JsonHttpResponseHandler - 方法不会覆盖超类的方法

Question

我一直在为这段代码苦苦挣扎了一段时间。为什么我会收到跟随错误？ 方法不自其超 这里重写方法是代码：

public void CanSendPassword() { 
    asyncHttpClientPassword = new AsyncHttpClient(); 
    requestParamsPassword = new RequestParams(); 

    requestParamsPassword.put("email", mEmail); 
    asyncHttpClientPassword.post(BASE_URL, requestParamsPassword, new JsonHttpResponseHandler() 
    { 
     @Override 
     public void onSuccess(int statusCode, Header[] headers, JSONObject response) { 
      super.onSuccess(statusCode, headers, response); 
      jsonResponse = response.toString(); 
     } 

     @Override 
     public void onFailure(int statusCode, Header[] headers, Throwable throwable, JSONObject errorResponse) { 
      super.onFailure(statusCode, headers, throwable, errorResponse); 
      jsonResponse = "failed"; 
     } 
    } 
    ); 
} 

@override两者都出现了同样的错误和的onSuccess和onFailure处太灰色？

Answer 1

终于解决了我的问题。这不是我正在做的一切正确的代码。我只是让旧的jar文件仍在app/libs文件夹中。

这是我看到的错误。一切正常，但Override和super.onSuccess是红色下划线。

删除Android平台的异步-1.4.3.jar文件和红线消失。

Answer 2

这是我正在工作的解决方案，但它不能解决我确切的问题。我发布这个，因为我不明白为什么这个代码工作，但我的代码不。

public class MainActivity extends AppCompatActivity implements View.OnClickListener { 

EditText etSearchTerms; 
Button btnSearch; 
TextView tvSearchResults; 
MyLoopjTask myLoopjTask; 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 

    etSearchTerms = (EditText) findViewById(R.id.etSearchTerms); 
    btnSearch = (Button) findViewById(R.id.btnSearch); 
    tvSearchResults = (TextView) findViewById(R.id.tvSearchResults); 

    btnSearch.setOnClickListener(this); 

    myLoopjTask = new MyLoopjTask(); 
} 

@Override 
public void onClick(View v) { 
    String searchTerm = etSearchTerms.getText().toString(); 
    etSearchTerms.setText(""); 
    // make loopj http call 
    myLoopjTask.executeLoopjCall(searchTerm); 
} 
} 

这里是其它类：

public class MyLoopjTask { 


AsyncHttpClient asyncHttpClient; 
RequestParams requestParams; 

String BASE_URL = "https://team.mycompany.com/teambeta/LoginApp/Recovery/"; 
String jsonResponse; 

public MyLoopjTask() { 
    asyncHttpClient = new AsyncHttpClient(); 
    requestParams = new RequestParams(); 
} 

public void executeLoopjCall(final String queryTerm) { 

    requestParams.put("email", queryTerm); 
    asyncHttpClient.post(BASE_URL, requestParams, new JsonHttpResponseHandler() { 
     @Override 
     public void onSuccess(int statusCode, Header[] headers, JSONObject response) { 
      super.onSuccess(statusCode, headers, response); 
      jsonResponse = response.toString(); 
     } 

     @Override 
     public void onFailure(int statusCode, Header[] headers, Throwable throwable, JSONObject errorResponse) { 
      super.onFailure(statusCode, headers, throwable, errorResponse); 
      jsonResponse = "failed"; 
     } 
    }); 
} 
} 

那么，为什么此代码的工作，但不是在我的oringinal的问题吗？

Answer 3

这里是我的代码

TwitterRestClient

import android.content.Context; 
import com.loopj.android.http.*; 
import cz.msebera.android.httpclient.entity.StringEntity; 

public class TwitterRestClient { 
    private static final String BASE_URL = "https://www.example.com/api/"; 

    private static AsyncHttpClient client = new AsyncHttpClient(); 

    public static void get(String url, RequestParams params, AsyncHttpResponseHandler responseHandler) { 
    client.get(getAbsoluteUrl(url), params, responseHandler); 
    } 

    public static void post(String url, RequestParams params, AsyncHttpResponseHandler responseHandler) { 
    client.post(getAbsoluteUrl(url), params, responseHandler); 
    } 

    public static void post(Context ctx, String url, StringEntity entity, java.lang.String contentType, AsyncHttpResponseHandler responseHandler){ 
    client.post(ctx,getAbsoluteUrl(url),entity,contentType,responseHandler); 
    } 

    private static String getAbsoluteUrl(String relativeUrl) { 
     return BASE_URL + relativeUrl; 
    } 
} 

此方法在我LoginAcitivity

public void testPost(StringEntity entity) throws JSONException { 
    TwitterRestClient.post(getApplicationContext(),"api-auth/", entity,"application/json", new JsonHttpResponseHandler() { 

     @Override 
     public void onSuccess(int statusCode, cz.msebera.android.httpclient.Header[] headers, org.json.JSONArray response) { 
      // If the response is JSONObject instead of expected JSONArray 
      GlobalFunctions.ShowToast(getApplicationContext(),"test"); 
     } 

     @Override 
     public void onFailure(int statusCode, cz.msebera.android.httpclient.Header[] headers, java.lang.Throwable throwable, org.json.JSONArray errorResponse){ 
      GlobalFunctions.ShowToast(getApplicationContext(),"test1123"); 
     } 

     @Override 
     public void onFailure(int statusCode, cz.msebera.android.httpclient.Header[] headers, java.lang.Throwable throwable, org.json.JSONObject errorResponse){ 
      GlobalFunctions.ShowToast(getApplicationContext(),errorResponse.toString()); 
     } 

     @Override 
     public void onFailure(int statusCode, cz.msebera.android.httpclient.Header[] headers, java.lang.String responseString, java.lang.Throwable throwable){ 
      GlobalFunctions.ShowToast(getApplicationContext(),responseString); 
     } 
    }); 
} 

这就是我所说的，当用户点击这个按钮

public void signIn(View v){ 
    try { 

     String url = "/api-auth"; 

     JSONObject jsonParams = new JSONObject(); 

     jsonParams.put("username", "binsoi@gmail.com"); 
     jsonParams.put("password", "cornedbeef"); 

     StringEntity entity = new StringEntity(jsonParams.toString()); 
     client.post(context, url, entity, "application/json", 
       responseHandler); 

     testPost(entity); 


    } catch (Exception err) 
    { 
     GlobalFunctions.ShowToast(this, err.toString()); 
    } 
} 

希望这会帮助你，告诉我这是不是工作，因为我在发布之前编辑这个。