How to externalize properties from JPAs persistence.xml?

Question

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)

Romberg

@Romberg Let me know: Have you deployed a datasource whose global jndi address is jdbc/appDatasource ??? – Mime

@Arthur thanks for your response. Yes, I tried several JNDI names after the error occurred. I suppose it gets the persistenceUnits name "pu" wrong since it isn't complaining about the datasource. – Romberg

@Romberg I need additional info: post how your app is built ( As shown here: https://mcmap.net/q/244562/-jboss-seam-enabling-debug-page-on-weblogic-10-3-2-11g#2459795 ) Try to use SeamTest (Take a look at Seam examples directory - There are a lot of tests). – Mime

@Romberg How to set up a connection pool in JBoss: 1° Copy your database's JDBC JAR file to $JBOSS_HOME/server/default/lib 2° JBoss includes example database connection-pool files in the directory $JBOSS_HOME/docs/examples/jca. The name of each file ends in -ds.xml For instance, oracle-ds.xml. Copy your Target database -ds.xml file to $JBOSS_HOME/service/default/deploy and set up your datasource As follows – Mime

<datasources> <local-tx-datasource> <jndi-name>OracleDS</jndi-name> <connection-url>jdbc:oracle:thin:@localhost:1521:app</connection-url> <driver-class>oracle.jdbc.driver.OracleDriver</driver-class> <user-name>scott</user-name> <password>tiger</password> </local-tx-datasource> </datasources> – Mime

For the example above, JBoss will create a global jndi datasource called java:/OracleDS – Mime

Answer 1

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).

References

• Hibernate Entity Manager Reference Guide
  • Table 2.1. Hibernate Entity Manager specific properties

Worthen

Thank you Pascal, I will try that ASAP – Romberg

And what's the solution for javax.persistence/EclipseLink? – Tokay

I cannot archive this with hibernate 4.3.5. Many exceptions, e.g., java.lang.UnsupportedOperationException: The application must supply JDBC connections, Connection cannot be null when 'hibernate.dialect' not set occurred. Besides, I cannot find this property in the least hibernate document: docs.jboss.org/hibernate/orm/4.3/manual/en-US/html_single I finally use the solution here: #4548853 – Burgher

Answer 2

Just found a an alleged way for EclipseLink users. There is "eclipselink.persistencexml" which has a default value of

public static final String ECLIPSELINK_PERSISTENCE_XML_DEFAULT = "META-INF/persistence.xml";

but it can't be overridden although the docs say it can be...

/**
 * The <code>"eclipselink.persistencexml"</code> property specifies the full
 * resource name to look for the persistence XML files in. If not specified
 * the default value defined by {@link #ECLIPSELINK_PERSISTENCE_XML_DEFAULT}
 * will be used.
 * <p>
 * IMPORTANT: For now this property is used for the canonical model
 * generator but it can later be used as a system property for customizing
 * weaving and application bootstrap usage.
 * <p>
 * This property is only used by EclipseLink when it is locating the
 * configuration file. When used within an EJB/Spring container in container
 * managed mode the locating and reading of this file is done by the
 * container and will not use this configuration.
 */

Tokay

Answer 3

I used this mechanism, seems to work for most of the properties, had issues with non-jta-data-source.

http://www.eclipse.org/eclipselink/api/2.4/index.html?org/eclipse/persistence/config/PersistenceUnitProperties.html

Hypervitaminosis

Answer 4

If you are using Spring to manage and inject entity manager, then it is possible to implement org.springframework.orm.jpa.persistenceunit.PersistenceUnitPostProcessor and pass on external properties. I could successfully externalize all properties from persistence.xml using this.

Kirshbaum

Answer 5

**persistence.xml**
            
            <?xml version="1.0" encoding="UTF-8"?>
            <persistence version="2.2"
                xmlns="http://xmlns.jcp.org/xml/ns/persistence"
                xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_2.xsd">
                <persistence-unit name="jsf-maven-project">
                    <properties>
                       <property name="hibernate.ejb.cfgfile" value="/hibernate.cfg.xml"/>
                    </properties>
                </persistence-unit>
            </persistence>
           
**hibernate.cfg.xml**
            <?xml version="1.0" encoding="UTF-8"?>
        <!DOCTYPE hibernate-configuration PUBLIC 
          "-//Hibernate/Hibernate Configuration DTD 3.0//EN" 
          "http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
          <hibernate-configuration>
            <session-factory>
              <property name="connection.driver_class">org.postgresql.Driver</property>
              <property name="connection.url">jdbc:postgresql://localhost:5432/testdb</property>
              <property name="connection.username">postgres</property>
              <property name="connection.password">Test@123</property>
              <property name="connection.pool_size">10</property>
              <property name="dialect">org.hibernate.dialect.PostgreSQLDialect</property>
              <property name="current_session_context_class">thread</property>
              <property name="show_sql">true</property>
              <property name="format_sql">true</property>
              <property name="hbm2ddl.auto">update</property>
              <!-- mapping  class="com.sample.Book" / -->
          </session-factory>
        </hibernate-configuration>
    
**HibernateUtil.java**
            public static EntityManager getEntityManager() {
                    if (entityManagerFactory == null) {
                        entityManagerFactory = Persistence.createEntityManagerFactory("jsf-maven-project");
                    }
                    return entityManagerFactory.createEntityManager();
                }
            

I was createing JSF project with hibernate JPA and this solution worked for me.

Cinerary