Java如何修复org.hibernate.LazyInitializationException-无法初始化代理-没有会话

Question

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)

当我尝试从以下几行拨打电话时：

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

我getModelByModelGroup(int modelgroupid)首先实现了这样的方法：

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);

        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 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;
    }
}

并得到了例外。然后一位朋友建议我始终测试该会话并获取当前会话，以避免出现此错误。所以我这样做：

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);

        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;        
        }
    }
}

但仍然会出现相同的错误。我已经阅读了很多有关此错误的内容，并找到了一些可能的解决方案。其中之一是将lazyLoad设置为false，但是不允许这样做，因此建议我控制会话

Answer 1

这是错误的，因为你在提交事务时将会话管理配置设置为关闭会话。检查是否有以下内容：

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

在你的配置中。

为了克服此问题，你可以更改会话工厂的配置或打开另一个会话，而仅要求提供那些延迟加载的对象。但是我在这里建议的是在getModelByModelGroup本身中初始化此惰性集合并调用：

Hibernate.initialize(subProcessModel.getElement());

当你仍处于活动状态时。

最后一件事。一个友好的建议。你的方法中有以下内容：

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

请安装此代码，只需在上面几行的查询语句中过滤ID等于3的那些模型。