杰克逊真的无法将json反序列化为泛型？ [英] Is Jackson really unable to deserialize json into a generic type?

问题描述

这是一个重复的问题，因为以下问题要么混乱，要么根本没有答案：

反序列化 - 通用类型与杰克逊

jackson-deserialize-into-runtime-specified-class

jackson-deserialize-using-generic-class

jackson-deserialize-generic-class-变量

我希望这个问题最终能够找到答案，使其变得清晰。

<有一个模型：

  public class AgentResponse< T> {
 
私人T的结果; 
 
 public AgentResponse（T结果）{
 this.result = result; 
} 
 public T getResult（）{
 return result; 
 
 
  

JSON输入：

  {result：{first-client-id：3，test-mail-module：3，third-client-id： 3，second-client-id：3}} 
  

以及两种推荐的反序列化类型：

$ b

  mapper.readValue（out，new TypeReference< AgentResponse< Map< String，Integer>>>（）{} ）; 
  

或

  JavaType javaType = mapper.getTypeFactory（）。constructParametricType（AgentResponse.class，Map.class）; 
 mapper.readValue（out，javaType）; 
  

杰克逊永远不能处理泛型类型T，它认为它是JavaType中的Map，但它会因类型擦除而发现Object type构造函数参数并引发错误。那么这是一个杰克逊错误，还是我做错了什么？什么是TypeReference或JavaType的明确规范？

  com.fasterxml.jackson.databind.JsonMappingException：找不到合适的构造函数类型[简单类型，类com.fg.mail.smtp.AgentResponse< java.util.Map< java.lang.String，java.lang.Integer>>]：无法从JSON对象实例化（需要添加/启用类型信息？）
 at [Source：java.io.InputStreamReader@4f2d26d; line：1，column：2] 
 at com.fasterxml.jackson.databind.JsonMappingException.from（JsonMappingException.java:164）
 at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromObjectUsingNonDefault（ BeanDeserializerBase.java:984）
 at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject（BeanDeserializer.java:276）
 at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize（ BeanDeserializer.java:121）
 at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose（ObjectMapper.java:2888）
 at com.fasterxml.jackson.databind.ObjectMapper.readValue（ObjectMapper.java： 2064）
  

解决方案

您需要在构造函数中添加一些注释告诉杰克逊如何构建这个对象。以下方法适用于我：

  public class AgentResponse< T> {
 
私人T的结果; 
 
 @JsonCreator 
 Public AgentResponse（@JsonProperty（result）T result）{
 this.result = result; 
} 
 public T getResult（）{
 return result;如果没有 @JsonCreator  
 $ b  

/ code>注释，杰克逊不知道调用这个构造函数。没有 @JsonProperty 注解，Jackson不知道构造函数的第一个参数映射到结果属性。


This is a duplicate question because the following questions are either messy or they are not answered at all:

deserializing-a-generic-type-with-jackson

jackson-deserialize-into-runtime-specified-class

jackson-deserialize-using-generic-class

jackson-deserialize-generic-class-variable

I hope that this question will finally find an answer that makes this clear for good.

Having a model :

public class AgentResponse<T> {

    private T result;

    public AgentResponse(T result) {
        this.result = result;
    }
    public T getResult() {
        return result;
    }
}

JSON input:

{"result":{"first-client-id":3,"test-mail-module":3,"third-client-id":3,"second-client-id":3}}

and two recommended ways of deserializing generic types :

mapper.readValue(out, new TypeReference<AgentResponse<Map<String, Integer>>>() {}); 

or

JavaType javaType = mapper.getTypeFactory().constructParametricType(AgentResponse.class, Map.class);
mapper.readValue(out, javaType);

Jackson is never able to deal with the generic type T, it figures it's a Map from JavaType, but it finds Object type constructor argument because of type erasure and throws an error. So is this a Jackson bug, or am I doing something wrong? What else is explicit specification of TypeReference or JavaType for?

com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class com.fg.mail.smtp.AgentResponse<java.util.Map<java.lang.String,java.lang.Integer>>]: can not instantiate from JSON object (need to add/enable type information?)
at [Source: java.io.InputStreamReader@4f2d26d; line: 1, column: 2]
at com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:164)
at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromObjectUsingNonDefault(BeanDeserializerBase.java:984)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:276)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:121)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:2888)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2064)

解决方案

You need to add some annotations on the constructor to tell Jackson how to build the object. The following worked for me:

public class AgentResponse<T> {

    private T result;

    @JsonCreator
    public AgentResponse(@JsonProperty("result") T result) {
        this.result = result;
    }
    public T getResult() {
        return result;
    }
}

Without the @JsonCreator annotation, Jackson cannot know to call this constructor. And without the @JsonProperty annotation, Jackson does not know that the first argument of the constructor maps to the result property.

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