Question

我有一个Student表，其中一个自动生成的id作为主键，一对多映射到Phone表。

My Phone表有一个复合键PhonePK，其中包含电话号码和Student表的外键ID作为主键。

如果我使用phonePk.setStudent它可以工作，但我真的不想这样做。我怎样才能做到这一点？如果它无法实现，我想安排设置学生ID但不是整个学生对象。我该怎么做呢？如果有人能够对此有所了解，我们将非常感激。

student.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;
}

}

Phone.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;
}



}

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;
}



}

Main.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();
    session.save(student);
    transaction.commit();
} catch (HibernateException e) {
    transaction.rollback();
    e.printStackTrace();
} finally {
    session.close();
}

}
}

hibernate.cfg.xml中

<?xml version="1.0" encoding="UTF-8"?>  
<!DOCTYPE hibernate-configuration PUBLIC  
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"  
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">  
<hibernate-configuration>  
<session-factory>  
<property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>  
<property name="hibernate.connection.password">cosmo99</property>  
<property    
name="hibernate.connection.url">jdbc:mysql://localhost:3306/ops_suite</property>  
<property name="hibernate.connection.username">cosmo-security</property>  
<property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>  
<property name="hbm2ddl.auto">create</property>  
 </session-factory>  
</hibernate-configuration>

Hibernate复合键与外键到另一个表 - 如何避免循环引用？

0 个答案: