使用 Oracle 实现唯一性

Question

例如，说我有类似的东西

CREATE TABLE COMPANY(id int not null primary key, ...);

CREATE TABLE DEPARTMENT(id int not null primary key, company_id int not null,
 CONSTRAINT FK_DEP_COMPANY_ID FOREIGN KEY(company_id) REFERENCES COMPANY(id),...);

CREATE TABLE EMPLOYEE(id int not null primary key, department_id int not null, 
  username varchar(30) NOT NULL, ...,
 CONSTRAINT FK_EMPLOYEE_DEP_ID FOREIGN KEY(department_id) REFERENCES DEPARTMENT(id));

我必须EMPLOYEE.username在一家公司内实现唯一性。

我看到 2 种方式。
1.AFTER在EMPLOYEE表上使用语句级触发器（类似于http://asktom.oracle.com/pls/asktom/ASKTOM.download_file?p_file=6551198119097816936 , case1）
2. 只是添加company_id到EMPLOYEE

它们对我来说都不理想，我想知道在这种情况下实现唯一性的正确方法是什么。

谢谢。

Answer 1

主要只是为了我自己的娱乐，您可以使用具有唯一索引的物化视图来完成此操作：

CREATE MATERIALIZED VIEW LOG ON COMPANY
WITH PRIMARY KEY, ROWID
INCLUDING NEW VALUES;

CREATE MATERIALIZED VIEW LOG ON DEPARTMENT
WITH PRIMARY KEY, ROWID (company_id)
INCLUDING NEW VALUES;

CREATE MATERIALIZED VIEW LOG ON EMPLOYEE
WITH PRIMARY KEY, ROWID (department_id, username)
INCLUDING NEW VALUES;

CREATE MATERIALIZED VIEW xcheck_mv
BUILD IMMEDIATE
REFRESH FAST ON COMMIT
AS SELECT c.id as company_id, lower(e.username) as username,
    c.rowid as c_rowid, d.rowid as d_rowid, e.rowid as e_rowid
from company c, department d, employee e
where d.company_id = c.id
and e.department_id = d.id;

CREATE UNIQUE INDEX xcheck_ind ON xcheck_mv(company_id, username);

然后尝试使用employee相同的用户名插入不同部门的记录会导致唯一约束违规，尽管在您提交之前不会发生：

insert into company (id) values(1);

1 row created.

SQL> insert into department (id, company_id) values(1, 1);

1 row created.

SQL> insert into department (id, company_id) values(2, 1);

1 row created.

SQL> insert into employee (id, department_id, username) values(1,1,'Joe Bloggs');

1 row created.

SQL> insert into employee (id, department_id, username) values(2,2,'Joe Bloggs');

1 row created.

SQL> commit;
commit
*
ERROR at line 1:
ORA-12008: error in materialized view refresh path
ORA-00001: unique constraint (STACKOVERFLOW.XCHECK_IND) violated

只是为了好玩，我放入lower()MV 定义来捕获最基本的解决方法 - 因此尝试插入'joe bloggs'也会失败 - 但这个模型永远不会非常健壮。

我并不是说这是一个好主意，只是说这是可能的......