3

我有一个学生表，其中自动生成的 id 作为主键，并且一对多映射到电话表。

My Phone 表有一个复合主键 PhonePK，其中包含电话号码和 Student 表的外键 id。

如果我只做 student.setPhones 而不是 phonepk.setStudent，它对id 的抱怨不能是 null。所以我正在设置 student.setPhones 和 phonePk.setStudent。但是现在我在 toString 上遇到了 stackoverflow 错误。

我真的不喜欢一开始就在两种方式上都设置它，但不知道如何绕过 id cannot be null 错误。我一直在问很多人，但他们无能为力。有人可以看看吗？

学生.java

import java.io.Serializable;
import java.util.Set;

import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToMany;

@Entity
@SuppressWarnings("serial")
public class Student implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;

private String fName;

private String lName;

private String mName;

@OneToMany(cascade = CascadeType.ALL)
@JoinColumn(name = "id")
private Set<Phone> phones;

/**
 * @return the fName
 */
public String getfName() {
    return fName;
}

/**
 * @return the id
 */
public int getId() {
    return id;
}

/**
 * @return the lName
 */
public String getlName() {
    return lName;
}

/**
 * @return the mName
 */
public String getmName() {
    return mName;
}

/**
 * @return the phones
 */
public Set<Phone> getPhones() {
    return phones;
}

/**
 * @param fName
 *            the fName to set
 */
public void setfName(final String fName) {
    this.fName = fName;
}

/**
 * @param id
 *            the id to set
 */
public void setId(final int id) {
    this.id = id;
}

/**
 * @param lName
 *            the lName to set
 */
public void setlName(final String lName) {
    this.lName = lName;
}

/**
 * @param mName
 *            the mName to set
 */
public void setmName(final String mName) {
    this.mName = mName;
}

/**
 * @param phones
 *            the phones to set
 */
public void setPhones(final Set<Phone> phones) {
    this.phones = phones;
}

/**
 * {@inheritDoc}
 */
@Override
public String toString() {
    return String.format("Student [id=%s, fname=%s, lname=%s, mname=%s, phones=%s]",      
id,
        fName, lName, mName, phones);
}

}

电话.java

import java.io.Serializable;

import javax.persistence.EmbeddedId;
import javax.persistence.Entity;

@Entity
@SuppressWarnings("serial")
public class Phone implements Serializable {

@EmbeddedId
private PhonePK PK;

private String color;

/**
 * @return the color
 */
public String getColor() {
    return color;
}

public PhonePK getPK() {
    return PK;
}

/**
 * @param color
 *            the color to set
 */
public void setColor(final String color) {
    this.color = color;
}

public void setPK(final PhonePK pK) {
    PK = pK;
}

/**
 * {@inheritDoc}
 */
@Override
public String toString() {
    return String.format("Phone [PK=%s, color=%s]", PK, color);
}

}

PhonePK.java

import java.io.Serializable;

import javax.persistence.Embeddable;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;

@Embeddable
@SuppressWarnings({ "serial" })
public class PhonePK implements Serializable {

@ManyToOne
@JoinColumn(name = "id", insertable = false, updatable = false)
private Student student;

private String phoneNumber;

public String getPhoneNumber() {
    return phoneNumber;
}

public Student getStudent() {
    return student;
}

public void setPhoneNumber(final String phoneNumber) {
    this.phoneNumber = phoneNumber;
}

public void setStudent(final Student student) {
    this.student = student;
}

/**
 * {@inheritDoc}
 */
@Override
public String toString() {
    return String.format("PhonePK [student=%s, phoneNumber=%s]", student, phoneNumber);
}

}

主.java

import java.util.LinkedHashSet;
import java.util.Set;

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

public class Main {

 public static void main(final String args[]) {

    Configuration configuration = new Configuration();
    Transaction transaction = null;

    configuration.addAnnotatedClass(Student.class);
    configuration.addAnnotatedClass(Phone.class);
    configuration.configure("hibernate.cfg.xml");
    SessionFactory sessionFactory = configuration.buildSessionFactory();
    Session session = sessionFactory.openSession();

    Student student = new Student();
    student.setfName("Bob");
    student.setlName("Buster");

    Set<Phone> phones = new LinkedHashSet<Phone>();
    Phone phone = new Phone();
    phone.setColor("Black");
    PhonePK phonePK = new PhonePK();
    phonePK.setPhoneNumber("1111111111");
    phonePK.setStudent(student); // Do not do this? But won't work (id cannot be null  
    error) if
                                 // commented out??
    phone.setPK(phonePK);
    phones.add(phone);

    student.setPhones(phones);

    try {
        transaction = session.beginTransaction();
        System.out.println(student.toString()); // stackoverflow error!
        session.save(student);
        transaction.commit();
    } catch (HibernateException e) {
        transaction.rollback();
        e.printStackTrace();
    } finally {
        session.close();
    }

}
}
4

1 回答 1

2

发生这种情况是因为您定义 toString() 方法的方式

学生toString()正在调用电话toString()，电话正在调用电话PK ，而电话PKtoString()又在调用学生toString()……导致无限循环。


让我们详细看看它是如何发生的

在 Student中toString()，因为其中的phones实例变量。它将遍历每个电话并调用 PhonetoString()

public String toString() {
    return String.format("Student [id=%s, fname=%s, lname=%s, mname=%s, phones=%s]",      
id,
        fName, lName, mName, phones);
}



在电话中toString()，因为其中的PK实例变量。它将调用 PhonePKtoString() 

public String toString() {
    return String.format("Phone [PK=%s, color=%s]", PK, color);
}



在PhonePK中toString()，因为其中的phoneNumber实例变量。它将调用PhonetoString() 

public String toString() {
    return String.format("PhonePK [student=%s, phoneNumber=%s]", student, phoneNumber);
}
于 2013-07-02T17:36:44.217 回答