java - 如何使用 RETROFIT 库发布原始 JSON 对象

Question

我尝试使用 Postman 并使用相同的 JSON 对象进行测试，它工作得很好，但当我使用改造库时它不起作用。

我尝试将 JSON 对象放入 Pojo 类中，但仍然不起作用。

通知服务类

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

}

API 常量类

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 Activity

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");
            }
        });
    }

我正在尝试在按下按钮时向某些特定用户发送通知。据我所知，唯一不起作用的是我在使用 Call<ResponseBody> sendNotification(@Body String body) 时为正文提供了错误的输入。

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改造调用方法的参数

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