Произошло PersistenceException: org.hibernate.Exception.ConstraintViolationException: невозможно вставить: [models.User] ⇐ JAVA

[Anonymous]

У меня есть это исключение!
вот мой класс модели

@Entity
public class User extends Model {

@OneToMany(mappedBy="email", cascade=CascadeType.ALL)
public List
photo;

@Email
@Required
@Unique
public String email;

@Required
public String passwordHash;

@Required
public String education;

@Required
public String fname;

@Required
public String lname;

@Required
public Date dob;

@Required
public String gender;

@Required
public String country;

@Required
public Long phone;

@Required
public String status;

@Required
public String jobtitle;

@Required
public String company;

@Required
public String industry;

@Required
public Date addDate;

public String needConfirmation;

public User(String email, String password, String fname,
String lname, Date dob, String gender, String country,
Long phone, String status, String education, String jobtitle, String company, String industry) {
//all initialization here
}
}


Можете ли вы сказать мне, где я ошибаюсь?
playframework× 2289



this is my photo class, and please tell me how can I allow JPA to generate my database

/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package models;

import controllers.Users;
import javax.persistence.Entity;
import javax.persistence.ManyToOne;
import play.db.jpa.Model;

/**
*
* @author nPandey
*/
@Entity
public class Photo extends Model{

public String photoname;

@ManyToOne
public User email;

public String owner;

public Photo(){

}

public Photo(User email, String photo){
this.photoname=photo;
this.email=email;
this.owner=email.email;
}

}


Подробнее здесь: https://stackoverflow.com/questions/106 ... tionexcept