org.hibernate.exception.SQLGrammarException:无法在线程" main"中执行语句异常。 javax.persistence.PersistenceException

时间:2017-08-13 20:17:49

标签: mysql hibernate nhibernate-mapping hibernate-mapping

package Org.Ajay.hibernate;

import org.hibernate.Session;

import org.hibernate.SessionFactory;

import org.hibernate.cfg.Configuration;

import AjayBraindto.UserDetails;

public class Hibernatetest {

public static void main(String[] args) {

    // TODO Auto-generated method stub
    UserDetails userdetail=new UserDetails();

    userdetail.setUserid(5);

    userdetail.setUsername("Ajay pal");

    SessionFactory sessionfactory=new 

    Configuration().configure().buildSessionFactory();

    Session session=sessionfactory.openSession();

    session.beginTransaction();

    session.save(userdetail);

    session.getTransaction().commit();
}
}

<session-factory>

<property name="hibernate.connection.username">root</property>

<property name="hibernate.connection.password">mcatopper</property>

<property name="show_sql">true</property>

<property name="hibernate.current_session_context_class">thread</property>


<!-- Drop and re create schema on start up -->
<property name="hbm2dll.auto">create</property>

<mapping class="AjayBraindto.UserDetails"/>

@Entity
@Table(name="UserDetails")

public class UserDetails {

@Id
private int Userid;

private String Username;


public int getUserid() {
    return Userid;
}

public void setUserid(int userid) {
    Userid = userid;
}

public String getUsername() {
    return Username;
}

public void setUsername(String username) {
    Username = username;
}



}

0 个答案:

没有答案