Question

我正在做一个小项目，我有2个表：User和Application。一个用户可以有多个应用程序，并且一个应用程序可能被多个用户使用，因此它们之间是多对多关系。每个表都有一些字段（ID，名称，密码，技术等），我还在User和Application类中用@ManyToMany注释声明了2个数组列表。问题是，在我的业务层中，我编写了一种应向用户添加应用程序的方法，当我尝试执行user.getListOfApplications（）。add（app）时，它给了我该异常...

公共类管理员休眠 { 私有SessionFactory sessionFactory;

public void setup()
{
     sessionFactory = new Configuration().configure().buildSessionFactory();
}

public void exit()
{
    sessionFactory.close();
}

public void create(Object obj)
{
    Session session = sessionFactory.openSession();
    session.beginTransaction();

    session.save(obj);

    session.getTransaction().commit();
    session.close();
}

public Object read(Class<?> c, int idObj)
{
    Session session = sessionFactory.openSession();
    session.beginTransaction();

    Object obj = session.get(c, idObj);

    System.out.println(obj);

    session.getTransaction().commit();
    session.close();
    return obj;
}

public void update(Object obj)
{
    Session session = sessionFactory.openSession();
    session.beginTransaction();

    session.update(obj);

    session.getTransaction().commit();
    session.close();
}

public void delete(Object obj)
{
    Session session = sessionFactory.openSession();
    session.beginTransaction();

    session.delete(obj);

    session.getTransaction().commit();
    session.close();
}

public <T> List<T> loadAllData(Class<T> type)
{
    Session session = sessionFactory.openSession();
    session.beginTransaction();

    CriteriaBuilder builder = session.getCriteriaBuilder();
    CriteriaQuery<T> criteria = builder.createQuery(type);
    criteria.from(type);
    List<T> data = session.createQuery(criteria).getResultList();

    session.getTransaction().commit();
    session.close();
    return data;
}

}

public Boolean addNewApplicationToUser(String userUserName, String applicationName)
{
    int okUser = 0;
    int okApp = 0;

     listOfApplications = managerHibernate.loadAllData(Application.class);
     listOfUsers = managerHibernate.loadAllData(User.class);
     User user = null;
     Application app = null;

    for(Application index: listOfApplications)
    {
        if(index.getApplicationName().equals(applicationName))
            {
                okApp = 1;
                app = index;
            }
    }

    for(User index: listOfUsers)
    {
        if(index.getUserUserName().equals(userUserName))
            {
                okUser = 1;
                user = index;
            }
    }

    if(okUser == 0  || okApp == 0)
        return false;
    else
    {   
        user.getListOfApplications().add(app);
        //app.getUserList().add(user);
        return true;
    }
}

方法addNewApplicationToUser是在另一个称为ControllerHibernate的类中编写的。只有else分支很重要，其余的是检查参数是否确实存在于数据库中

Answer 1

当您使用以下方法 managerHibernate.loadAllData

加载数据时，问题开始

public <T> List<T> loadAllData(Class<T> type)
{
        // New session was opened here
        Session session = sessionFactory.openSession();
        session.beginTransaction();

        CriteriaBuilder builder = session.getCriteriaBuilder();
        CriteriaQuery<T> criteria = builder.createQuery(type);
        criteria.from(type);
        List<T> data = session.createQuery(criteria).getResultList();

        session.getTransaction().commit();
        session.close();
        //session is close here
        return data;
    }

因此，当您加载数据时，休眠框架将仅加载用户对象。由于您已选择在模型类中使用延迟加载，因此仅当您尝试访问列表时才加载应用程序值。由于您已经已经关闭会话，因此框架无法再获取导致延迟加载异常的应用程序列表。

listOfApplications = managerHibernate.loadAllData(Application.class);
//loading user data and close the session associated with it
listOfUsers = managerHibernate.loadAllData(User.class);
User user = null;
Application app = null;

for(Application index: listOfApplications)
{
  if(index.getApplicationName().equals(applicationName))
      {
          okApp = 1;
          app = index;
      }
}

for(User index: listOfUsers)
{
  if(index.getUserUserName().equals(userUserName))
      {
          okUser = 1;
          user = index;
      }
}

if(okUser == 0  || okApp == 0)
  return false;
else
{   
  // when you run this line the hibernate framework will try to retrieve the application data.Since you have the closed session lazy load exception occurs 
  user.getListOfApplications().add(app);
  return true;
}

解决此问题的方法

1）尝试保持会话打开，以便框架可以获取应用程序数据

2）将延迟加载更改为模型pojo类中的急切加载（因为您不建议使用这种多对多关系）

Answer 2

由于没有事务可用于获取用户的Application的惰性列表，因此您需要先获取它。为此，您可以按以下方式更改loadAllData：

public interface CriteriaSpec 
{
public void joinFetch(CriteriaBuilder builder, CriteriaQuery criteria, Root root);
}

public <T> List<T> loadAllData(Class<T> type, Optional<CriteriaSpec> spec)
{
Session session = sessionFactory.openSession();
session.beginTransaction();

CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery<T> criteria = builder.createQuery(type);
Root root = criteria.from(type);
if(spec.isPresent())
    spec.joinFetch(builder, criteria, root);
List<T> data = session.createQuery(criteria).getResultList();

session.getTransaction().commit();
session.close();
return data;
}

然后使用它：

managerHibernate.loadAllData(Application.class, Optional.empty());
listOfUsers = managerHibernate.loadAllData(User.class, (rootEntity, query, 
criteriaBuilder) -> {
            rootEntity.fetch("listOfApplications", JoinType.Left_OUTER_JOIN);

        });

如何在Java中修复LazyInitializationException？

2 个答案: