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Java JSON:子对象列表的UnrecognizedPropertyException_Java_Json_Jackson - Fatal编程技术网

Java JSON:子对象列表的UnrecognizedPropertyException

Java JSON:子对象列表的UnrecognizedPropertyException,java,json,jackson,Java,Json,Jackson,这似乎是一个简单的实现,但不知何故对我来说并不奏效 public class ParentEntity { private List<ChildEntity> childFields; public List<ChildEntity> getChildFields() { return childFields; } public void setChildFields(List<ChildEntity> childFi

这似乎是一个简单的实现,但不知何故对我来说并不奏效

public class ParentEntity {

   private List<ChildEntity> childFields;

   public List<ChildEntity> getChildFields() {
      return childFields;
   }

   public void setChildFields(List<ChildEntity> childFields) {
     this.childFields = childFields;
   }

}

关于您添加的异常消息,您在JSON中为ChildEntity和ChildEntity属性指定的属性不匹配

如果存在不匹配,并且希望在JSON中指定比ChildEntity类中更多的属性,那么可以使用Jackson的

@JsonIgnoreProperties
注释。它将忽略POJO中未定义的所有属性

您也可以选择使用:

ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

它将忽略所有未声明的属性。

在我看来是正确的,您可以在读取JSON的地方共享您的代码吗?
private List childFields=new ArrayList()
@JsonIgnoreProperties
ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);