org.hibernate.exception.GenericJDBCException

[anarkali perera]

this is my config file

<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">

<hibernate-configuration>
<session-factory>
      <property name="hibernate.connection.driver_class">

com.mysql.jdbc.Driver</property>
      <property name="hibernate.connection.url">

jdbc:mysql://localhost/hibernatetutorial</property>
      <property name="hibernate.connection.username">root</property>
      <property name="hibernate.connection.password"></property>
      <property name="hibernate.connection.pool_size">10</property>
      <property name="show_sql">true</property>
      <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
      <property name="hibernate.hbm2ddl.auto">update</property>
      <!-- Mapping files -->
      <mapping resource="contact.hbm.xml"/>
</session-factory>
</hibernate-configuration>

this is contact.java file

package roseindia.tutorial.hibernate;

/**
 * @author Deepak Kumar
 *
 * Java Class to map to the datbase Contact Table
 */
public class Contact {
  private String firstName;
  private String lastName;
  private String email;
  private long id;

  /**
   * @return Email
   */
  public String getEmail() {
    return email;
  }

  /**
   * @return First Name
   */
  public String getFirstName() {
    return firstName;
  }

  /** 
   * @return Last name
   */
  public String getLastName() {
    return lastName;
  }

  /**
   * @param string Sets the Email
   */
  public void setEmail(String string) {
    email = string;
  }

  /**
   * @param string Sets the First Name
   */
  public void setFirstName(String string) {
    firstName = string;
  }

  /**
   * @param string sets the Last Name
   */
  public void setLastName(String string) {
    lastName = string;
  }

  /**
   * @return ID Returns ID
   */
  public long getId() {
    return id;
  }

  /**
   * @param l Sets the ID
   */
  public void setId(long l) {
    id = l;
  }

}

this is hbm file

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC 
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping>
  <class name="roseindia.tutorial.hibernate.Contact" table="CONTACT">
   <id name="id" type="long" column="ID" >
   <generator class="assigned"/>
  </id>

  <property name="firstName">
     <column name="FIRSTNAME" />
  </property>
  <property name="lastName">
    <column name="LASTNAME"/>
  </property>
  <property name="email">
    <column name="EMAIL"/>
  </property>
 </class>
</hibernate-mapping>

this is main class

package roseindia.tutorial.hibernate;

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;


/**
 * @author Deepak Kumar
 *
 * http://www.roseindia.net
 * Hibernate example to inset data into Contact table
 */
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.setId(3);
        contact.setFirstName("Deepak");
        contact.setLastName("Kumar");
        contact.setEmail("[e-mail]");
        session.save(contact);
        System.out.println("Done");
    }catch(Exception e){
 e.printStackTrace();
      System.out.println(e.getMessage());
    }finally{
      // Actual contact insertion will happen at this step
      session.flush();
      session.close();

      }
    
  }
}

but when i run this it gives some ERROR

log4j:WARN No appenders could be found for logger (org.hibernate.cfg.Environment).
log4j:WARN Please initialize the log4j system properly.
Inserting Record
Done
Hibernate: insert into CONTACT (FIRSTNAME, LASTNAME, EMAIL, name) values (?, ?, ?, ?)
Exception in thread "main" org.hibernate.exception.GenericJDBCException: Could not execute JDBC batch update
	at org.hibernate.exception.ErrorCodeConverter.handledNonSpecificException(ErrorCodeConverter.java:92)
	at org.hibernate.exception.ErrorCodeConverter.convert(ErrorCodeConverter.java:80)
	at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:43)
	at org.hibernate.jdbc.AbstractBatcher.executeBatch(AbstractBatcher.java:161)
	at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:226)
	at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:136)
	at org.hibernate.event.def.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:274)
	at org.hibernate.event.def.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:27)
	at org.hibernate.impl.SessionImpl.flush(SessionImpl.java:669)
	at roseindia.tutorial.hibernate.FirstExample.main(FirstExample.java:38)
Caused by: java.sql.BatchUpdateException: General error message from server: "Field 'ID' doesn't have a default value"
	at com.mysql.jdbc.PreparedStatement.executeBatch(PreparedStatement.java:1492)
	at org.hibernate.jdbc.BatchingBatcher.doExecuteBatch(BatchingBatcher.java:57)
	at org.hibernate.jdbc.AbstractBatcher.executeBatch(AbstractBatcher.java:154)
	... 6 more

[Paul Clapham]

Here's the important line from the stack trace:

Caused by: java.sql.BatchUpdateException: General error message from server: "Field 'ID' doesn't have a default value"

[anarkali perera]

when i making database i make the ID field as a VARCHAR field.Is it wrong?