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