Subject Re: [Firebird-Java] JPA, Hibernate and Firebird
Author Serge Bogatyrjov
You don't need to use TABLE. This can help:

@Id @GeneratedValue(strategy = GenerationType.AUTO)
Integer id;

Also you can use GenerationType.SEQUENCE, but it is dangerous. Then
@SequenceGenerator can be used to set explisitly the used generator name.

To get the answer on the first question you should provide database url (and
check file permissions at first).

2007/12/6, Marcelo Siqueira <mdm-listas@...>:
> Laurent Duperval escreveu, em 5/12/2007 20:15:
> > * I also get an error that says:
> > javax.persistence.PersistenceException:
> > org.hibernate.MappingException: Dialect does not support identity
> > key generation. That, I understand. Is there a way to make this
> > work without resorting to using a UUID (32 char) generator? Is
> > there a generator that can be used/defined specifically for
> > running under Firebird?
> Hi Laurent,
> I found myself with the same problem. So far I decided to use the TABLE
> identifier strategy and allow JPA to use a table to hold the primary
> keys counters. I know this is not the optimal solution, but it will
> allow me to make the same code runs with Firebird and MySQL.
> Marcelo

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