休眠一對多連接表 - 重複插入

Question

我確信這個問題之前已經被問過了，但我似乎無法找到答案。我正在使用Hibernate 4.0做一個基本的關係。我有一個Person對象可以包含可能的地址（是的，從另一個網站得到的代碼認爲它會工作，但事實並非如此）。我想要使​​用連接表。我覺得我已經嘗試了所有可能的組合，但不斷提出相同的錯誤：「對於鍵1，重複輸入」1-3「。基本上，當我在設置雙方關係（人員/地址）後嘗試提交事務時發生Hibernate錯誤。很明顯，它試圖插入相同的記錄兩次，這是連接表上不允許的，因爲personId/addressId組合應該只出現一次。任何幫助將不勝感激，一個有效的Junit測試也許。在下面我手動插入一個人和三個地址，前兩個已經與該人有關。

1）我是否必須明確設置雙方的關係（人員和地址）？似乎不正確，因爲其他對象之一會不同步。 2）爲什麼這個工作正常，如果我使用List而不是Set？

單元測試：

@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的堆棧跟蹤

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錯誤：1062 SQLSTATE：23000
2011年12月26日10點四十分27秒PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 錯誤：對於密鑰1重複條目'1-3'

代碼

@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

一個在發展中的關鍵原則是幹：不要重複自己。 JPA將其應用於此，因此，當您有雙向關聯時，您應該只在關聯的一側聲明該關聯的映射一次。另一方面應該說：我使用mappedBy屬性映射爲在另一端聲明。

由於您將雙向關聯映射了兩次，Hibernate認爲您有兩個不同的關聯，因此會生成兩個插入。

下面是如何設定的地址應該聲明：

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

其他說明：

• OO的另一個重要原則是封裝。您的字段應該是私人的
• 在toOne關係中使用級聯= ALL始終是錯誤的：您真的希望在刪除某個地址時刪除某個人嗎？
• 如果關聯是雙向的，爲什麼要使用連接表？爲什麼不直接在地址上使用外鍵？