULL,
type_id INT NOT NULL,
PRIMARY KEY (coy_id, type_id),
FOREIGN KEY (coy_id) REFERENCES ins_coy (coy_id)
);
这里的复合主键非常适合subscribe
表上的复合FK约束。
CREATE TABLE subscribe (
naf_no INT NOT NULL,
coy_id INT NOT NULL,
type_id INT NOT NULL,
PRIMARY KEY (naf_no, type_id),
FOREIGN KEY (naf_no, coy_id) REFERENCES offr (offr_id, coy_id),
FOREIGN KEY (coy_id, type_id) REFERENCES provide (coy_id, type_id)
);
重叠的复合FK约束将确保官员只能订阅他/她注册的公司提供的保险。coy_id
从逻辑上讲是冗余的,但对于完整性而言是必需的,并且不会因FK约束而存在更新异常的风险。
或者,您可以使用触发器来检查值是否通过内部联接相关:
CREATE TRIGGER check_subscribe BEFORE INSERT OR UPDATE ON subscribe
FOR EACH ROW
WHEN NOT EXISTS (
SELECT 1
FROM offr
INNER JOIN provide ON offr.coy_id = provide.coy_id
WHERE offr.offr_id = new.naf_no AND provide.type_id = new.type_id
)
RAISE_APPLICATION_ERROR (num => -20000, msg => 'Officers can only subscribe to types provided by their company');
免责声明:我无法在sqlfiddle上进行测试,也没有安装Oracle,但希望它将为您指明正确的方向。