org.hibernate.LazyInitializationException - 无法初始化代理 - 没有会话

问题

我得到以下异常:

Exception in thread "main" org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:167)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:215)
    at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:190)
    at sei.persistence.wf.entities.Element_$$_jvstc68_47.getNote(Element_$$_jvstc68_47.java)
    at JSON_to_XML.createBpmnRepresantation(JSON_to_XML.java:139)
    at JSON_to_XML.main(JSON_to_XML.java:84)

当我尝试从main调用以下行时:

Model subProcessModel = getModelByModelGroup(1112);
System.out.println(subProcessModel.getElement().getNote());

我首先实现了getModelByModelGroup(int modelgroupid)方法,如下所示:

public static Model getModelByModelGroup(int modelGroupId, boolean openTransaction) {

        Session session = SessionFactoryHelper.getSessionFactory().getCurrentSession();     
        Transaction tx = null;

        if (openTransaction)
            tx = session.getTransaction();

        String responseMessage = "";

        try {
            if (openTransaction)            
                tx.begin();
            Query query = session.createQuery("from Model where modelGroup.id = :modelGroupId");
            query.setParameter("modelGroupId", modelGroupId);
            @SuppressWarnings("unchecked")
            List<Model> modelList = (List<Model>)query.list(); 
            Model model = null;
            // Cerco il primo Model che è in esercizio: idwf_model_type = 3
            for (Model m : modelList)
                if (m.getModelType().getId() == 3) {
                    model = m;
                    break;
                }

            if (model == null) {
                Object[] arrModels = modelList.toArray();
                if (arrModels.length == 0) 
                    throw new Exception("Non esiste ");

                model = (Model)arrModels[0];
            }

            if (openTransaction)
                tx.commit();
            return model;

        } catch(Exception ex) {
            if (openTransaction)
                tx.rollback();
            ex.printStackTrace();
            if (responseMessage.compareTo("") == 0)
                responseMessage = "Error" + ex.getMessage();
            return null;        
        }

得到了例外。然后一位朋友建议我总是测试会话并获取当前会话以避免此错误。所以我这样做了:

public static Model getModelByModelGroup(int modelGroupId) {

        Session session = null;
        boolean openSession = session == null;
        Transaction tx = null;
        if (openSession){
          session = SessionFactoryHelper.getSessionFactory().getCurrentSession();   
            tx = session.getTransaction();
        }
        String responseMessage = "";

        try {
            if (openSession)            
                tx.begin();
            Query query = session.createQuery("from Model where modelGroup.id = :modelGroupId");
            query.setParameter("modelGroupId", modelGroupId);
            @SuppressWarnings("unchecked")
            List<Model> modelList = (List<Model>)query.list(); 
            Model model = null;
            for (Model m : modelList)
                if (m.getModelType().getId() == 3) {
                    model = m;
                    break;
                }

            if (model == null) {
                Object[] arrModels = modelList.toArray();
                if (arrModels.length == 0) 
                    throw new RuntimeException("Non esiste");

                model = (Model)arrModels[0];

            if (openSession)
                tx.commit();
            return model;

        } catch(RuntimeException ex) {
            if (openSession)
                tx.rollback();
            ex.printStackTrace();
            if (responseMessage.compareTo("") == 0)
                responseMessage = "Error" + ex.getMessage();
            return null;        
        }

    }

但仍然得到相同的错误。我一直在阅读这个错误,并找到了一些可能的解决方案。其中一个是将lazyLoad设置为false但我不允许这样做,这就是为什么我被建议控制会话


#1 热门回答(91 赞)

你可以尝试设置

<property name="hibernate.enable_lazy_load_no_trans">true</property>

在hibernate.cfg.xml或persistence.xml中

要记住这个属性的问题很好地解释了here


#2 热门回答(62 赞)

如果你使用Spring标记类为@Transactional,那么Spring将处理会话管理。

@Transactional
public class My Class {
    ...
}

通过使用@Transactional,可以自动处理许多重要方面,例如事务传播。在这种情况下,如果调用另一个事务方法,则该方法可以选择加入正在进行的事务,避免"无会话"异常。


#3 热门回答(59 赞)

这里的错误是你的会话管理配置设置为在提交事务时关闭会话。检查你是否有类似的东西:

<property name="current_session_context_class">thread</property>

在你的配置中。

为了克服这个问题,你可以更改会话工厂的配置或打开另一个会话,而不仅仅是要求那些延迟加载的对象。但我在这里建议的是在getModelByModelGroup本身初始化这个惰性集合并调用:

Hibernate.initialize(subProcessModel.getElement());

当你还在活跃的会话中。

最后一件事。友好的建议。你的方法中有这样的东西:

for (Model m : modelList)
            if (m.getModelType().getId() == 3) {
                model = m;
                break;
            }

请注意这个代码只是在查询语句中过滤那些类型id等于3的模型,只需要几行。

更多阅读:
session factory configurationproblem with closed session