Hibernate:OneToMany 映射失败

Question

我是 Hibernate 新手，正在尝试一些应该很容易的事情，但我无法让它工作。

有两个表，一个人员和一个地址。一个人可以有一个或多个地址，即:一对多映射。当我尝试将相同的地址添加给两个不同的人时，出现异常。这几乎就像在“join”表内的 foriegn_key 上强制执行 Unique 一样。

我的源代码:

人

package testing.com.hibernate.entities;

import java.util.HashSet;
import java.util.Set;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.OneToMany;
import javax.persistence.Table;

import org.hibernate.annotations.GenericGenerator;

@Entity
@Table(name="Person")
public class Person {
private long personID; 
private String personName;
private Set<Address> addresses = new HashSet<Address>(0);

public Person() {}

public Person(String personName, Set<Address> addresses) {
    setPersonName(personName);
    setAddresses(addresses);
}

@Id
@GeneratedValue(generator="increment")
@GenericGenerator(name="increment", strategy="increment")
@Column(name="Person_ID")
public long getPersonID() {
    return personID;
}
@Column(name="Person_Name")
public String getPersonName() {
    return personName;
}
@OneToMany
@JoinTable(name = "PersonAddresses", joinColumns = {@JoinColumn(name="PersonID", unique=false)}, inverseJoinColumns = {@JoinColumn(name="AddressID", unique=false)})
public Set<Address> getAddresses() {
    return addresses;
}

public void setPersonID(long personID) {
    this.personID = personID;
}
public void setPersonName(String personName) {
    this.personName = personName;
}
public void setAddresses(Set<Address> addresses) {
    this.addresses = addresses;
}
}

地址

package testing.com.hibernate.entities;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;

import org.hibernate.annotations.GenericGenerator;

@Entity
@Table(name="Address")
public class Address {
private long addressID;
private String address; 

public Address() {}

public Address(String address) {
    setAddress(address);
}

@Id
@GeneratedValue(generator="increment")
@GenericGenerator(name="increment", strategy="increment")
@Column(name="Address_ID")
public long getAddressID() {
    return addressID;
}
@Column(name="Address")
public String getAddress() {
    return address;
}

public void setAddressID(long addressID) {
    this.addressID = addressID;
}
public void setAddress(String address) {
    this.address = address;
}
}

主要

package testing.com.hibernate.tests;

import java.util.Date;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

import testing.com.hibernate.entities.Address;
import testing.com.hibernate.entities.EntityCategory;
import testing.com.hibernate.entities.EntityNetflixFilm;
import testing.com.hibernate.entities.Person;
import testing.com.hibernate.sessionmanager.HibernateUtil;

import org.hibernate.Session;

public class Test {

public static void main(String[] args) {
    Session session = HibernateUtil.getSessionFactory().openSession();
    // Create multiple addresses
    Address addr1 = new Address("Address one");
    Address addr2 = new Address("Address two");
    Address addr3 = new Address("Address three");
    Address addr4 = new Address("Address four");

    // Add all addresses to the database
    session.beginTransaction();
    session.save(addr1);
    session.save(addr2);
    session.save(addr3);
    session.save(addr4);
    session.getTransaction().commit();
    session.close();

    /*
     * At this point in the code there will be four addresses 
     * in the table 'Address'. 
     * 
     * 1 - Address one
     * 2 - Address two
     * 3 - Address three
     * 4 - Address four
     * 
     * Now we want to create some people and add addresses to them
     */

    // Pull out a list of addresses from the database
    List<Address> addresses = null;
    session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();
    addresses = session.createQuery("from Address").list();
    session.getTransaction().commit();
    session.close();

    /*
     * We now have a List of Address objects.
     * 
     * This is replicating a Database with existing
     * Addresses and querying them from the table
     * to attach to a new Person being added. 
     */

    // Create person with Address 1 and 2
    Set<Address> addressSet1 = new HashSet<Address>();
    addressSet1.add(addresses.get(0));
    addressSet1.add(addresses.get(1));
    Person george = new Person("George", addressSet1);

    // Create person with Address 3 and 4
    Set<Address> addressSet2 = new HashSet<Address>();
    addressSet2.add(addresses.get(2));
    addressSet2.add(addresses.get(3));
    Person robert = new Person("Robert", addressSet2);

    // Create person with Address 1 and 2. The same as George
    Person harry = new Person("Harry", addressSet1);

    // Attempt to add them to the Person table
    session = HibernateUtil.getSessionFactory().openSession();
    session.beginTransaction();
    session.save(george);
    session.save(robert);
    session.save(harry);
    session.getTransaction().commit();
    session.close();
}
}

代码在保存 Person 对象“Harry”时崩溃。这是因为在 PersonAddress 表中，带有引用“1”的地址已经存在并分配给“George”。我怎样才能将“Harry”和“George”映射到地址“1”？

日志

Hibernate:插入到 Person (Person_Name, Person_ID) 值 (?, ?) 22:39:42,284 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [VARCHAR] - George 22:39:42,284 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 1 Hibernate:插入到Person(Person_Name，Person_ID)值(？，？) 22:39:42,285 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [VARCHAR] - 罗伯特 22:39:42,286 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 2 Hibernate:插入到Person(Person_Name，Person_ID)值(？，？) 22:39:42,287 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [VARCHAR] - Harry 22:39:42,287 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 3

Hibernate:插入PersonAddresses(PersonID，AddressID)值(？，？) 22:39:42,289 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [BIGINT] - 1 22:39:42,292 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 1

Hibernate:插入PersonAddresses(PersonID，AddressID)值(？，？) 22:39:42,294 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [BIGINT] - 1 22:39:42,295 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 2

Hibernate:插入PersonAddresses(PersonID，AddressID)值(？，？) 22:39:42,296 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [BIGINT] - 2 22:39:42,296 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 3

Hibernate:插入PersonAddresses(PersonID，AddressID)值(？，？) 22:39:42,297 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [BIGINT] - 2 22:39:42,298 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 4

Hibernate:插入PersonAddresses(PersonID，AddressID)值(？，？) 22:39:42,298 TRACE BasicBinder:83 - 绑定(bind)参数 [1] 作为 [BIGINT] - 3 22:39:42,300 TRACE BasicBinder:83 - 绑定(bind)参数 [2] 作为 [BIGINT] - 1

22:39:42,320 警告 SqlExceptionHelper:143 - SQL 错误:1062，SQLState:23000 22:39:42,320 错误 SqlExceptionHelper:144 - 键“AddressID”重复条目“1”

线程“main”org.hibernate.exception.ConstraintViolationException中的异常:键“AddressID”的重复条目“1”

Answer 1

好的。所以我自己解决了这个问题。

事实证明，Hibernate 将一对多的“多”部分视为连接表中的唯一实体。一个例子是汽车/服务历史记录情况，您有一辆记录其独特服务历史记录的汽车。如果您尝试再次将服务历史记录添加到另一辆汽车，则会失败，因为此服务历史记录对于汽车来说已经是唯一的。

如果您想共享信息，在本例中是一个可以在多个人之间共享的地址，您需要多对多关系。要修复此问题，请更新源代码，以便 Person 具有 @ManyToMany 并提供 @JoinTable 注释。