Question

我想将变量dataSourceString（可能的值：HR，FINANCE：我通过jsp动态获取dataSourceString值）绑定到DataSource。当dataSourceString值为HR然后连接到TESTDS，当dataSourceString值为FINANCE时，根据我要连接到datasource的dataSourceString值连接到TESTDS1.Means。 Enviornment：EJB3，weblogic10.3.3，JPA

注意：还有一件事我不想在sessionbean中编写if-else循环，比如当dataSourceString是HR时，然后连接到此EnityManage，不同的是EntityManager。当前有10-15个dataSourceString的值。我想编写代码，如果将来添加新的dataSourceString，那么只需要更改persistence.xml。

经过研究后我得到了以下代码，但得到了一些错误。

错误： -

No persistence unit named 'em' is available in scope test.jar. Available persistence units: [HR, FINANCE]


    at weblogic.ejb.container.deployer.EJBModule.prepare(EJBModule.java:467)
    at weblogic.application.internal.flow.ModuleListenerInvoker.prepare(ModuleListenerInvoker.java:199)
    at weblogic.application.internal.flow.DeploymentCallbackFlow$1.next(DeploymentCallbackFlow.java:507)
    at weblogic.application.utils.StateMachineDriver.nextState(StateMachineDriver.java:41)
    at weblogic.application.internal.flow.DeploymentCallbackFlow.prepare(DeploymentCallbackFlow.java:149)
    Truncated. see log file for complete stacktrace

      Caused By: java.lang.IllegalArgumentException: No persistence unit named 'em' is available in scope test.jar. Available persistence units: [HR, FINANCE]
    at weblogic.deployment.ModulePersistenceUnitRegistry.getPersistenceUnit(ModulePersistenceUnitRegistry.java:132)
    at weblogic.deployment.BasePersistenceContextProxyImpl.<init>(BasePersistenceContextProxyImpl.java:38)
    at weblogic.deployment.TransactionalEntityManagerProxyImpl.<init>(TransactionalEntityManagerProxyImpl.java:35)
    at weblogic.deployment.BaseEnvironmentBuilder.createPersistenceContextProxy(BaseEnvironmentBuilder.java:974)
    at weblogic.deployment.BaseEnvironmentBuilder.addPersistenceContextRefs(BaseEnvironmentBuilder.java:855)
    Truncated. see log file for complete stacktrace

错误很明显，persistence.xml中没有em的持久性单元但是如何动态地使用jpa实现数据源的查找。以下是我的代码

会话Bean

package entity.library;
import java.util.Collection;
import javax.ejb.Stateless;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import javax.persistence.PersistenceContext;
import javax.persistence.PersistenceUnit;
import java.io.Serializable;
import javax.ejb.*;

@Remote(TestInterface.class)
@Stateless(mappedName="ejb3/TestBeans")

public class TestSessionBean implements Serializable, TestInterface 
{
 protected TestJPA test;
protected Collection <TestJPA> list;

@PersistenceContext
private EntityManager em;
@PersistenceUnit
private EntityManagerFactory emf; 


   public Collection <TestJPA> getAllList(String dataSourceString) {

emf = Persistence.createEntityManagerFactory(dataSourceString);
em = emf.createEntityManager(); 

list=em.createQuery("SELECT test FROM TestJPA test").getResultList();
   return list;
  }
  }

的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="HR" transaction-type="JTA">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <jta-data-source>TESTDS</jta-data-source>
    <non-jta-data-source>TESTDS</non-jta-data-source>
    <properties>
    <property name="eclipselink.target-server" value="WebLogic_10"/>
    <property name="eclipselink.logging.level" value="FINEST"/>
    </properties>
    </persistence-unit>

    <persistence-unit name="FINANCE" transaction-type="JTA">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <jta-data-source>TESTDS1</jta-data-source>
    <non-jta-data-source>TESTDS1</non-jta-data-source>
    <properties>
    <property name="eclipselink.target-server" value="WebLogic_10"/>
    <property name="eclipselink.logging.level" value="FINEST"/>
    </properties>
    </persistence-unit>

    </persistence>

Answer 1

如果在persistence.xml中定义了一个持久性单元，则它是应用程序＆amp;的默认单位。同样是由注释注入。

您可以在运行时手动查找特定的持久性上下文。

 javax.persistence.EntityManager entityManager =
                (javax.persistence.EntityManager)initCtx.lookup(
                    "java:comp/env/" + persistenceContext);

Answer 2

更好......

而不是： @PersistenceContext（名称= “myPU”）
私有EntityManager em;

指定：

@PeristenceContext（的unitName = “myPU”）
私有EntityManager em;

使用JPA进行动态数据源查找

2 个答案: