使用自定义JsonSerializer.

public class Response {
  private String status;
  private String error;

  @JsonProperty("p")
  @JsonSerialize(using = CustomSerializer.class)
  private Object data;

  // ...
}

public class CustomSerializer extends JsonSerializer<Object> {
  public void serialize(Object value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
    jgen.writeStartObject();
    jgen.writeObjectField(value.getClass().getName(), value);
    jgen.writeEndObject();
  }
}

然后，假设您要序列化以下两个对象:

And then, suppose you want to serialize the following two objects:

public static void main(String... args) throws Exception {
  ObjectMapper mapper = new ObjectMapper();
  Response r1 = new Response("Error", "Some error", 20);
  System.out.println(mapper.writeValueAsString(r1));
  Response r2 = new Response("Error", "Some error", "some string");
  System.out.println(mapper.writeValueAsString(r2));
}

第一个将打印:

{"status":"Error","error":"Some error","p":{"java.lang.Integer":20}}

第二个:

{"status":"Error","error":"Some error","p":{"java.lang.String":"some string"}}

我使用名称 p 作为包装器对象，因为它仅用作 placeholder.如果要删除它，则必须为整个 类编写自定义序列化程序，即JsonSerializer.

I have used the name p for the wrapper object since it will merely serve as a placeholder. If you want to remove it, you'd have to write a custom serializer for the entire class, i.e., a JsonSerializer<Response>.