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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

这里有几个很好的答案，可以在广泛的范围内处理这个错误。我使用Spring Security遇到了一个特定的情况，它有一个快速(尽管可能不是最佳的)修复。

在用户授权期间(登录并通过身份验证后立即)，我在扩展SimpleUrlAuthenticationSuccessHandler的自定义类中测试用户实体的特定权限。

我的用户实体实现了UserDetails，并有一组惰性加载的角色，这些角色抛出了“org.hibernate. properties”。LazyInitializationException -不能初始化代理-没有会话”异常。将Set从“fetch=FetchType.”LAZY" to "fetch=FetchType。“渴望”帮我解决了这个问题。

2016-03-15 01:55:36

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

这意味着您正在尝试访问的对象没有加载，因此编写一个查询，对您正在尝试访问的对象进行连接获取。

Eg:

如果你试图从ObjectA中获取ObjectB，其中ObjectB是ObjectA中的外键。

查询:

SELECT objA FROM ObjectA obj JOIN FETCH obj.objectB objB
2018-04-23 11:36:11

springBootVersion = '2.6.7' hibernate = 5.6.8.Final'

对我来说，我得到的错误是:

MyEntity myEntity = myEntityRepository.getById(id);

我改成这样:

MyEntity myEntity = myEntityRepository.findById(id).orElse(null);

和我添加@ManyToOne(fetch = FetchType.EAGER)在实体

2022-04-26 13:13:57

你可以试着设置

<property name="hibernate.enable_lazy_load_no_trans">true</property>

在hibernate.cfg.xml或persistence.xml中

这里已经很好地解释了这个性质需要记住的问题

2014-12-04 20:33:43

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