Question

我一直在努力使用这段代码。为什么我收到以下错误？ 方法不会覆盖其超类中的方法这是代码：

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都显示相同的错误，onSuccess和onFailure也显示为灰色？

Answer 1

这是我的代码

<强> 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;
    }
}

这是我的LoginAcitivity中的方法

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);
        }
    });
}

这是用户点击按钮时调用的内容

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());
    }
}

希望这会对你有所帮助，告诉我这是否有效，因为我在发布之前编辑了这个。

Answer 2

这是我工作的解决方案，但它无法解决我的确切问题。我发布这个，因为我无法理解为什么这段代码有效但我的代码没有。

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);
}
}

这是另一个班级：

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";
        }
    });
}
}

那么为什么这段代码有效，但不是我的oringinal问题？

Answer 3

终于解决了我的问题。这不是代码，因为我正在做正确的事情。我只是将旧的jar文件保存在app / libs文件夹中。

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

删除android-async-1.4.3.jar文件，红线消失。

JsonHttpResponseHandler - 方法不会覆盖其超类中的方法

3 个答案: