UserType. How to do whithout a table?

TimeLord · **Joined:** Tue Dec 26, 2006 2:17 am **Posts:** 1

Hi everybody!

I have some question about a UserType.

I need these Type:

Code:

public class A { 
  private int id;
  private int name;
}

and it uses:

Code:

public class B {
  private int id;
  private A a;
}

If I kept A in a separate table a with the fields: "ID, Name", I would do the following:

Code:

<hibernate-mapping>
  <class name="B" table="b">
    <id name="id">
      <generator class="native"/>
    </id>
    <many-to-one name="a" column="a_id" class="A"/>
  </class>
<hibernate-mapping>

But I want to do that without a table a and with a static consts in A, like this:

Code:

public final class A {
    public static final A A_1 = new A(1,"str1");
    public static final A A_2 = new A(2,"str2");
    private int id;
    private String name;
    ...
}

and when a hibernate loading some B he uses well-defined staic const from A

Thx a lot for help!

...and sorry for my english :(

evdelst · **Joined:** Wed Aug 24, 2005 11:49 am **Posts:** 63

You need to create usertype and map the property as a 'normal' column (no many-to-one is needed).

In the usertype, you translate between the db-value and your constant (and the other way around)

Edwin