Question

我确定之前已经问过这个问题，但我似乎找不到答案。我正在使用Hibernate 4.0来建立基本关系。我有一个可以包含地址的Person对象（是的，从另一个站点获得代码认为它可以工作，但事实并非如此）。我想使用连接表。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：“重复输入'1-3'为键1”。基本上是在我设置双方关系（Person / Address）之后尝试提交事务时发生的Hibernate错误。显然，它试图在连接表上插入两次相同的记录，因为personId / addressId组合应该只出现一次。任何帮助将不胜感激，可能是一个有效的Junit测试。在下面我同样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）？看起来不对，因为其他一个对象会不同步。 2）如果我使用List而不是Set，为什么这样可以正常工作？

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 Stack Trace

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日下午10:40:27 org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 错误：密钥1的重复条目“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

发展的关键原则之一是DRY：不要重复自己。 JPA应用它，因此当您具有双向关联时，您应该在关联的一侧声明此关联的映射仅一次。另一方应该说：我使用mappedBy属性映射为另一方声明的。

由于您将双向关联映射了两次，因此Hibernate认为您有两个不同的关联，因此会进行两次插入。

以下是如何声明地址集：

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

其他评论：

OO的另一个关键原则是封装。您的字段应该是私有的
在toOne关系中使用cascade = ALL总是错误的：当你的某个地址被删除时，你真的想要删除一个人吗？
如果关联是双向的，为什么要使用连接表？为什么不在地址中使用外键？

使用连接表进行一对多休眠 - 重复插入

1 个答案: