what is the property name for registering an user type in hibernate

Question

Accoriding to hibernate document: section 6.5. Type registry: http://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/types.html

We can create a new user type and override existing hibernate's basic types. To use the new user type we need to register it using (see above link):

Configuration cfg = ...;
cfg.registerTypeOverride( new SuperDuperStringType() );

But never in document mentioned how we can register it in hibernate.cfg.xml? I can't believe they forgot to add this to hibernate.cfg.xml, Does anybody know about this? Thanks

Answer 1

Take a look at that post: Mapping A Custom Type In Hibernate

It worked for me with the following little modification:

the hibernate-mapping closing tag is wrongly formed, it misses the / (slash).

Appart from that, it explains everything.

Answer 2

Sorry, for the omission. The documentation describes both the registry and the keys, but to an extent expects users to be able to deduce that 1+1=2. I'll make that more explicit.

http://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/types.html#types-value-basic Describes the built-in registration keys.

As you can see in that section, the default mapping in the registry for java.lang.String is to org.hibernate.type.StringType. So when Hibernate sees an attribute whose java type is java.lang.String it looks into this registry using java.lang.String as the "registration key". Again, this is all by default; you can still give explicit type information on each attribute if you wish. That is described elsewhere in the documentation. Basically you would use @Type or <type/>.

If you want Hibernate to use your SuperDuperStringType anytime there is no explicit type information supplied, you'd use "java.lang.String" as SuperDuperStringType's getRegistrationKeys():

public class SuperDuperStringType implements BasicType {
    ...

    @Override
    public String[] getRegistrationKeys() {
        // lets use delegation and register ourselves under all of StringType's keys
        return org.hibernate.type.StringType.INSTANCE.getRegistrationKeys();
    }
}

In that case nothing explicit is needed in the mappings.