JPA一對多集合返回空

做一個以前保存的1對多的實體的查找。授權是：JPA一對多集合返回空

Book.java：

@Entity 
@Table(name = "Book") 
public class Book { 

    @Id 
    @GeneratedValue 
    @Column(name = "id") 
    private Long id; 

    @OneToMany(fetch=FetchType.LAZY, cascade=CascadeType.ALL, mappedBy="book", orphanRemoval=true) 
    private Collection<Author> authors; 

    public Collection<Author> getAuthors() { 
     return authors; 
    } 
     ... 

     @Override 
    public String toString() { 
     return "Book [id=" + id + ", authors=" + authors + ", title=" + title 
       + ", isbn=" + isbn + "]"; 
    } 
}

Author.java：

@Entity 
@Table(name = "Authors") 
public class Author { 

    @Id 
    @GeneratedValue 
    @Column(name="id") 
    private Long id; 

    @ManyToOne(fetch=FetchType.LAZY) 
    @JoinColumn(name="book_id") 
    private Book book; 

     ... 

    @Override 
    public String toString() { 
     return "Author [id=" + id + ", book=" + book + ", firstName=" 
      + firstName + ", lastName=" + lastName + "]"; 
    } 
}

忽略數據模型的可疑基數，查找碼是

long bookID = 10; // saved entity ID 
entityManager.getTransaction().begin(); 
Book savedBook = entityManager.find(Book.class, bookID); 
System.out.println("Book " + savedBook); 
entityManager.getTransaction().commit();

其給出輸出

Book Book [id=10, authors=[], title=Treasure Island, isbn=123456]

任何想法，爲什麼作者是空的？

Retrieving 11... 

Exception in thread "main" java.lang.StackOverflowError 
    at java.lang.Long.toString(Long.java:240) 
    at java.lang.Long.toString(Long.java:100) 
    at java.lang.String.valueOf(String.java:2946) 
    at java.lang.Long.toString(Long.java:733) 
    at java.lang.String.valueOf(String.java:2827) 
    at java.lang.StringBuilder.append(StringBuilder.java:115) 
    at org.hibstore.domain.Book.toString(Book.java:64) 
    at java.lang.String.valueOf(String.java:2827) 
    at java.lang.StringBuilder.append(StringBuilder.java:115) 
    at org.hibstore.domain.Author.toString(Author.java:67) 
    at java.lang.String.valueOf(String.java:2827) 
    at java.lang.StringBuilder.append(StringBuilder.java:115) 
    at java.util.AbstractCollection.toString(AbstractCollection.java:422) 
    at org.hibernate.collection.internal.PersistentBag.toString(PersistentBag.java:501) 
    at java.lang.String.valueOf(String.java:2827) 
    at java.lang.StringBuilder.append(StringBuilder.java:115) 
    at org.hibstore.domain.Book.toString(Book.java:64) 
    at java.lang.String.valueOf(String.java:2827)

來源

2014-01-15 user1882491

您可能需要顯式調用'getAuthors（）'在本書中，你嘗試打印之前。 –

試過了，沒有工作:( – user1882491

見JPA生成的查詢，並確保兒童查詢被解僱，你在DB有子記錄。希望你已經嘗試了這種選擇，但一切在這裏看起來很不錯。 –