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2024-09-06 07:00:00

如何修复org.hibernate.LazyInitializationException -不能初始化代理-没有会话

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)

当我试图从主要呼叫以下线路:

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);

        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;
    }
}

得到了异常。然后一个朋友建议我总是测试会话并获取当前会话以避免这种错误。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;        
        }
    }
}

但还是得到相同的错误。 我已经阅读了很多关于这个错误的文章，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

当前回答

在不同的用例中遇到相同的异常。

用例:尝试用DTO投影从DB读取数据。

解决方案:使用get方法代替load方法。

通用操作

public class HibernateTemplate {
public static Object loadObject(Class<?> cls, Serializable s) {
    Object o = null;
    Transaction tx = null;
    try {
        Session session = HibernateUtil.getSessionFactory().openSession();
        tx = session.beginTransaction();
        o = session.load(cls, s); /*change load to get*/
        tx.commit();
        session.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
    return o;
}

}

持久化类

public class Customer {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "Id")
private int customerId;

@Column(name = "Name")
private String customerName;

@Column(name = "City")
private String city;

//constructors , setters and getters

}

CustomerDAO接口

public interface CustomerDAO 
     {
   public CustomerTO getCustomerById(int cid);
     }

实体传输对象类

public class CustomerTO {

private int customerId;

private String customerName;

private String city;

//constructors , setters and getters

}

工厂类

public class DAOFactory {

static CustomerDAO customerDAO;
static {
    customerDAO = new HibernateCustomerDAO();
}

public static CustomerDAO getCustomerDAO() {
    return customerDAO;
}

}

实体特定的DAO

public class HibernateCustomerDAO implements CustomerDAO {

@Override
public CustomerTO getCustomerById(int cid) {
    Customer cust = (Customer) HibernateTemplate.loadObject(Customer.class, cid);
    CustomerTO cto = new CustomerTO(cust.getCustomerId(), cust.getCustomerName(), cust.getCity());
    return cto;
}

}

检索数据:测试类

CustomerDAO cdao = DAOFactory.getCustomerDAO();
CustomerTO c1 = cdao.getCustomerById(2);
System.out.println("CustomerName -> " + c1.getCustomerName() + " ,CustomerCity -> " + c1.getCity());

显示数据

Hibernate系统生成的查询和输出

Hibernate:选择customer0_。Id为Id1_0_0_， customer0_。City as city2_0_， customer0_。Name为Name3_0_0_ from CustomerLab31 customer0_ where customer0_. id =?

客户名:>科迪，客户名:>洛杉矶

2019-07-14 19:18:36

其他回答

如果您使用Spring将类标记为@Transactional，那么Spring将处理会话管理。

@Transactional
public class MyClass {
    ...
}

通过使用@Transactional，可以自动处理事务传播等许多重要方面。在这种情况下，如果调用了另一个事务方法，则该方法可以选择加入正在进行的事务，以避免出现“无会话”异常。

警告如果你使用@Transactional，请注意产生的行为。有关常见陷阱，请参阅本文。例如，即使没有显式调用save，对实体的更新也会被持久化

2015-08-28 17:45:37

使用session.get(*.class, id);但不加载函数

2015-08-25 21:21:21

所有关于添加JOIN FETCH(或左JOIN FETCH)的答案都是正确的，我只想添加这个: 如果你有转换器，请确保getAsObject子使用一个“查找”，而不是包括在sql Join Fetch太。 我花了很多时间来修复一个类似的问题，问题是在转换器。

2022-02-16 17:34:59

这个异常是因为当你调用session. getentitybyid()时，会话将被关闭。因此，您需要将实体重新附加到会话。或者简单的解决方案是配置default-lazy="false"到你的entity.hbm.xml，或者如果你使用注释，只需添加@Proxy(lazy=false)到你的实体类。

2016-09-07 14:17:24

如果你正在使用JPQL，使用JOIN FETCH是最简单的方法: http://www.objectdb.com/java/jpa/query/jpql/from#LEFT_OUTER_INNER_JOIN_FETCH_

2017-03-23 21:55:05

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