How to annotate hibernate collections with generic types?

Question

I have a one-to-many class member of type List< GenericType < T > >. What is the recommended way to annotate this member in Hibernate?

Morbid details:

@Entity
public class DvOrdered extends DataValue
{
    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    @JoinTable(name="dv_ordered_other_reference_ranges", 
        joinColumns = @JoinColumn(name = "reference_range_id"),
        inverseJoinColumns = @JoinColumn(name = "dv_ordered_id"))
    private List<ReferenceRange<T>> otherReferenceRanges;
}

I get

Caused by: org.hibernate.AnnotationException: Property com.safehis.ehr.rm.datatypes.quantity.DvOrdered.otherReferenceRanges has an unbound type and no explicit target entity. Resolve this Generic usage issue or set an explicit target attribute (eg @OneToMany(target=) or use an explicit @Type
at org.hibernate.cfg.PropertyContainer.assertTypesAreResolvable(PropertyContainer.java:140)
at org.hibernate.cfg.PropertyContainer.getProperties(PropertyContainer.java:118)
at org.hibernate.cfg.AnnotationBinder.addElementsOfClass(AnnotationBinder.java:1554)
at org.hibernate.cfg.InheritanceState.getElementsToProcess(InheritanceState.java:236)
at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:775)
at org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3845)
at org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3799)
at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1412)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1846)

Thanks in advance for any reply.

What have you tried and whats the error that you are getting? As a side note do you really have oneToMany with a generic type !!!! — Shahzeb
– Shahzeb, Commented May 5, 2015 at 22:28
Error makes perfect sense . It does not know which concrete entity the relationship is with . — Shahzeb
– Shahzeb, Commented May 5, 2015 at 23:27
I'm sure the error makes sense. The question however asks how to write this type of idiom. — adolfoc
– adolfoc, Commented May 5, 2015 at 23:32

Vlad Mihalcea · Accepted Answer · 2015-05-06 11:17:32Z

1

The generic information is lost at compile-time due, to type erasure so Hibernate can't figure out what type T is in your case.

You can change your code to this:

private List<ReferenceRange<T>> otherReferenceRanges;

And then use inheritance so that each ReferenceRange subclass embeds a certain type.

public abstract class ReferenceRange<T> {

    public abstract T getReference();
}

public class IntegerReferenceRange extends ReferenceRange<Integer> {

    private Integer reference;

    @Override
    public Integer getReference() {
        return reference;
    }
}

answered May 6, 2015 at 11:17

Vlad Mihalcea

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2 Comments

adolfoc Over a year ago

Thank you for your reply, but I don't understand how this helps. Is there a way to use Hibernate AnyMetaDef with a OneToMany association and then use, say, ReferenceRange < Integer > as you suggest as a MetaValue?

Vlad Mihalcea Over a year ago

Using @AnyMetaDef gives you the same effect as if using inheritance, but you won;t be able to use nested generic types.

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