我遇到了Java Persistence API和Hibernate的问题。 我的项目情况是:
我的persistence.xml文件是:
<persistence
xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="JPA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>com.david.Libro</class>
<class>com.david.Categoria</class>
<properties>
<property name="hibernate.show_sql" value="true" />
<property name="javax.persistence.transactionType" value="RESOURCE_LOCAL" />
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost/arquitecturaJava" />
<property name="javax.persistence.jdbc.user" value="root" />
<property name="javax.persistence.jdbc.password" value="root" />
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5Dialect" />
</properties>
</persistence-unit>
</persistence>
我在:
创建了EntityManagerFactoryprivate static EntityManagerFactory buildEntityManagerFactory()
{
try
{
return Persistence.createEntityManagerFactory("JPA");
}
catch (Throwable ex)
{
ex.printStackTrace();
//throw new RuntimeException("Error al crear la factoria de JPA:->"+ ex.getMessage());
}
}
我的错误是关于创建EntityManagerFactory:
javax.persistence.PersistenceException: [PersistenceUnit: JPA] Unable to build EntityManagerFactory
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:924)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:899)
at org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory(HibernatePersistence.java:59)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:63)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:47)
at com.david.JPAHelper.buildEntityManagerFactory(JPAHelper.java:14)
at com.david.JPAHelper.<clinit>(JPAHelper.java:8)
at com.david.Categoria.buscarTodos(Categoria.java:93)
at com.david.FormularioInsertarLibroAccion.ejecutar(FormularioInsertarLibroAccion.java:25)
at com.david.ControladorLibros.doGet(ControladorLibros.java:38)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:621)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:222)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:472)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:171)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:99)
at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:947)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1009)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:589)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:310)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)
Caused by: org.hibernate.DuplicateMappingException: Duplicate class/entity mapping com.david.Libro
at org.hibernate.cfg.Configuration$MappingsImpl.addClass(Configuration.java:2638)
at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:706)
at org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3512)
at org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3466)
at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1355)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1756)
at org.hibernate.ejb.EntityManagerFactoryImpl.<init>(EntityManagerFactoryImpl.java:96)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:914)
... 27 more
Libro和Categoria类代码的一部分是:
@Entity
@Table(name = "Categorias")
public class Categoria implements Serializable
{
private static final long serialVersionUID = 1L;
@Id
@JoinColumn(name = "categoria")
private String id;
private String descripcion;
....
和
@Entity
@Table(name="Libros")
public class Libro implements Serializable
{
private static final long serialVersionUID = 1L;
@Id
private String isbn;
private String titulo;
@ManyToOne
@JoinColumn (name="categoria")
private Categoria categoria;
....
我的Hibernate配置文件是:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC "-//Hibernate/Hibernate Configuration DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost/arquitecturajava</property>
<property name="connection.username">root</property>
<property name="connection.password">root</property>
<property name="connection.pool_size">5</property>
<property name="dialect">org.hibernate.dialect.MySQL5Dialect</property>
<property name="show_sql">true</property>
<mapping class="com.david.Categoria"></mapping>
<mapping class="com.david.Libro"></mapping>
</session-factory>
</hibernate-configuration>
任何想法!! Thank's !!
答案 0 :(得分:13)
在这种情况下,您不需要hibernate.cfg.xml
和persistence.xml
。您是否尝试删除hibernate.cfg.xml
并仅在persistence.xml
中映射所有内容?
但正如另一个答案也指出的那样,这不是这样的:
@Id
@JoinColumn(name = "categoria")
private String id;
您不想使用@Column
吗?
答案 1 :(得分:1)
取消@JoinColumn(name="categoria")
课程的ID字段上的Categoria
,我认为它会有效。
答案 2 :(得分:0)
在pom中添加以下依赖项后,它对我有用,
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-validator</artifactId>
<version>4.3.0.Final</version>
</dependency>
答案 3 :(得分:0)
如果有人面临以下情况,请使用上述注释: - org.hibernate.jpa.HibernatePersistenceProvider持久性提供程序,当它尝试为paymentenginePU持久性单元创建容器实体管理器工厂时。发生以下错误:[PersistenceUnit:paymentenginePU]无法构建Hibernate SessionFactory **如果您使用的是Audit表,这是一个解决方案。@ Audit
使用: - 超类上的@Audited(targetAuditMode = RelationTargetAuditMode.NOT_AUDITED)。