如何修复 org.hibernate.LazyInitializationException - 无法初始化代理 - 没有会话 [英] How to fix org.hibernate.LazyInitializationException - could not initialize proxy - no Session

问题描述

我收到以下异常:

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 调用以下几行时:

when I try to call from main the following lines:

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

我首先像这样实现了 getModelByModelGroup(int modelgroupid) 方法:

I implemented the getModelByModelGroup(int modelgroupid) method firstly like this :

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:

and got the exception. Then a friend suggested me to always test the session and get the current session to avoid this error. So I did this:

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;        
        }
    }
}

但是仍然出现相同的错误.我已经阅读了很多关于这个错误的信息，并找到了一些可能的解决方案.其中之一是将 lazyLoad 设置为 false，但我不允许这样做，这就是为什么我被建议控制会话的原因

but still, get the same error. I have been reading a lot for this error and found some possible solutions. One of them was to set lazyLoad to false but I am not allowed to do this that's why I was suggested to control the session

推荐答案

这里的错误是您的会话管理配置设置为在您提交事务时关闭会话.检查你是否有类似的东西:

What is wrong here is that your session management configuration is set to close session when you commit transaction. Check if you have something like:

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

在您的配置中.

为了克服这个问题，您可以更改会话工厂的配置或打开另一个会话，然后只请求那些延迟加载的对象.但是我在这里建议的是在 getModelByModelGroup 本身中初始化这个惰性集合并调用:

In order to overcome this problem you could change the configuration of session factory or open another session and only than ask for those lazy loaded objects. But what I would suggest here is to initialize this lazy collection in getModelByModelGroup itself and call:

Hibernate.initialize(subProcessModel.getElement());

当您仍处于活动会话中时.

when you are still in active session.

还有最后一件事.一个友好的建议.你的方法中有这样的东西:

And one last thing. A friendly advice. You have something like this in your method:

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

请插入这段代码，只需在上面几行的查询语句中过滤那些类型 id 等于 3 的模型.

Please insted of this code just filter those models with type id equal to 3 in the query statement just couple of lines above.

更多阅读:

会话工厂配置

关闭会话的问题

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