Hibernate enum mapping

Question

I need to map the enums which didn't implement the interface beforehand to the existing database, which stores enums in the same table as the owner class using the @Enumerated(EnumType.STRING).

class A {
    HasName name;
}

interface HasName {
    String getName();
}

enum X implements HasName {
    John, Mary;

    public String getName() { return this.name(); }
}

enum Y implements HasName {
    Tom, Ann;

    public String getName() { return this.name(); }
}

How the mapping should be handled in this case? Persisting to the database doesn't change as all of the enums implementing the interface will have different values, but I'm not sure how the objects should be retrieved from the DB (do I need a custom mapper, which will try to instantiate an enum using the specified enum classes? Does Hibernate natively support this functionality?).

Answer 1

It's possible to create a custom UserType (e.g. this one) and use it from your mappings

<property name="type" not-null="true">
  <type name="at.molindo.util.hibernate.EnumUserType">
    <param name="enumClass">
      com.example.MyEnum
    </param>
  </type>
</property>

Answer 2

Hibernate provides org.hibernate.type.EnumType to map Enumerated types. For instance,

package com.igalia.enumerates;

public enum Status {
   BUSY,
   AVAILABLE;
}

package com.igalia.entities;

class MyClass {
   private Status status;
}

Then, do your Hibernate mapping as follows:

<class name="MyClass">
   <id name="id">
      <generator class="native"/>
   </id>

   <property name="status">
      <type name="org.hibernate.type.EnumType">
         <param name="enumClass">com.igalia.enumerates.Status</param>
      </type>
   </property>
</class>

And that's it. If you prefer to use JPA annotations instead of hbm.xml, use @Enumerated(EnumType.STRING). Check it here:

http://stackoverflow.com/questions/417062/enumerations-in-hibernate