Hibernate Cannot insert values

[Steve James]

Tried for 3 hours can't find a answer , posting it

After running the program printing Inserting Record and Done inserting values , but couldn't find values in db . Any help what to check and where to check ?

public class FirstExample {
	public static void main(String[] args) {
		Session session = null;
		try{
			// This step will read hibernate.cfg.xml and prepare hibernate for use
			SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory();
			 session =sessionFactory.openSession();
				//Create new instance of Contact and set values in it by reading them from form object
			 	System.out.println("Inserting Record");
				Contact contact = new Contact();
                contact.setEmpid("106");
                contact.setAge("103");
                contact.setName("RaviKiran");
				session.save(contact);
				System.out.println("Done inserting values ");
		}catch(Exception e){
			System.out.println(e.getMessage());
		}finally{
			// Actual contact insertion will happen at this step
           // session.evict(contact);
         //   session.clear();
			session.flush();
			session.close();
			}
	}
}

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC "-//Hibernate/Hibernate Configuration DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
  <session-factory>
    <property name="hibernate.dialect">org.hibernate.dialect.OracleDialect</property>
    <property name="hibernate.connection.driver_class">oracle.jdbc.OracleDriver</property>
    <property name="hibernate.connection.url">jdbc:oracle:thin:@localhost:1521:oracle</property>
    <property name="hibernate.connection.username">scott</property>
    <property name="hibernate.connection.password">tiger</property>
    <property name="hibernate.hbm2ddl.auto">create</property>
    <property name="hibernate.show_sql">true</property>
    <mapping resource="mine/Contact.hbm.xml"/>
  </session-factory>
</hibernate-configuration>

Pojo class :

package mine;
// Generated Apr 18, 2009 10:56:39 AM by Hibernate Tools 3.2.1.GA

/**
 * Contact generated by hbm2java
 */
public class Contact  implements java.io.Serializable {


     private String empid;
     private String age;
     private String name;

    public Contact() {
    }

	
    public Contact(String empid) {
        this.empid = empid;
    }
    public Contact(String empid, String age, String name) {
       this.empid = empid;
       this.age = age;
       this.name = name;
    }
   
    public String getEmpid() {
        return this.empid;
    }
    
    public void setEmpid(String empid) {
        this.empid = empid;
    }
    public String getAge() {
        return this.age;
    }
    
    public void setAge(String age) {
        this.age = age;
    }
    public String getName() {
        return this.name;
    }
    
    public void setName(String name) {
        this.name = name;
    }




}

[Joe carco]

hi,

I'm not an Hibernate expert, but there is one very important thing that I think your code is missing:
you're neither beginning, closing a Transaction nor committing your insert. Therefore Hibernate keeps the object you created in its object cache.
Try something like this:

 private static void listHoney() {
    Transaction tx = null;
    Session session = SessionFactoryUtil.getInstance().getCurrentSession();
    try {
      tx = session.beginTransaction();
      List honeys = session.createQuery("select h from Honey as h")
              .list();
      for (Iterator iter = honeys.iterator(); iter.hasNext();) {
        Honey element = (Honey) iter.next();
        logger.debug("{}", element);
      }
      tx.commit();
    } catch (RuntimeException e) {
      if (tx != null && tx.isActive()) {
        try {
// Second try catch as the rollback could fail as well
          tx.rollback();
        } catch (HibernateException e1) {
          logger.debug("Error rolling back transaction");
        }
// throw again the first exception
        throw e;
      }


    }
  }   

except that you should obviously do your object creating instead of doing a select in lines 6-11

(taken from laliluna)

[Steve James]

joe that worked . Thanks