Integrating JPA instances in the spring framework

Hibernate 3.2 supports JPA annotations, and spring integration JPA is optional.

The following is an example of the spring integration JPA.

Entity class MyUser, using JPA annotations to implement mappings to database table MyUser, as follows:

Package org.shirdrn.entity;
Import Java.util.Date;
Import Javax.persistence.Column;
Import javax.persistence.Entity;
Import Javax.persistence.GeneratedValue;
Import Javax.persistence.GenerationType;
Import Javax.persistence.Id;
Import javax.persistence.Transient;
@Entity
public class MyUser {
Private Long ID;
Private String UserName;
private String password;
Private String gender;
Private Integer age;
Private Integer birthyear;
Private String addr;
Private String Email;
@Id
@GeneratedValue (strategy = generationtype.identity)
Public Long getId () {
return ID;
}
public void SetId (Long id) {
This.id = ID;
}
@Column (name= "UserName")
Public String GetUserName () {
return userName;
}
public void Setusername (String userName) {
This.username = UserName;
}
@Column (name= "password")
Public String GetPassword () {
return password;
}
public void SetPassword (String password) {
This.password = password;
}
@Column (name= "gender")
Public String Getgender () {
return gender;
}
public void Setgender (String gender) {
This.gender = gender;
}
@Column (name= "age")
Public Integer Getage () {
return age;
}
public void Setage (Integer age) {
This.age = age;
}
@Column (name= "addr")
Public String getaddr () {
return addr;
}
public void setaddr (String addr) {
this.addr = addr;
}
@Column (name= "email")
Public String Getemail () {
return email;
}
public void Setemail (String email) {
This.email = email;
}
@Transient
Public Integer Getbirthyear () {
return new Integer (2008-age);
}
public void Setbirthyear (Integer age) {
This.birthyear = new Integer (2008-age);
}
}