Unit Testing your JPA entities

If you want to avoid exceptions like this at runtime after adding properties to your JPA entities, it might be worthwhile to invest some time in writing unit tests for them.

Caused by: org.hibernate.MappingException: Could not determine type for: be.ecs.validation.entities.City, at table: Person, for columns: [org.hibernate.mapping.Column(city)]
at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:291)
at org.hibernate.tuple.PropertyFactory.buildStandardProperty(PropertyFactory.java:143)
at org.hibernate.tuple.component.ComponentMetamodel.(ComponentMetamodel.java:68)
at org.hibernate.mapping.Component.buildType(Component.java:183)
at org.hibernate.mapping.Component.getType(Component.java:176)
at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:275)
at org.hibernate.mapping.Property.isValid(Property.java:217)
at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:464)
at org.hibernate.mapping.RootClass.validate(RootClass.java:236)
at org.hibernate.cfg.Configuration.validate(Configuration.java:1193)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1378)
at org.hibernate.cfg.AnnotationConfiguration.buildSessionFactory(AnnotationConfiguration.java:954)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:891)
... 43 more

The Spring TestContext Framework provides generic, annotation-driven unit and integration testing support. More information can be found on http://static.springsource.org/spring/docs/3.0.3.RELEASE/spring-framework-reference/html/testing.html

The following needs to be done

1. Create a persistence.xml specific to your unit tests
2. Create a Spring application context to bootstrap the entitymanager
3. Create the unit test that

• constructs an entity
• persists it
• retrieves it

1. Create a persistence.xml specific to your unit tests
We'll create a dedicated persistence.xml file in our package (src/test/java/META-INF/persistence.xml). In this example, we're using an in-memory HSQL database.

<persistence>
 <persistence-unit name="pu" transaction-type="RESOURCE_LOCAL">
  <provider>org.hibernate.ejb.HibernatePersistence</provider>
  <class>be.ecs.validation.entities.Person</class>
  <class>be.ecs.validation.entities.City</class>
  <class>be.ecs.validation.entities.Country</class>
     <properties>
         <property name="hibernate.connection.url" value="jdbc:hsqldb:mem:unit-testing-jpa"/>
         <property name="hibernate.connection.driver_class" value="org.hsqldb.jdbcDriver"/>
         <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect"/>
         <property name="hibernate.hbm2ddl.auto" value="create-drop"/>
         <property name="hibernate.connection.username" value="sa"/>
         <property name="hibernate.connection.password" value="x"/>
   <property name="hibernate.archive.autodetection" value="class, hbm"/>
     </properties>
 </persistence-unit>
</persistence>

2. Create a Spring application context

xxx

<beans>

 <bean id="entityManagerFactory"
  class="org.springframework.orm.jpa.LocalEntityManagerFactoryBean">
  <property name="persistenceUnitName" value="pu" />
 </bean>
 
 <bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
  <property name="entityManagerFactory" ref="entityManagerFactory" />
 </bean>
</beans>

3. Create the unit test

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations={"/applicationContext.xml"})
public class PersonPersistenceTest {

@PersistenceContext
private EntityManager entityManager;

public EntityManager getEntityManager() {
return entityManager;
}

@Test
public void createAndPersistPerson() {
Person person = new Person();
getEntityManager().persist(person);
Person retrievedPerson = getEntityManager().find(Person.class, person.getId());
Assert.assertNotNull(retrievedPerson);
}

}