![]() Starting with version 6.2, Hibernate supports record classes as embeddables. And with Hibernate 6.2, you don’t even need to implement that instantiator yourself. If you only need to support Hibernate in at least version 6.0, you can implement a proprietary EmbeddableInstantiator and use a record as an embeddable. So, if you want to create a specification-compliant entity model, you can’t use records. But the JPA specification also requires it to be a non-final class and define a public, parameter-less constructor. If the information stored in the embeddable is immutable, modeling it as a record class seems obvious. The embeddable object then becomes part of the entity, doesn’t have its own lifecycle, and gets mapped to the same database table as the entity class. As a side-effect, this also enables you to map a record as an embeddable.Īn embeddable is a reusable mapping component that you can use as an attribute type. In version 6, Hibernate introduced the EmbeddableInstantiator feature that makes the instantiation of an embeddable more flexible. But you can use them to model embeddable classes. Until now, Hibernate also doesn’t provide any proprietary extensions that enable you to use a record as an entity class. That makes it impossible to use it as a JPA entity class. It has to declare an identifier that consists of at least 1 attribute.Īs mentioned earlier, a record class is implicitly final and doesn’t provide a parameter-less constructor.It has to provide a public or protected, parameter-less constructor.It has to be a top-level class annotated with It can’t be final.So, it’s understandable that everyone wants to use them to model their entities.īut you might have already recognized that the definition of a record doesn’t fulfill JPA’s requirements of an entity. ![]() an implementation of the toString method that includes all component names and values.Īs you can see, a record automatically provides you with all the boilerplate code you previously let your IDE generate. ![]()
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