oracle - 使用具有父类(super class)型属性的类型创建对象表

Question

我正在尝试创建Oracle对象类型的Oracle表。

这是我的对象结构的样子

CREATE OR REPLACE TYPE PERS_T AS OBJECT 
( 

 empno number(4)
, ename varchar2(10)
, job varchar2(9)
, hiredate DATE
, sal number(7,2)
, comm number(7,2)
, deptno number(2)
)NOT FINAL;

CREATE OR REPLACE TYPE EMP_T FORCE UNDER pers_t (
  mgr pers_t
);

所有这些都很好，但是当我尝试使用创建一个 EMP_T类型的表时该怎么办

CREATE TABLE table_name(emp_type EMP_T);

我遇到错误

SQL Error: ORA-30756: cannot create column or table of type that contains a supertype attribute

在oracle中可以创建这样的表吗？

Answer 1

我不这么认为。根据Oracle自己的支持数据库，以下内容适用于ora-30756

Error Text, Cause and Action from Message File/s for ORA-30756

Versions 9.2, 10.1, 10.2, 11.1, 11.2, 12.1

Error: ORA-30756 cannot create column or table of type that contains a supertype attribute

Cause: The user tried to create a column or table of an object type that contains a supertype attribute. This is not supported because it leads to infinite recursion in our current storage model. Note that creating a column of a type implies that we create columns corresponding to all subtype attributes as well. Action: Change the type definition to contain a supertype REF attribute instead of the supertype object attribute.

您已经创建了一个 super 类型PERS_T，所以我相信这是您的问题。如文章所述，您可以创建引用属性而不是对象属性，如@tbone所解释的那样-创建表table_name(emp_type REF EMP_T);