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());
我首先像这样实现了 getModelByModelGroup(int modelgroupid)
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);
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;
}
}
}
但仍然得到同样的错误。我已经为这个错误阅读了很多内容,并找到了一些可能的解决方案。其中之一是将lazyLoad设置为false,但我不允许这样做,这就是为什么我被建议控制会话
原文由 Blerta Dhimitri 发布,翻译遵循 CC BY-SA 4.0 许可协议
这里的问题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:
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,而只要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:
当您仍处于活动会话中时。
最后一件事。一个友好的建议。你的方法中有这样的东西:
请插入此代码,只需在上面几行的查询语句中过滤类型 id 等于 3 的模型。
更多阅读:
会话工厂配置
封闭会话的问题