Hi,

This question has been asked a few times before with no solution, but that was a while ago so I'll give it a shot anyway.

I need to register a user-defined type as the default when no explicit type mapping exists. Same as the example shown in the Hibernate 3.6 documentation (section 6.4.1)

• http://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/types.html#types-custom-type

The catch is that I'm using JPA around Hibernate so don't have obvious access to the "org.hibernate.cfg.Configuration" instance. I've dug through the code a bit, but haven't found an obvious way to do this yet. Is there any way to:

• Get access to the "org.hibernate.cfg.Configuration" or "org.hibernate.type.TypeResolver" instance in a JPA-initialised application; or
• Register types using a system property of some kind in the "persistence.xml"

Old versions of this question:

• https://forum.hibernate.org/viewtopic.php?f=1&t=1012460
• https://forum.hibernate.org/viewtopic.php?f=1&t=1011906
• https://forum.hibernate.org/viewtopic.php?f=1&t=1011659

So for the time being I've decided to forego the type override and manually annotate my properties with @Type.

For anybody where this is not an option, the only approach I could see for registering type overrides through JPA is:

• Subclass HibernatePersistence and override the create[Container]EntityManagerFactory(...) methods
• Get the Configuration instance through the Ejb3Configuration.getHibernateConfiguration() method
• Ignore the big scary warning on that method and register your types through that Configuration instance anyway

Far too ugly for my liking, and I have relatively few instances of the type in question.

If anybody has a better solution for achieving what the example in section 6.4.1 does using JPA, I (and I'm sure others) would really love to hear it.