Jackson真的无法将json反序列化为泛型类型吗？

Question

这是一个重复问题，因为下面的问题要么乱七八糟，要么根本没有回答：

public class AgentResponse<T> {

    private T result;

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

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

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

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

Jackson永远无法处理泛型类型T，它认为它是JavaType的映射，但它会发现对象类型构造函数参数，因为类型擦除并抛出错误。这是杰克逊的错误，还是我做错了什么？TypeReference或JavaType的显式规范还用于什么？

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)

Answer 1

您需要在构造函数上添加一些注释来告诉Jackson如何构建对象。以下几点对我起到了作用：

public class AgentResponse<T> {

    private T result;

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

如果没有@jsonCreator注释，Jackson就不知道调用这个构造函数。如果没有@JSONProperty注释，Jackson不知道构造函数的第一个参数映射到result属性。