休眠一对多连接表 - 重复插入

Question

我确信这个问题之前已经被问过了，但我似乎无法找到答案。我正在使用Hibernate 4.0做一个基本的关系。我有一个Person对象可以包含可能的地址（是的，从另一个网站得到的代码认为它会工作，但事实并非如此）。我想要使​​用连接表。我觉得我已经尝试了所有可能的组合，但不断提出相同的错误：“对于键1，重复输入”1-3“。基本上，当我在设置双方关系（人员/地址）后尝试提交事务时发生Hibernate错误。很明显，它试图插入相同的记录两次，这是连接表上不允许的，因为personId/addressId组合应该只出现一次。任何帮助将不胜感激，一个有效的Junit测试也许。在下面我手动插入一个人和三个地址，前两个已经与该人有关。

1）我是否必须明确设置双方的关系（人员和地址）？似乎不正确，因为其他对象之一会不同步。 2）为什么这个工作正常，如果我使用List而不是Set？

单元测试：

@Test 
public void tryAgain(){ 
Person p = em.find(Person.class, 1);<br> 
Address a = em.find(Address.class, 1);<br> 
Address b= em.find(Address.class,2);<br> 
Address c= em.find(Address.class, 3);<br> 
em.getTransaction().begin();<br> 
p.addresses.add(c);<br> 
c.person=p;<br> 
em.getTransaction().commit();<br> 
assertTrue(p.addresses.size()==3);<br> 
} 

JUnit的堆栈跟踪

Hibernate: select person0_.personId as personId3_0_ from PERSON person0_ where person0_.personId=? 
<br>Hibernate: select address0_.addressId as addressId6_1_, address0_1_.personId as personId7_1_, person1_.personId as personId3_0_ from ADDRESS address0_ left outer join PersonAddress address0_1_ on address0_.addressId=address0_1_.addressId left outer join PERSON person1_ on address0_1_.personId=person1_.personId where address0_.addressId=? 
<br>Hibernate: select address0_.addressId as addressId6_1_, address0_1_.personId as personId7_1_, person1_.personId as personId3_0_ from ADDRESS address0_ left outer join PersonAddress address0_1_ on address0_.addressId=address0_1_.addressId left outer join PERSON person1_ on address0_1_.personId=person1_.personId where address0_.addressId=? 
<br>Hibernate: select address0_.addressId as addressId6_1_, address0_1_.personId as personId7_1_, person1_.personId as personId3_0_ from ADDRESS address0_ left outer join PersonAddress address0_1_ on address0_.addressId=address0_1_.addressId left outer join PERSON person1_ on address0_1_.personId=person1_.personId where address0_.addressId=? 
<br>Hibernate: select addresses0_.personId as personId3_2_, addresses0_.addressId as addressId2_, address1_.addressId as addressId6_0_, address1_1_.personId as personId7_0_, person2_.personId as personId3_1_ from PersonAddress addresses0_ inner join ADDRESS address1_ on addresses0_.addressId=address1_.addressId left outer join PersonAddress address1_1_ on address1_.addressId=address1_1_.addressId left outer join PERSON person2_ on address1_1_.personId=person2_.personId where addresses0_.personId=? 
<b>Hibernate: insert into PersonAddress (personId, addressId) values (?, ?) 
<br>Hibernate: insert into PersonAddress (personId, addressId) values (?, ?)</b> 
<br>Dec 26, 2011 10:40:27 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 

警告：SQL错误：1062 SQLSTATE：23000
2011年12月26日10点四十分27秒PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 错误：对于密钥1重复条目'1-3'

代码

@Entity<br> 
@Table(name="PERSON")<br> 
public class Person {<br> 
@Id<br> 
@GeneratedValue(strategy = GenerationType.AUTO)<br> 
@Column(name = "personId")<br> 
public int id;<br> 
@OneToMany()<br> 
@JoinTable(name = "PersonAddress", 
joinColumns = { 
@JoinColumn(name="personId", unique = true)}, 
inverseJoinColumns = { 
@JoinColumn(name="addressId")})<br> 
public Set<Address> addresses = new HashSet<Address>();<br> 
} 


@Entity 
@Table(name = "ADDRESS") 
public class Address { 
@Id 
@GeneratedValue(strategy = GenerationType.AUTO) 
@Column(name = "addressId") 
public int id; 

@ManyToOne(optional=true,cascade={CascadeType.ALL}) 
@JoinTable(name = "PersonAddress", 
joinColumns = {@JoinColumn(name="addressId",insertable=false,updatable=true)}, 
inverseJoinColumns = { 
@JoinColumn(name="personId") 
}) 
public Person person; 
} 

Answer 1

一个在发展中的关键原则是干：不要重复自己。 JPA将其应用于此，因此，当您有双向关联时，您应该只在关联的一侧声明该关联的映射一次。另一方面应该说：我使用mappedBy属性映射为在另一端声明。

由于您将双向关联映射了两次，Hibernate认为您有两个不同的关联，因此会生成两个插入。

下面是如何设定的地址应该声明：

@OneToMany(mappedBy = "person") 
private Set<Address> addresses = new HashSet<Address>(); 

其他说明：

• OO的另一个重要原则是封装。您的字段应该是私人的
• 在toOne关系中使用级联= ALL始终是错误的：您真的希望在删除某个地址时删除某个人吗？
• 如果关联是双向的，为什么要使用连接表？为什么不直接在地址上使用外键？