问题
Is there way to map immutable Value objects like email address using JPA?
@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();
}
}
Now I get exception on entity save
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)
I want use final fields and hibernate as JPA implementation.
回答1:
You won't be able to do that using standard JPA annotations and an embeddable object, because the object will have to be created using a default constructor, and the value set via reflection.
You could however use a Hibernate custom type. Read this part of the Hibernate reference documentation, where there is an example Money
type, which is instantiated using a constructor with arguments, and could thus be immutable.
回答2:
Probably the easiest solution which works with a little older versions like 3.5 of Hibernate is to implement org.hibernate.usertype.UserType. There is quite a few methods in it but for immutable types you can extract most of them to common superclass:
package com.acme;
import java.io.Serializable;
import org.hibernate.usertype.UserType;
public abstract class AbstractImmutableType
implements UserType {
public AbstractImmutableType() {
super();
}
public boolean isMutable() {
return false;
}
public Serializable disassemble(Object value) {
return (Serializable) value;
}
public Object assemble(Serializable cached, Object owner) {
return cached;
}
public Object deepCopy(Object value) {
return value;
}
public Object replace(Object original, Object target,
Object owner) {
return original;
}
public boolean equals(Object x, Object y) {
if (x != null && y != null) {
return x.equals(y);
}
// Two nulls are equal as well
return x == null && y == null;
}
public int hashCode(Object x) {
if (x != null) {
return x.hashCode();
}
return 0;
}
}
And you can use it like this:
package com.acme;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Types;
public class CurrencyType extends AbstractImmutableType {
public static final String TYPE = "com.acme.CurrencyType";
private static final int[] SQL_TYPES = {
Types.VARCHAR
};
public CurrencyType() {
super();
}
public Object nullSafeGet(ResultSet rs, String[] names,
Object owner) throws SQLException {
String value = rs.getString(names[0]);
if (rs.wasNull()) {
return null;
}
return Currency.valueOf(value);
}
public void nullSafeSet(PreparedStatement st, Object value,
int index) throws SQLException {
if (value != null) {
st.setString(index, ((Currency)value).getCode());
} else {
st.setNull(index, SQL_TYPES[0]);
}
}
public Class<?> returnedClass() {
return Currency.class;
}
public int[] sqlTypes() {
return SQL_TYPES;
}
}
Longer explanation for this code you can find here
回答3:
For JPA to be able to create objects via reflection, you have to have a default constructor, but it doesn't have to be public. I also like to keep my fields final, but this might be too restrictive for reflection -- you'll have to try.
I'd suggest dropping the final field modifier and adding a private default constructor with a short comment (so you still know why that no-op constructor is there next week):
public final class EmailAddress {
private String value; // no final modifier
private EmailAddress() {
// for JPA
}
public EmailAddress(String value) {
this.value = value;
}
...
}
来源:https://stackoverflow.com/questions/7222366/immutable-value-objects-and-jpa