有没有办法使用 JPA 映射电子邮件地址等不可变值对象？

@Immutable
@Embeddable
public final class EmailAddress {
    private final String value;

    public EmailAddress(String value) {
        this.value = value;
    }

    public String getValue() {
        return value;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        EmailAddress that = (EmailAddress) o;
        return value.equals(that.value);
    }

    @Override
    public int hashCode() {
        return value.hashCode();
    }
}

现在我得到实体保存的异常

org.hibernate.InstantiationException: No default constructor for entity: com.domain.EmailAddress
    org.hibernate.tuple.PojoInstantiator.instantiate(PojoInstantiator.java:107)
    org.hibernate.tuple.component.AbstractComponentTuplizer.instantiate(AbstractComponentTuplizer.java:102)
    org.hibernate.type.ComponentType.instantiate(ComponentType.java:515)
    org.hibernate.type.ComponentType.deepCopy(ComponentType.java:434)
    org.hibernate.type.TypeHelper.deepCopy(TypeHelper.java:68)
    org.hibernate.event.def.AbstractSaveEventListener.performSaveOrReplicate(AbstractSaveEventListener.java:302)
    org.hibernate.event.def.AbstractSaveEventListener.performSave(AbstractSaveEventListener.java:203)
    org.hibernate.event.def.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:129)
    org.hibernate.ejb.event.EJB3PersistEventListener.saveWithGeneratedId(EJB3PersistEventListener.java:69)
    org.hibernate.event.def.DefaultPersistEventListener.entityIsTransient(DefaultPersistEventListener.java:179)
    org.hibernate.event.def.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:135)
    org.hibernate.event.def.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:61)
    org.hibernate.impl.SessionImpl.firePersist(SessionImpl.java:808)
    org.hibernate.impl.SessionImpl.persist(SessionImpl.java:782)
    org.hibernate.impl.SessionImpl.persist(SessionImpl.java:786)
    org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:672)
    sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    java.lang.reflect.Method.invoke(Method.java:597)
    org.springframework.orm.jpa.SharedEntityManagerCreator$SharedEntityManagerInvocationHandler.invoke(SharedEntityManagerCreator.java:240)
    $Proxy25.persist(Unknown Source)
    org.springframework.data.jpa.repository.support.SimpleJpaRepository.save(SimpleJpaRepository.java:360)
    sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    java.lang.reflect.Method.invoke(Method.java:597)
    org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.executeMethodOn(RepositoryFactorySupport.java:368)
    org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.invoke(RepositoryFactorySupport.java:349)
    org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:172)
    org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:110)
    org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:172)
    org.springframework.dao.support.PersistenceExceptionTranslationInterceptor.invoke(PersistenceExceptionTranslationInterceptor.java:155)
    org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:172)
    org.springframework.aop.framework.JdkDynamicAopProxy.invoke(JdkDynamicAopProxy.java:202)
    $Proxy26.save(Unknown Source)
    com.controller.UserController.create(UserController.java:64)

我想使用 final 字段和 hibernate 作为 JPA 实现。

最佳答案

使用标准 JPA 注释和可嵌入对象将无法做到这一点，因为必须使用默认构造函数创建对象，并通过反射设置值。

但是您可以使用 Hibernate 自定义类型。阅读this part of the Hibernate reference documentation ，其中有一个示例 Money 类型，它是使用带参数的构造函数实例化的，因此可能是不可变的。

关于java - 不可变值对象和 JPA，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7222366/