覆盖JPA持久性单元

Question

我的jar文件中有一个空的persistenceUnit：

 <persistence-unit transaction-type="JTA" name="base1">
    </persistence-unit>

    <persistence-unit transaction-type="JTA" name="base2">
    </persistence-unit>

我的想法是在我的主项目中用带有属性和类的完整persistenceUnit替换空persistenceUnit，如下所示：

<persistence-unit name="base1" transaction-type="JTA">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <jta-data-source>java:jboss/datasources/myDS</jta-data-source>
        <class>br.com.myproject.MyClass</class>
        <exclude-unlisted-classes>true</exclude-unlisted-classes>
        <shared-cache-mode>NONE</shared-cache-mode>
        <properties>
            <property name="hibernate.hbm2ddl.auto" value="none" />
            <property name="hibernate.show_sql" value="true" />
            <property name="hibernate.format_sql" value="false" />
            <property name="hibernate.cache.use_second_level_cache"
                      value="false" />
        </properties>
    </persistence-unit>

但是当我尝试启动服务器时，我得到了以下错误：

 Caused by: org.jboss.as.server.deployment.DeploymentUnitProcessingException: WFLYJPA0038: Falha ao adicionar o serviço da unidade de persistência para base1
    Caused by: org.jboss.msc.service.DuplicateServiceException: Service jboss.persistenceunit.myproject#base1.__FIRST_PHASE__ is already registered"}}

有什么方法可以覆盖persistenceUnit吗？

Answer 1

如果您确实需要动态覆盖您的persistence.xml，最好在构建过程中完成。

我个人的警告:这听起来像是一种配置--对我来说是地狱，我建议在这里使用容器管理的JNDI方法。

但不管怎样:使用2个maven配置文件。如果您激活了profile1，那么来自profile1的persistence.xml将被添加到正确的位置。如果你激活了profile2，来自profile2的persistence.xml将被占用。

因此，对maven使用copy-resources-mojo。https://maven.apache.org/plugins/maven-resources-plugin/examples/copy-resources.html

如果只是参数的值改变了，而不是整个结构改变了，那么你也可以在maven过程中只“过滤”和替换字符串，然后你可以在概要文件中定义属性。https://maven.apache.org/plugins/maven-resources-plugin/examples/filter.html

您还可以将基本persistence.xml作为默认文件添加到项目中。因此，如果没有激活maven-profile，将使用此配置文件。(即使可能发生这种情况，如果数据资源配置不正确，应用程序也不会按预期工作)

Answer 2

Spring提供了一个接口JpaVendorAdapter，它允许在应用程序启动时通过Spring Java config或XML配置插入任何特定于JPA供应商的配置。

您可以使用LocalContainerEntityManagerFactoryBean以及HibernateJpaVendorAdapter、EclipseLinkJpaVendorAdapter或OpenJpaVendorAdapter等任何JpaVendorAdapter实现类来创建EntityManagerFactory实例。

我相信如果您的应用程序使用Spring，您甚至不需要在persistence.xml中定义空的持久化单元。

下面是如何使用Spring Java配置创建EntityManagerFactory的示例：

    @Inject 
    private DataSource base1DataSource;

    @Inject 
    private DataSource base2DataSource;

    @Bean
    public EntityManagerFactory base1EntityManagerFactory()
            throws IOException, NamingException {
        HibernateJpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();

        LocalContainerEntityManagerFactoryBean containerEntityManagerFactoryBean = new LocalContainerEntityManagerFactoryBean();
        containerEntityManagerFactoryBean.setJpaVendorAdapter(vendorAdapter);
        containerEntityManagerFactoryBean.setPackagesToScan("YOUR_PACKAGE_NAMES");
        containerEntityManagerFactoryBean.setJtaDataSource(base1DataSource);
        containerEntityManagerFactoryBean.setJpaProperties(loadBase1JpaProperties());
        containerEntityManagerFactoryBean.setSharedCacheMode(SharedCacheMode.ENABLE_SELECTIVE);
        containerEntityManagerFactoryBean.setPersistenceUnitName("base1");
        containerEntityManagerFactoryBean.afterPropertiesSet();

        return containerEntityManagerFactoryBean.getObject();
    }

    @Bean
    public EntityManagerFactory base2EntityManagerFactory()
            throws IOException, NamingException {
        HibernateJpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();

        LocalContainerEntityManagerFactoryBean containerEntityManagerFactoryBean = new LocalContainerEntityManagerFactoryBean();
        containerEntityManagerFactoryBean.setJpaVendorAdapter(vendorAdapter);
        containerEntityManagerFactoryBean.setPackagesToScan("YOUR_PACKAGE_NAMES");
        containerEntityManagerFactoryBean.setJtaDataSource(base2DataSource);
        containerEntityManagerFactoryBean.setJpaProperties(loadBase2JpaProperties());
        containerEntityManagerFactoryBean.setSharedCacheMode(SharedCacheMode.ENABLE_SELECTIVE);
        containerEntityManagerFactoryBean.setPersistenceUnitName("base2");
        containerEntityManagerFactoryBean.afterPropertiesSet();

        return containerEntityManagerFactoryBean.getObject();
    }

    @Bean
    public Properties loadBase1JpaProperties() throws IOException {
        ClassPathResource resource = new ClassPathResource("base1-persistence.properties");

        return PropertiesLoaderUtils.loadProperties(resource);
    }

    @Bean
    public Properties loadBase2JpaProperties() throws IOException {
        ClassPathResource resource = new ClassPathResource("base2-persistence.properties");

        return PropertiesLoaderUtils.loadProperties(resource);
    }

有关可以覆盖到persistence.xml的内容的其他信息，请参阅以下网址：http://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/orm/jpa/LocalContainerEntityManagerFactoryBean.html

Answer 3

我假设你想在一个超类中声明你的持久化单元，并且你想在一个显式的项目中定义持久化单元。如果是真的，你可以使用如下的JNDI方法：

<persistence-unit name="MyPersistenceUnit"
    transaction-type="JTA">

    <jta-data-source>java:/myDS</jta-data-source>

    <mapping-file>META-INF/orm.xml</mapping-file>


    <jar-file>Persistence.jar</jar-file>

    <properties>

        <property name="jboss.entity.manager.jndi.name" value="java:app/applicationEntitymanager"/>  
        <!-- Properties for Hibernate -->
        <property name="hibernate.hbm2ddl.auto" value="update" />
        <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect" />
        <property name="hibernate.show_sql" value="false" />
    </properties>

</persistence-unit>

在另一端，您可以使用以下命令访问Entitymanager：

@Resource(mappedName = "java:app/applicationEntitymanager")
protected EntityManager em;