Kotlin 中带有 val 的 @Embeddable 类

Question

我正在研究 Gradle-Kotlin-Hibernate 项目。我想保持我的一些类不可变，这在 Kotlin 中特别容易。这对于@Entity来说效果很好

@Entity(name = "SOMETHING")
class MeetingKeychain(
        val immutableProp: String
) {

// ID and others

}

感谢kotlin-jpa插件的使用。但相同的插件不适用于@Embeddable。

@Embeddable
class MeetingKeychain(
        val immutableProp: String
) {

// ID and others

}

抛出以下异常：

Caused by: org.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.component.PojoComponentTuplizer]
    at org.hibernate.tuple.component.ComponentTuplizerFactory.constructTuplizer(ComponentTuplizerFactory.java:98)
    at org.hibernate.tuple.component.ComponentTuplizerFactory.constructDefaultTuplizer(ComponentTuplizerFactory.java:119)
    at org.hibernate.tuple.component.ComponentMetamodel.<init>(ComponentMetamodel.java:64)
    at org.hibernate.mapping.Component.getType(Component.java:169)
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:398)
    at org.hibernate.mapping.Property.isValid(Property.java:225)
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:595)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:265)
    at org.hibernate.boot.internal.MetadataImpl.validate(MetadataImpl.java:329)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:443)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:879)
    ... 46 more
Caused by: java.lang.reflect.InvocationTargetException
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:423)
    at org.hibernate.tuple.component.ComponentTuplizerFactory.constructTuplizer(ComponentTuplizerFactory.java:95)
    ... 56 more
Caused by: org.hibernate.PropertyNotFoundException: Could not locate setter method for property [com.scherule.calendaring.domain.entities.ParticipantId#id]
    at org.hibernate.internal.util.ReflectHelper.findSetterMethod(ReflectHelper.java:540)
    at org.hibernate.property.access.internal.PropertyAccessBasicImpl.<init>(PropertyAccessBasicImpl.java:44)
    at org.hibernate.property.access.internal.PropertyAccessStrategyBasicImpl.buildPropertyAccess(PropertyAccessStrategyBasicImpl.java:27)
    at org.hibernate.mapping.Property.getGetter(Property.java:299)
    at org.hibernate.tuple.component.PojoComponentTuplizer.buildGetter(PojoComponentTuplizer.java:143)
    at org.hibernate.tuple.component.AbstractComponentTuplizer.<init>(AbstractComponentTuplizer.java:46)
    at org.hibernate.tuple.component.PojoComponentTuplizer.<init>(PojoComponentTuplizer.java:42)
    ... 61 more

我的问题是有什么办法可以解决这个问题吗？我希望我的@Embeddable类是不可变的。

Answer 1

事实上，这被证明是可行的，并且与 Kotlin 无关。唯一相关的方式是您需要生成kotlin-jpa插件的默认构造函数。你需要使用

@Access(AccessType.FIELD)

尽管

@Access(AccessType.PROPERTY)

是默认值。然后它寻找设置器而不是使用字段注入。