How to externalize properties from JPAs persistence.xml?

stacker picture stacker · Oct 14, 2010 · Viewed 28.3k times · Source

I would like to put some of the hibernate configuration in a property file to make it editable without build and deploy.

I tried to solve my problem by following the instructions from Create JPA EntityManager without persistence.xml configuration file

app.properties:

hibernate.show_sql=true 
hibernate.dialect=org.hibernate.dialect.MySQLDialect
hibernate.hbm2ddl.auto=validate 
hibernate.show_sql=true
hibernate.format_sql=true
hibernate.default_schema=myschema

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<!-- Persistence deployment descriptor for dev profile -->
<persistence xmlns="http://java.sun.com/xml/ns/persistence" 
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd" 
             version="1.0">

   <persistence-unit name="pu">
      <provider>org.hibernate.ejb.HibernatePersistence</provider>
      <jta-data-source>jdbc/appDatasource</jta-data-source>
      <properties>
         <property name="jboss.entity.manager.factory.jndi.name" value="java:/appEntityManagerFactory"/>
      </properties>
   </persistence-unit>

</persistence>

In the initialization code the application executes the following sequence, (which finds the properties),

Properties props = new Properties();
InputStream is = ClassLoader.getSystemResourceAsStream( "app.properties" );
props.load( is );
Persistence.createEntityManagerFactory( "pu", props );

but fails with the error message:

 INFO  [SessionFactoryImpl] building session factory
 INFO  [SessionFactoryObjectFactory] Not binding factory to JNDI, no JNDI name configured
ERROR [STDERR] javax.persistence.PersistenceException: [PersistenceUnit: pu] Unable to build EntityManagerFactory

Does anyone have an idea what could be wrong with my configuration?

Versions: JBoss 4.3 Seam: 2.1.2

EDIT:

JBoss JNDI enlists "pu" as persistence unit:

persistence.units:ear=app.ear,jar=app.jar,unitName=pu (class: org.hibernate.impl.SessionFactoryImpl)

Answer

Pascal Thivent picture Pascal Thivent · Oct 16, 2010

As an alternative to your current approach and since you're using Hibernate, you could use Hibernate to configure JPA by declaring a hibernate.cfg.xml file using the hibernate.ejb.cfgfile property, like this:

<persistence>
 <persistence-unit name="manager1" transaction-type="JTA">
    <jta-data-source>java:/DefaultDS</jta-data-source>
    <properties>
       <property name="hibernate.ejb.cfgfile" value="/hibernate.cfg.xml"/>
    </properties>
 </persistence-unit>
</persistence>

My understanding is that the hibernate.cfg.xml is just supposed to be on the classpath (so it could be outside the packaged archive).

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