注入两个@PersistenceContext中的一个

Question

考虑拥有两个实体经理工厂:

<bean id="writeEntityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">...</bean>
<bean id="readOnlyEntityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">...</bean>

然后我想要两个Beans,我将注入正确的持久化上下文:

<bean id="readOnlyManager" class="..MyDatabaseManager">
<bean id="writeManager" class="..MyDatabaseManager">

这个bean看起来像:

public class MyDatabaseManager {

    private javax.persistence.EntityManager em;

    public EntityManager(javax.persistence.EntityManager em) {
        this.em = em;
    }
    ...
}

这显然不起作用,因为EntityManager不是bean,不能以这种方式注入:

No qualifying bean of type 'javax.persistence.EntityManager' available: expected at least 1 bean which qualifies as autowire candidate. Dependency annotations: {}

如何在bean中限定正确的EntityManager？我以前使用@PersistenceContext注释,但这不可用,因为我需要注入它.

如何为这样的Bean指定PersistenceContext？

更新:我的问题是如何通过XML而不是通过注释注入带有限定符的PersistenceContext.

Answer 1

假设您使用 spring 来管理事务，我会做的是 2 个不同的事务管理器，然后在我的服务中我将使用最合适的事务管理器，如下所示：

配置部分

@Bean
public LocalContainerEntityManagerFactoryBean writeEntityManagerFactory() {

    LocalContainerEntityManagerFactoryBean factory = new LocalContainerEntityManagerFactoryBean();
    //Your configuration here
    return factory;
}

@Bean(name={"writeTx"})
public PlatformTransactionManager writeTransactionManager() {

    JpaTransactionManager txManager = new JpaTransactionManager();
    txManager.setEntityManagerFactory(writeEntityManagerFactory().getObject());
    return txManager;
}

@Bean
public LocalContainerEntityManagerFactoryBean readEntityManagerFactory() {

    LocalContainerEntityManagerFactoryBean factory = new LocalContainerEntityManagerFactoryBean();
    //Your configuration here
    return factory;
}

@Bean(name={"readTx"})
public PlatformTransactionManager readTransactionManager() {

    JpaTransactionManager txManager = new JpaTransactionManager();
    txManager.setEntityManagerFactory(readEntityManagerFactory().getObject());
    return txManager;
}

服务层

@Transactional(value="readTx")
public List<Object> read(){
    //Your read code here
}

@Transactional(value="writeTx")
public void write(){
    //Your write code here
}

更新的答案 我误解了这个问题。

在您的配置类中您可以定义：

@Bean
    public LocalContainerEntityManagerFactoryBean writeEntityManagerFactory() {

        LocalContainerEntityManagerFactoryBean em  = new LocalContainerEntityManagerFactoryBean();
        em.setDataSource(dataSource());
        em.setPackagesToScan(new String[] { "models" });
        JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
        em.setJpaVendorAdapter(vendorAdapter);
        em.setJpaProperties(hibProps());
        em.setPersistenceUnitName("writer");
        return em;
    }
    @Bean
    public LocalContainerEntityManagerFactoryBean readEntityManagerFactory() {

        LocalContainerEntityManagerFactoryBean em  = new LocalContainerEntityManagerFactoryBean();
        em.setDataSource(dataSource());
        em.setPackagesToScan(new String[] { "models" });
        JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
        em.setJpaVendorAdapter(vendorAdapter);
        em.setJpaProperties(hibProps());
        em.setPersistenceUnitName("reader");
        return em;
    }   

请查看PersistenceUnitName数值

然后你可以通过执行以下操作来注入它们：

@PersistenceContext(unitName="writer")
private EntityManager emWriter;

@PersistenceContext(unitName="reader")
private EntityManager emReader;

我刚刚测试过，一切都很好

安吉洛