In my application I have unidirectional associations between entities where the source is marked as @IndexedEmbedded.
To update the index of the source entity, if the target of the assosiation is changed, I use JPA Callbacks on target entities to track the changes.
In the JPA Callbacks I want to find generically all JPA entitiy candidate classes whose index needs to be udated manually.

I believe, the simplest way to find such entities would be to analyse the Hibernate Search Metamodel and JPA Metamodel by traversing the entity metamodel graph.

I could implement code which analyses the Search annotations on entities. But if the Search Mapping is provided programatically by custom SearchMapping @Factory method,
I have probably no chance to reconstruct the mapping correctly at runtime.

Another problem is to find bi-directional associations in the JPA metamodel. But this is not releated to Search.

I didnt't find anything about Hibernate Search Metamodel in reference documentation.

Is there any safe (maybe non-API) way to access the Search Metamodel at runtime?

Hello,
no sorry the metamodel is not accessible and also the way it's stored today might not make it too easy for you to hack around it.
If you want to take a look, it's all in a large and overgrown class named DocumentBuilder (with two extensions).

Coincidentally we just decided last week that it's time to refactor this class and expose the metadata, as there are several other projects which could better integrate with it. If you want to take this task, we'll be glad to advise you and do some rounds of code review, and apply your patches upstream.

This looks like a good starting point:
https://hibernate.onjira.com/browse/HSEARCH-918

As a simpler solution, wouldn't you prefer to change your domain model to use bi-directional relations?

Hello Sanne,

Thank you for the advice. I will at least try it in the next time.


Unfortunately it's not possible by component based design of our software.
If two entities belong to different components, only one of them can reference the other one.