Jackson序列化过程中带有Hibernate FetchType.LAZY的Java Spring HttpMessageNotWritableException/JsonMappingException

Question

我使用Java Spring 网络控制器和jackson 来解析响应。我还有四个JavaHibernate实体：UserFilterMakeModel，在 Make -> Model(s) 中有双向关系：

GeUser.getFilters().get(0).getModel().getMake().getModels().get(0).getMake()...

我尝试在Model 类中添加@JsonIgnoreProperties(value = "models")，但仍然出现异常：

WARNING: Failed to write HTTP message: org.springframework.http.converter.HttpMessageNotWritableException: Could not write content: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: com.gecars.objects.GeUser["filters"]->org.hibernate.collection.internal.PersistentSet[0]->com.gecars.objects.Filter["model"]->com.gecars.objects.models.Model_$$_jvste81_2["handler"]); nested exception is com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: com.gecars.objects.GeUser["filters"]->org.hibernate.collection.internal.PersistentSet[0]->com.gecars.objects.Filter["model"]->com.gecars.objects.models.Model_$$_jvste81_2["handler"])

实体 GeUser：

@Entity
@Table(name="users")
public class GeUser {
...
  @OneToMany(fetch = FetchType.LAZY)
  private Set<Filter> filters = new HashSet<Filter>(0);
..

过滤器：

@Entity
@Table(name="filters")
public class Filter {
...
  @ManyToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "fk_model") 
  private Model model;
..

型号：

@Entity
@Table(name="models")
public class Model {
...
  @ManyToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "fk_make") 
  @JsonIgnoreProperties(value = "models") //Not working (if I correct understand it should tell jackson to skip models from Make class )
  private Make make;
... 

制作：

@Entity
@Table(name="auto")
public class Make {
....
   @OneToMany(fetch = FetchType.LAZY, mappedBy = "make")
   @JsonIgnoreProperties(value = "make")
   private List<Model> models = new ArrayList<Model>(0);
...

Answer 1

要告诉杰克逊在将实体解析为 Json 时忽略属性，您可以使用 @JsonIgnore 注释：

@JsonIgnore
private PropertyClass property;

另一方面，所有实体类都必须实现java.io.Serializable