Question

我面临有关使用翻新2在android中的数组中发布数据的问题。我必须在个人资料设置活动中发布近14个字段的数据... 像这样...

name="basics[first_name] , name="basics[last_name]" , name="basics[phone_number]"

我必须以这种格式发送数据。我不知道该怎么做需要帮助。我不了解如何在接口中调用api，因为我必须将数据放入数组中。目前，我正在这样做，但是我知道它不正确...

@FormUrlEncoded
    @POST("profile_setting/basic_setting")
    Call<ResponseBody> UpdateBasics(
            @Query("user_id") int user_id ,
            @Field("nickname") String nickname ,
            @Field("first_name") String  first_name ,
            @Field("last_name") String last_name ,
            @Field("phone_number") String phone_number ,
            @Field("fax") String fax
    );

Answer 1

以这种方式使用Json将Retrofit 2对象作为请求参数发送

    @Headers("Content-Type: application/json")
    @POST("profile_setting/basic_setting")
    Call<ResponseBody> UpdateBasics(@Body String body);

这就是您使用上述方法发送json对象的方式

try {
                    JSONObject paramObject = new JSONObject();
                    paramObject.put(value_one, "field_one"));
                    paramObject.put(value_second, "field_second"));
                    Call<ResponseBody> userCall = apiInterface.UpdateBasics(paramObject.toString());
                    userCall.enqueue(new Callback<ResponseBody>() {
                        @Override
                        public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                            //handle your result here

                        }

                        @Override
                        public void onFailure(Call<LoginResponse> call, Throwable t) {
                           //handle failure
                        }
                    });
                } catch (JSONException e) {
                    e.printStackTrace();
                }

Answer 2

上课

public class Basic {
  public final int user_id;
  public final String nickname;
  ....

  public Basic(int user_id, ...) {

  }
}

然后将此类对象的列表传递给此接口

public interface MyService {
  @POST("/basic")
  Response void sendData(@Body List<Basic> basic);
}

或者您可以对JSONObject进行相同的操作。只需列出jsonobjects

JSONObject paramObject = new JSONObject();
                    paramObject.put(value_one, "field_one"));
                    paramObject.put(value_second, "field_second"));

将对象放入列表

list.add(paramObject);

然后通过改造

public interface MyService {
  @POST("/basic")
  Response void sendJsonObjectData(@Body List<JSONObject> basic);
}

Answer 3

您可以按照此操作。（发布我的操作方式）

@POST("Users.json")
Call<UploadData> uploadToken(@Body UploadData uploadData);

UploadData.class

public class UploadData {

private String DeviceToken, DeviceIMEI;

public UploadData(String deviceToken, String deviceIMEI) {
    DeviceToken = deviceToken;
    DeviceIMEI = deviceIMEI;
}

public String getDeviceToken() {
    return DeviceToken;
}

public String getDeviceIMEI() {
    return DeviceIMEI;
}
}

然后在您的活动中

private void uploadToken() {
     ApiInterface apiInterface = ApiClient.getClient().create(ApiInterface.class);
     UploadData uploadToken = new UploadData(deviceToken, imei);
     final Call<UploadData> callUpload = apiInterface.uploadToken(uploadToken);

callUpload.enqueue(new Callback<UploadData>() {
        @Override
        public void onResponse(Call<UploadData> call, Response<UploadData> response) {
             if (response.isSuccessful()) {
                Toasty.success(Main.this, "Token Uploaded !! ", Toast.LENGTH_LONG).show();
              }
      }

       @Override
       public void onFailure (Call < UploadData > call, Throwable t){
              call.cancel();
              Toasty.error(Main.this, "Error: " + t.getLocalizedMessage(),Toast.LENGTH_SHORT).show();
        }
    });

 }

如何使用改造2在android中的数组中发布数据

3 个答案: