JSON序列化程序中的惰性Loadng错误 [英] Lazy Loadng error in JSON serializer

问题描述

我有这样的@OneToOne Hibernate relationShip

  public class Address implements Serializable {
 
私人字符串ID; 
私人字符串城市; 
私人字符串国家; 
 // setter getters ommitted 
} 
 
 public class Student实现Serializable {
 
 private String id; 
 private String firstName; 
 private String lastName; 
私人地址; 
} 
  

地址项目被映射为LAZY。

现在我想使用

  session.load（Student.class，id）获取用户及其地址。 ; 
  

在我的daoService中。

然后我将它作为JSON从我的Spring MVC控制器返回：

  @RequestMapping（value =/ getStudent.do，method = RequestMethod。 POST）
 @ResponseBody 
 public Student getStudent（@RequestParam（studentId）String id）{
 Student student = daoService.getStudent（id）; 
回报学生; 
} 
  

不幸的是，由于Lazy clasees， p>

  org.codehaus.jackson.map.JsonMappingException：找不到类org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer的序列化程序，没有（通过引用链：com.vanilla.objects.Student _ $$ _ javassist_1 [address]  - > com.vanilla.objects.Address _ $$）创建BeanSerializer的属性（以避免异常，禁用SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS） _javassist_0 [handler]）
 at org.codehaus.jackson.map.ser.StdSerializerProvider $ 1.serialize（StdSerializerProvider.java:62）
  

我使用OpenSessionInViewInterceptor，它工作得很好。
我明白我可以让用户左键加入HQL查询并以这种方式检索学生和地址并解决问题。我也明白，改变与EAGER的关系将解决它。

但是，我怎样才能使用标准杰克逊消息转换器序列化为JSON懒惰类，这是因为我添加到了我的XML文件中。

解决方案

最简单的解决方案：不要序列化实体，使用值对象。

如果这不是您的选择，请确保实体Object已分离。

使用JPA（2），您可以使用 EntityManager.detach （实体） ，与普通的Hibernate相当的是 Session.evict（entity） 。

I have such kind of @OneToOne Hibernate relationShip

public class Address implements Serializable {

    private String id;
    private String city;
    private String country;
//setter getters ommitted
}

public class Student implements Serializable {

    private String id;
    private String firstName;
    private String lastName;    
    private Address address;
}

address Item is mapped as LAZY.

Now I want to fetch user and it's address using

session.load(Student.class,id);

In my daoService.

Then I return it as JSON from my Spring MVC controller:

@RequestMapping(value="/getStudent.do",method=RequestMethod.POST)
    @ResponseBody
    public Student getStudent(@RequestParam("studentId") String id){
        Student student = daoService.getStudent(id);
        return student;
    }

Unfortunately, it's not working because of Lazy clasees and I fails with:

org.codehaus.jackson.map.JsonMappingException: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: com.vanilla.objects.Student_$$_javassist_1["address"]->com.vanilla.objects.Address_$$_javassist_0["handler"])
    at org.codehaus.jackson.map.ser.StdSerializerProvider$1.serialize(StdSerializerProvider.java:62)

I do use OpenSessionInViewInterceptor and it works just fine. I understand that I can user left join HQL query and retrieve student and address that way and solve the problem. I also understand that changing relation to EAGER will solve it.

But how can I serialize to JSON lazy classes using standard jackson message converter which of cause I added to my XML file.

解决方案

The easiest solution: Don't serialize entities, use Value Objects.

If that is not an option for you, make sure that the entity Object is detached.

With JPA (2), you would use EntityManager.detach(entity), with plain Hibernate the equivalent is Session.evict(entity).

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