I have faced problem while getting the hibernate session, So as suggested by Scott I have tried for EntityManager to work with database for persistence.
With below entries in persistence.xml, I am able to get the EntityManager Object
<persistence version="2.0" 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_2_0.xsd ">
<persistence-unit name="ldm_pu">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>java:/ldmDS</jta-data-source>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect" />
<property name="hibernate.transaction.jta.platform" value="org.hibernate.service.jta.platform.internal.JBossAppServerJtaPlatform" />
</properties>
</persistence-unit>
</persistence>
The code which I used for getting session using entity manager is as follows
EntityManagerFactory emf = javax.persistence.Persistence.createEntityManagerFactory("ldm_pu");
EntityManager em = emf.createEntityManager();
Session session = (Session)em.unwrap(Session.class);
Using this approach I am able to proceed for persistence of information using JPA.
Thank You,
Jilani