Question

我必须准备要发送给模型的模型。

这是接收到的示例JSON格式：

{
\\ rest of the model
"Model": {
    "name": "name_01",
    "description": "name_01",
    "features": {
      "name": "feature 01",
      "description": "feature 01"
    }
  }
\\ Rest of the model
}

要发送到fronEnd的响应：

{
\\ rest of the model

"model_name":"name_01",
"feature_name":"feature_01"
}
\\ rest of the model

这是我到目前为止实现的：

代码：

@SuppressWarnings("unchecked")
    @JsonProperty("model")
    private void unpackNestedModel(Map<String,Object> model) {
        this.model_name = (String) model.get("name");
        Map<String, Object> model2 = (Map<String, Object>) model.get("features");
        this.feature_name = (String) model2.get("name");
    }

有更好的方法吗？

Answer 1

您可以通过JSONObject解析结果，但建议通过POJO Object操作json结果。

@Test
public void test02() {
    String receivedJson = "{\"Model\":{\"name\":\"name_01\",\"description\":\"name_01\",\"features\":{\"name\":\"feature 01\",\"description\":\"feature 01\"}}}";
    JSONObject receivedObj = JSONObject.parseObject(receivedJson);

    JSONObject model = (JSONObject) receivedObj.get("Model");
    JSONObject features = (JSONObject) model.get("features");

    JSONObject responseObj = new JSONObject();
    responseObj.put("model_name", model.getString("name"));
    responseObj.put("feature_name", features.getString("name"));

    System.out.println(responseObj);
    //output in console
    //{"model_name":"name_01","feature_name":"feature 01"}
}

Answer 2

假设您正在使用Jackson，并且您的请求有效载荷是这样的：

@Data
public class RequestData {

    @JsonProperty("Model")
    private Model model;
}

您可以使用以下类来表示它：

@Data
public class Model {

    private String name;
    private String description;
    private Features features;
}

@Data
public class Features {

    private String name;
    private String description;
}

ObjectMapper

如果您在@Data @JsonRootName("Model") public class Model { private String name; private String description; private Features features; }中启用了UNWRAP_ROOT_VALUE反序列化功能，则可以简单地进行以下操作：

@Data
public class Features {

    private String name;
    private String description;
}

{
  "model_name": "name_01",
  "feature_name": "feature_01"
}

如果您的响应有效载荷是：

@Data
public class ResponseData {

    @JsonProperty("model_name")
    private String modelName;

    @JsonProperty("feature_name")
    private String featureName;
}

您可能会：

if (modal.classList.contains('is-active')) {
   body.classList.toggle('is-modal-open');
}

如何在没有POJOS的情况下为接收的JSON格式定义数据模型？

2 个答案: