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eclipse - JPA 与 hibernate 错误 : [PersistenceUnit: JPA] Unable to build EntityManagerFactory

转载 作者:行者123 更新时间:2023-12-02 09:28:00 25 4
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我在使用 Java Persistence API 和 Hibernate 时遇到问题。我的项目情况是:

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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>

我在以下位置创建 EntityManagerFactory:

private 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 Category 类的部分代码是:

@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;

....

@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!谢谢!!

最佳答案

在这种情况下,您不需要同时使用 hibernate.cfg.xmlpersistence.xml。您是否尝试过删除 hibernate.cfg.xml 并仅映射 persistence.xml 中的所有内容?

但正如其他答案也指出的那样,这样是不行的:

@Id
@JoinColumn(name = "categoria")
private String id;

您不想使用@Column吗?

关于eclipse - JPA 与 hibernate 错误 : [PersistenceUnit: JPA] Unable to build EntityManagerFactory,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16922314/

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