> Can we assume that, after 1) and 2) execute, all information (in the form of
> hbm/annotations source) is available to resolve all existing associations?
>
> If anything is left unresolved, should it be an exception?
I think it is reasonable to make that a rule.
>
> Can we assume that after 3)
gt; Sent: Friday, January 25, 2013 9:45:26 AM
> Subject: Re: [hibernate-dev] MetadataSourceProcessor and associations
>
> ** DISCLAIMER : Its been a long time since I have been on the
> metamodel
> branch, so please excuse any misuse of class/interface names...
>
> Comments inl
y as envers currently does (creating hbm in memory and
> ask hibernate to do the binding stuff) -- 4)
> 2. create/change metamodel directly -- 3)
>
>
>>
>>
>> Regards,
>> Gail
>>
>> - Original Message -
>>>
>>> From: "Steve
Regards,
> Gail
>
> - Original Message -
>> From: "Steve Ebersole"
>> To: "Gail Badner"
>> Cc: "Hibernate hibernate-dev"
>> Sent: Wednesday, January 23, 2013 11:16:23 AM
>> Subject: Re: [hibernate-dev] Metad
Hi,
as Steve is saying, you can basically mix and match hbm and annotation, however
there is a caveat.
Within one entity hierarchy you have to stick to one mapping approach. You
cannot map the base class
Animal in hbm and then use annotations for the subclass entity Dog.
If the entities are n
uot;Steve Ebersole"
> To: "Gail Badner"
> Cc: "Hibernate hibernate-dev"
> Sent: Wednesday, January 23, 2013 11:16:23 AM
> Subject: Re: [hibernate-dev] MetadataSourceProcessor and associations
>
> What you mention is in fact possible.
>
> And I have
It seems that JoinColumnResolutionDelegate impls can have the same problem.
- Original Message -
> From: "Steve Ebersole"
> To: "Gail Badner"
> Cc: "Hibernate hibernate-dev"
> Sent: Wednesday, January 23, 2013 11:16:23 AM
> Subject: Re
Is it possible for an entity containing one side of an association be mapped
using annotations and the entity containing the opposite side of the
association be mapped using an hbm mapping?
For example, suppose Order has annotations:
@OneToMany( mappedby="order" )
List Order.orderLines
OrderLi
