Can I use Spring Data JPA Auditing without the orm.xml file (using JavaConfig instead)?

Eric B. picture Eric B. · Mar 12, 2014 · Viewed 11k times · Source

I'm trying to get Spring Data Auditing to work in my Spring 3.2.8 / Spring Data 1.5 / Hibernate 4 project.

As per the Spring Data Auditing docs, I've added the @CreatedBy, etc annotations to my entities, created by AuditorAware implementation, and instantiated it from within my JavaConfig. However, it never seems to fire.

I find the docs a little confusing. It appears that the JavaConfig entry replaces the xml entry, but I am not sure.

I don't currently have any orm.xml file in my application. To be entirely honest, I'm not even sure where/how to configure it, or why I need it. All my entities are using annotations. I have tried adding @EntityListeners(AuditingEntityListener.class) to the entity, but that has not helped.

My current entity manager is defined without a persistence.xml file:

    <!--  entity manager -->
    <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
        <property name="dataSource" ref="dataSource"/>
        <property name="jpaVendorAdapter" ref="hibernateJpaVendorAdapter"/>
        <property name="packagesToScan" value="com.ia.domain"/>
        <property name="jpaProperties">
            <props>
                <prop key="hibernate.dialect">org.hibernate.dialect.MySQL5InnoDBDialect</prop>
                <prop key="hibernate.query.substitutions">true '1', false '0'</prop>
                <prop key="hibernate.generate_statistics">true</prop>
                <prop key="hibernate.show_sql">false</prop>
                <prop key="hibernate.format_sql">true</prop>
                <prop key="hibernate.hbm2ddl.auto">update</prop>
                <prop key="hibernate.ejb.naming_strategy">org.hibernate.cfg.ImprovedNamingStrategy</prop>
                <prop key="hibernate.connection.charSet">UTF-8</prop>
            </props>
        </property>
    </bean>

JavaConfig:

@Configuration
@EnableJpaAuditing
public class AuditConfig {
    @Bean
    public AuditorAware<User> auditorProvider(){
        return new SpringSecurityAuditorAware();
    }
}

Entity:

@EntityListeners({AuditingEntityListener.class})
@Entity
public class User
{

  @TableGenerator(name="UUIDGenerator", pkColumnValue="user_id", table="uuid_generator", allocationSize=1)
  @Id
  @GeneratedValue(strategy=GenerationType.TABLE, generator="UUIDGenerator")
  @Column(name="id")
  private Long id;

  @NotNull
  private String username;

  @CreatedDate
  @NotNull
  @Temporal(TemporalType.TIMESTAMP)
  @Column(name="created_date", nullable=false)
  private Date createdDate;

  @LastModifiedDate
  @NotNull
  @Temporal(TemporalType.TIMESTAMP)
  @Column(name="last_modified_date", nullable=false)
  private Date lastModifiedDate;

  @CreatedBy
  @ManyToOne(fetch=FetchType.LAZY)
  @JoinColumn(name="created_by")
  private User createdBy;

  @LastModifiedBy
  @ManyToOne(fetch=FetchType.LAZY)
  @JoinColumn(name="last_modified_by")
  private User lastModifiedBy;
  private String password;
  private Boolean enabled;


...
}

I've put a breakpoint in my SpringSecurityAuditorAware class but it is never being hit.

Do I still need an orm.xml file? How/where is this referenced from the EntityManager?

Answer

Stephan picture Stephan · Oct 7, 2014

Short version: No

As of JPA 2.0, it is not possible to define such entity listener without an XML file (orm.xml).

JPA 2.0:

Default entity listeners—entity listeners that apply to all entities in the persistence unit—can be specified by means of the XML descriptor. (p.93)

Long version: The workaround...

If all entities in your project extends an AbstractAuditable superclass then you can put @EntityListeners({AuditingEntityListener.class}) on AbstractAuditable. Listeners attached to an entity class are inherited by its subclasses.

JPA 2.0:

Multiple entity classes and mapped superclasses in an inheritance hierarchy may define listener classes and/or lifecycle callback methods directly on the class. (p.93)

Note that a subclass can exclude explicitly an inherited listener using the @ExcludeSuperclassListeners annotation.

There is one last interesting footnote from the spec I'd like to quote:

JPA 2.0:

Excluded listeners may be reintroduced on an entity class by listing them explicitly in the EntityListeners annotation or XML entity-listeners element. (Footnote [45] p.97)


Here is some code for illustrating the workaround:

AbstractAuditableEntity.java

import java.util.Date;

import javax.persistence.EntityListeners;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.MappedSuperclass;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

import org.springframework.data.annotation.CreatedDate;
import org.springframework.data.annotation.LastModifiedDate;
import org.springframework.data.jpa.domain.support.AuditingEntityListener;

@MappedSuperclass
@EntityListeners({AuditingEntityListener.class}) // AuditingEntityListener will also audit any subclasses of AbstractAuditable...
public abstract class AbstractAuditableEntity {
    @Id
    @GeneratedValue
    private Long id;

    @CreatedDate
    @Temporal(TemporalType.TIMESTAMP)
    private Date createdDate;

    @LastModifiedDate
    @Temporal(TemporalType.TIMESTAMP)
    private Date lastModifiedDate;
}

MyEntity.java

@Entity
public abstract class MyEntity extends AbstractAuditableEntity {

}

I think an interface Auditable may be used (@EntityListeners can appear on an interface) instead of an AbstractAuditable class but I didn't try...


Reference: JSR-000317 Java Persistence 2.0 - Final Release