Hibernate保存数据的顺序

Question

我对hibernate很新。我的问题是为什么Hibernate没有按照我想要的顺序保存数据。

我想存储数据的方式是..... 姓名，年龄，地址，联系方式

虽然hibernate将其保存为..... 地址，年龄，联系人，姓名

实体

@Entity
public class PersonDetails implements Serializable{

   private static final long serialVersionUID = 1L;
   @Id
   @GeneratedValue(strategy=GenerationType.IDENTITY)
   private int personId;
   private String personName;
   private int personAge;
   private String personAddress;
   private String personContactNumber;


   public int getPersonId() {
      return personId;
   }
   public void setPersonId(int personId) {
      this.personId = personId;
   }
   public String getPersonName() {
      return personName;
   }
   public void setPersonName(String personName) {
      this.personName = personName;
   }
   public int getPersonAge() {
      return personAge;
   }
   public void setPersonAge(int personAge) {
      this.personAge = personAge;
   }
   public String getPersonAddress() {
      return personAddress;
   }
   public void setPersonAddress(String personAddress) {
      this.personAddress = personAddress;
   }
   public String getPersonContactNumber() {
      return personContactNumber;
   }
   public void setPersonContactNumber(String personContactNumber) {
      this.personContactNumber = personContactNumber;
   }

}

拯救人员Dao实施

public class PersonDatabaseDaoImpl implements PersonDatabaseDao{

private SessionFactory sessionFactory=new Configuration().configure().buildSessionFactory();
private Session session=sessionFactory.getCurrentSession();
@Override
public void addPerson(PersonDetails person) {
    session.beginTransaction();
    session.save(person);
    session.getTransaction().commit();
}

从人那里获取信息，这是在主

中写的

    System.out.print("Enter Name = ");
    person.setPersonName(input.next());

    System.out.print("\n Enter Age = ");
    person.setPersonAge(input.nextInt());

    System.out.print("\n Enter Address = ");
    person.setPersonAddress(input.next());

    System.out.print("\n Enter Contact = ");
    person.setPersonContactNumber(input.next());

    personData.addPerson(person);