Java持久性问题 [英] Java Persistence Issue

问题描述

我试图通过GlassFish在EJB中使用JPA创建并运行一个简单的示例。我有以下 persistence.xml

 < persistence version = 1.0xmlns =http://java.sun.com/xml/ns/persistencexmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexsi：schemaLocation =http：/ /java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd\"> 
< persistence-unit name =defaulttransaction-type =RESOURCE_LOCAL> 
< provider> org.eclipse.persistence.jpa.PersistenceProvider< / provider> 
< jta-data-source> jdbc / wms< / jta-data-source> 
< class> com.xxx.xxx.datamodel.MyTest< / class> 
< exclude-unlisted-classes /> 
<属性> 
< property name =eclipselink.target-servervalue =SunAS9/> 
< property name =eclipselink.logging.levelvalue =FINEST/> 
< property name =eclipselink.target-databasevalue =Oracle/> 
 
< property name =eclipselink.jdbc.drivervalue =oracle.jdbc.OracleDriver/> 
< property name =eclipselink.jdbc.urlvalue =[dbconnectionstring]/> 
< property name =eclipselink.jdbc.uservalue =user/> 
< property name =eclipselink.jdbc.passwordvalue =password/> 
< / properties> 
< / persistence-unit> 
< /余辉> 
  

一个简单的实体：

  package com.xxx.xxx.datamodel; 
 
 import javax.persistence.Entity; 
 import javax.persistence.Id; 
 import javax.persistence.Table; 
 
 @Entity 
 @Table（name =move_task）
 public class MyTest {
 
 private int key; 
私有字符串描述; 
 
 @Id 
 public int getKey（）{
 return key; 
} 
 
 public void setKey（int key）{
 this.key = key; 
} 
 
 public String getDescription（）{
 return this.description; 
} 
 
 public void setDescription（String description）{
 this.description = description; 
} 
 
 @Override 
 public String toString（）{
 returnKey：+ key +Description：+ description; 
} 
 
} 
  

最后下面的代码尝试使用上面的代码：

  private void jpaCall（）{
 try {
 emf = Persistence.createEntityManagerFactory 默认）; 
 em = emf.createEntityManager（）; 
 log.info（JPA init complete）; 
 
最终列表< MyTest> list = em.createQuery（从MyTest p中选择p）。getResultList（）; （MyTest current：list）
 
 {
 final String description = current.getDescription（）; 
 log.info（JPA：Desc：+ description）; 
 
 
 $ b catch（Exception e）{
 log.error（JPA错误，e）; 
} 
 
} 
  

当它作为我的EJB初始化出现以下错误：

$ p $ java.lang.IllegalArgumentException：在EntityManager中创建查询时发生异常：
异常说明：编译查询时出错[从MyTest p中选择p]。未知实体类型[MyTest]。
...
导致：异常[EclipseLink-8034]（Eclipse持久性服务 - 2.0.0.v20091127-r5931）：org.eclipse.persistence.exceptions.JPQLException
异常说明：编译查询时出错[从MyTest p选择p]。未知实体类型[MyTest]。


我不确定我做错了什么，并希望得到任何帮助。

干杯，

James

解决方案

<因此，正如上面的评论表明，这似乎是一个问题与eclipse插入glassfish。手动部署耳朵时，我没有任何问题。

感谢所有人的帮助。

p>

I am trying to get a simple example up and running using JPA in an EJB through GlassFish. I have the following persistence.xml

<persistence version="1.0" 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_1_0.xsd">
  <persistence-unit name="default" transaction-type="RESOURCE_LOCAL">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <jta-data-source>jdbc/wms</jta-data-source>
    <class>com.xxx.xxx.datamodel.MyTest</class>
    <exclude-unlisted-classes /> 
    <properties>
      <property name="eclipselink.target-server" value="SunAS9"/>
      <property name="eclipselink.logging.level" value="FINEST"/>
      <property name="eclipselink.target-database" value="Oracle"/>

      <property name="eclipselink.jdbc.driver" value="oracle.jdbc.OracleDriver" />  
        <property name="eclipselink.jdbc.url" value="[dbconnectionstring]" />
        <property name="eclipselink.jdbc.user" value="user" />  
        <property name="eclipselink.jdbc.password" value="password" />  
    </properties>
  </persistence-unit>
</persistence>

A simple entity:

package com.xxx.xxx.datamodel;

import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "move_task")
public class MyTest {

    private int key;
    private String description;

    @Id
    public int getKey(){
        return key;
    }

    public void setKey(int key){
        this.key = key;
    }

    public String getDescription(){
        return this.description;
    }

    public void setDescription(String description){
        this.description = description;
    }

    @Override
    public String toString(){
        return "Key: " + key + " Description: " + description;
    }

}

And finally the following code to try use the above:

private void jpaCall() {
        try{
            emf = Persistence.createEntityManagerFactory("default");
            em = emf.createEntityManager();
            log.info("JPA init complete");

            final List<MyTest> list = em.createQuery("select p from MyTest p").getResultList();

            for (MyTest current : list) {
                final String description = current.getDescription();
                log.info("JPA: Desc: " + description);
            }

        }
        catch(Exception e){
            log.error("Error on JPA", e);
        }

    }

When this is run as part of my EJB initialisation I get the following error:

java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Error compiling the query [select p from MyTest p]. Unknown entity type [MyTest].
...
Caused by: Exception [EclipseLink-8034] (Eclipse Persistence Services - 2.0.0.v20091127-r5931): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Error compiling the query [select p from MyTest p]. Unknown entity type [MyTest].

I am not sure what I am doing wrong and would appreciate any help.

Cheers,

James

解决方案

So as the comments above suggest this seems to be an issue with the eclipse plug in for glassfish. I am having no problems when deploying the ear manually.

Thanks all for the help.

James

这篇关于Java持久性问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！