android - 从改造 onResponse 返回数据

Question

我正在做一个基本的 Android 应用程序，它通过 API 执行简单的登录。我需要返回接收到的数据，但改进的 onResponse 方法具有 void 返回类型。我的代码是:

package com.example.deberesloginjava.data;

import android.util.Log;

import com.example.deberesloginjava.data.APIServices.Post;
import com.example.deberesloginjava.data.model.LoggedInUser;

import java.io.IOException;

import com.example.deberesloginjava.data.APIServices.APIService;
import com.example.deberesloginjava.data.APIServices.ApiUtils;

import retrofit2.Call;
import retrofit2.Callback;
import retrofit2.Response;

import static android.content.ContentValues.TAG;

/**
 * Class that handles authentication w/ login credentials and retrieves user information.
 */

public class LoginDataSource {

    private APIService mAPIService;

    public void getData(Callback<Result<LoggedInUser>> callback){
        apiClient.getData().enqueue(callback);
    }

    public Result<LoggedInUser> login(String username, String password) {

        try {

            mAPIService = ApiUtils.getAPIService();

           mAPIService.login(username, password, "password").enqueue(new Callback<Post>() {
                @Override
                public void onResponse(Call<Post> call, Response<Post> response) {
                    if(response.isSuccessful()) {
                        LoggedInUser loggedInUser;
                        loggedInUser = new LoggedInUser(response.body().getUserName(),response.body().getUserName(), response.body().getAccess_token());
                        return new Result.Success<>(loggedInUser);
                    }
                }



                @Override
                public void onFailure(Call<Post> call, Throwable t) {
                    Log.e(TAG, "Unable to submit post to API.");
                }
            });

        } catch (Exception e) {
            return new Result.Error(new IOException("Error logging in", e));
        }
    }
}

我发现至少有两个这样的问题，并且都具有几乎相同的回调解决方案，但我无法在我的案例中复制它。

提前致谢。

Answer 1

您必须添加自定义回调来处理这种情况。你不能在 onResponse() 和 onFailure() 中返回任何东西

public interface CustomCallback {

    void onSucess(Result<LoggedInUser> value);
    void onFailure();
}

像这样更新你的方法:

public void login(String username, String password,CustomCallback customCallback) {

        try {

            mAPIService = ApiUtils.getAPIService();

           mAPIService.login(username, password, "password").enqueue(new Callback<Post>() {
                @Override
                public void onResponse(Call<Post> call, Response<Post> response) {
                    if(response.isSuccessful()) {
                        LoggedInUser loggedInUser;
                        loggedInUser = new LoggedInUser(response.body().getUserName(),response.body().getUserName(), response.body().getAccess_token());
                        customCallback.onSucess(new Result.Success<>(loggedInUser));
                    }
                }


                @Override
                public void onFailure(Call<Post> call, Throwable t) {
                    Log.e(TAG, "Unable to submit post to API.");
                    customCallback.onFailure();
                }
            });

        } catch (Exception e) {
            e.printStactTrace();
        }
    }

您必须像下面这样调用您的 api:

login("user name","passwor",new CustomCallback(){
       @Override
       public void onSucess(Result<LoggedInUser> value){
      //do your success code here
     }
       @Override
       public void onFailure(){
      }

});