Pergunta

my persistence.xml

   <?xml version="1.0" encoding="UTF-8"?>
   <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="caronteemirJPAprj">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
       <non-jta-data-source>ERCOLE</non-jta-data-source>
       <class>model.Customer</class>
        <properties>
          <property name="hibernate.dialect" 
           value="org.hibernate.dialect.OracleDialect"/>
          <property name="javax.persistence.jdbc.url" value="jdbc:oracle:thin:@172.31.62.61:1521:ERCOLE"/>
          <property name="javax.persistence.jdbc.user" value="ELI"/>
          <property name="javax.persistence.jdbc.password" value="eli"/>
          <property name="javax.persistence.jdbc.driver" value="oracle.jdbc.OracleDriver"/>
        </properties>
     </persistence-unit>

 </persistence>

enter image description here

@Stateless
public class QueryBean implements Serializable {


 /* 
 * 
 */
 private static final long serialVersionUID = -7101661704004061600L;
 EntityManagerFactory emf = Persistence.createEntityManagerFactory("caronteemirJPAprj");

 @PersistenceContext(unitName="caronteemirJPAprj")
 private EntityManager em;

 public QueryBean(){
       em = emf.createEntityManager();
 }

 public List<Customer> findAll() {
  Query query = em.createQuery("Select h from Customer h");
  return (List<Customer>) query.getResultList();
 }

 public static void main(String[] args) {

     QueryBean q = new QueryBean();

     q.findAll();
 }

}
Foi útil?

Solução

I needed hibernate-entitymanager-4.1.4.Final

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