如何使用Raterofit库发布原始JSON对象

Question

我尝试使用邮递员并使用同一JSON对象进行测试，并且在使用Raterofit库时效果很好，但是它不起作用。

我尝试将JSON对象放入POJO类，但仍然不起作用。

notificationservice类

public interface NotificationService {
    @Headers({
            "Content-Type:"+ApiConstants.CONTENT_TYPE,
            "Authorization:"+ApiConstants.SERVER_KEY
    })
    @POST(ApiConstants.SEND)
    Call<ResponseBody> sendNotification(@Body String body);
}

apiconstants class

public class ApiConstants {
    public static final String SERVER_KEY = "key=mykey";
    public static final String SEND = "send";
    public static final String BASE_URL = "https://fcm.googleapis.com/fcm/";
    public static final String CONTENT_TYPE = "application/json";
    public static final String NOTIFICATION = "{n" +
            "    "to": "/topics/MISSING",n" +
            "    "data": {n" +
            "        "extra_information": "This is some extra information"n" +
            "    },n" +
            "    "notification": {n" +
            "        "title": "She pressed the button",n" +
            "        "text": "She misses you",n" +
            "        "click_action": "MAINACTIVITY"n" +
            "    }n" +
            "}";
}

主要活动

private void buttonIsPressed() {
        RetrofitClient retrofitClient;
        retrofitClient = RetrofitClient.getInstance();

        Call<ResponseBody> call = retrofitClient.getNotificationService().sendNotification(ApiConstants.NOTIFICATION);
        call.enqueue(new Callback<ResponseBody>() {
            @Override
            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                ResponseBody result = response.body();
                String gs = new Gson().toJson(result);
                Log.d("MainActivity", "response = " + gs);
            }
            @Override
            public void onFailure(Call<ResponseBody> call, Throwable t) {
                Log.d(TAG, "//onFailure");
            }
        });
    }

按下按钮时，我正在尝试将通知发送给某些特定用户。从我可以看出的唯一不起作用的是，我在使用Call<ResponseBody> sendNotification(@Body String body)

Answer 1

您应该创建一个模型类

public class MyRequestBodyModel {
    @SerializedName("id")
    private int id;
    @SerializedName("name")
    private String name;
}

创建MyRequestBodyModel的新实例并传递您的值

MyRequestBodyModel model = new MyRequestBodyModel();
model.id = 1;
model.name = "john";

将此模型的实例设置为要改造呼叫方法的参数

Call<ResponseBody> sendNotification(@Body MyRequestBodyModel model);