2014-04-09 45 views
0

我有一個屬性(MetaData中的creationTimestamp),我註釋了加載懶惰的屬性。不,我需要它爲一個實體加載渴望。我試圖與取圖沒有成功。有沒有像我可以用來在查詢中獲取它的JOIN FETCH功能?JPA 2.1渴望獲取屬性

我正在使用Wildfly捆綁休眠。

當我嘗試使用提取圖獲取它時,我得到了一個例外,即hibernate無法找到id字段。該字段位於實體基礎(受保護)中。

@NamedQueries({ @NamedQuery(name = ChatMessage.FIND_BY_CHAT_TIME, query = "SELECT m FROM ChatMessage m " 
     + "WHERE m.chat=:chat " 
     + "AND m.creationTimestamp > :index ORDER BY m.creationTimestamp") }) 
@NamedEntityGraph(
     name = ChatMessage.GRAPH_FETCH_ALL, 
     attributeNodes = {@NamedAttributeNode("id"), @NamedAttributeNode("player"), @NamedAttributeNode("text"), @NamedAttributeNode("creationTimestamp")} 
    ) 
public class ChatMessage extends EntityBase { 

@MappedSuperclass 
public abstract class EntityBase extends MetaData { 
    private static final long serialVersionUID = -5579667581450362176L; 

    public static final String FETCH_GRAPH_HINT = "javax.persistence.fetchgraph"; 

    @Id 
    @Column(length = 36) 
    protected String id = java.util.UUID.randomUUID().toString(); 

@MappedSuperclass 
public class MetaData implements Serializable { 
    private static final long serialVersionUID = 8870539357817188030L; 

    @Version 
    private long version; 

    @Temporal(TemporalType.TIMESTAMP) 
    @Column(nullable = false, updatable = false) 
    @Basic(fetch=FetchType.LAZY) 
    protected Calendar creationTimestamp; 

    @Temporal(TemporalType.TIMESTAMP) 
    @Column(nullable = false) 
    @Basic(fetch=FetchType.LAZY) 
    private Calendar updateTimestamp; 

導致這個例外:

Caused by: java.lang.IllegalArgumentException: Unable to locate Attribute with the the given name [id] on this ManagedType [ChatMessage] 
    at org.hibernate.jpa.internal.metamodel.AbstractManagedType.checkNotNull(AbstractManagedType.java:144) 
    at org.hibernate.jpa.internal.metamodel.AbstractManagedType.getDeclaredAttribute(AbstractManagedType.java:137) 
    at org.hibernate.jpa.graph.internal.EntityGraphImpl.resolveAttribute(EntityGraphImpl.java:139) 
    at org.hibernate.jpa.graph.internal.AbstractGraphNode.buildAttributeNode(AbstractGraphNode.java:119) 
    at org.hibernate.jpa.graph.internal.AbstractGraphNode.addAttribute(AbstractGraphNode.java:114) 
    at org.hibernate.jpa.internal.EntityManagerFactoryImpl.applyNamedAttributeNodes(EntityManagerFactoryImpl.java:279) 
    at org.hibernate.jpa.internal.EntityManagerFactoryImpl.applyNamedEntityGraphs(EntityManagerFactoryImpl.java:266) 
    at org.hibernate.jpa.internal.EntityManagerFactoryImpl.<init>(EntityManagerFactoryImpl.java:173) 
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:865) 
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:843) 
    at org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:399) 
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:842) 
    at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:73) 

有沒有一種方法來獲取像抓取連接的屬性?或者我必須以另一種方式引用抓取圖嗎?

最好的問候, 米

回答

1

貌似子類的屬性必須引用不同的方式:

@NamedEntityGraph(
     name = ChatMessage.GRAPH_FETCH_ALL, 
     attributeNodes = {@NamedAttributeNode("player"), @NamedAttributeNode("text")} 
     , subgraphs={@NamedSubgraph(name="MetaData.subgraph", attributeNodes={@NamedAttributeNode("id"), @NamedAttributeNode("creationTimestamp")})} 
    )