Howto check if a object is connected to another in hibernate

Imagine two domain object classes, A and B. A has a bidirectional one-to-many relationship to B. A is related to thousands of 开发者_运维百科B. The relations must be unique, it's not possible to have a duplicate.

To check if an instance of B is already connected to a given instance of A, we could perform an easy INNER JOIN but this will only ensure the already persisted relations.

What about the current transient relations?

class A {
   @OneToMany
   private List<B> listOfB;
}

If we access the listOfB and perform a check of contains() this will fetch all the connected instances of B lazy from the datasource. I only want to validate them by their primary-key.

Is there an easy solution where I can do things like "Does this instance of A is connected with this instance of B?" without loading all these data into memory and performing a lookup based on collections?

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Thank you for all the answers. The extra lazy collection did the trick for me. I configured the @OneToMany connection with the LazyCollection annotation.

@IndexColumn(name = "index", base = 1)
@LazyCollection(LazyCollectionOption.EXTRA)

The Hibernate: Extra-lazy collection fetching article helped me doing this. When you use this option, #size(), #contains(), #get(), etc. do not trigger collection initialization.

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i think it can be done in two steps. For transient B's, Add your transient B's to listOfB also add to a transient list too. And do your contains checks in this list.

For your persisted B's, use a query something like,

select count(*) from B b where b.a.id = :aId

If this query returns zero, you can say that there isn't a relation between A and B.

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Thinking about the last paragraph of your question some more, since the association is bidirectional, I would do a search on your specific instance of B with an association to A.

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Why can't you perform INNER JOIN?

With default flush mode Hibernate will flush the session before executing query, so that unsaved elements of the collection is not a problem.