ProcessDialog 没有正确显示？

Question

这是我在 LoginActivity.java 中的函数。所以点击按钮我正在调用这个函数。

public  void postHttpRequest(String userId,String pass,TextView error){
        RequestClient reqClient = new RequestClient(LoginActivity.this);
        String AppResponse = null;
        try {
            url = "myurl";
            Log.d("URL", url);
            AppResponse = reqClient.execute().get();
            String status = ValidateLoginStatus.checkLoginStatus(AppResponse);
            Log.d("Status recived", status);

            if(status.equals("200")){
                saveInformation(userId,pass);
                startingActivity(HOST_URL);
            }else{
                error.setText("Incorrect UserName or Password");
            }
        } catch (Exception e) {
            Log.e("Exception Occured", "Exception is "+e.getMessage());
        }
    }

从这个函数中，我正在调用一个 AsynkTask 进行 Http 通信。所以当我收到响应时单击按钮，然后我的 processDialog 只打开一秒钟。我希望当我单击按钮时我的 processDialog 应该打开直到我得到回应

public class RequestClient extends AsyncTask<String, Void, String>{
    ProgressDialog pDialog;
    Context context;

    public RequestClient(Context c) {
        context = c;
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(context);
        pDialog.setMessage("Authenticating user...");
        pDialog.show();

    }

    @Override
    protected String doInBackground(String... aurl){
    String responseString="";
    DefaultHttpClient httpClient=new DefaultHttpClient();
    try {
         HttpClient client = new DefaultHttpClient();  
         HttpGet get = new HttpGet(LoginActivity.url);
         HttpResponse responseGet = client.execute(get);  
         HttpEntity resEntityGet = responseGet.getEntity();  
         if (resEntityGet != null) {  
             responseString = EntityUtils.toString(resEntityGet);
             Log.i("GET RESPONSE", responseString);
         }
    } catch (Exception e) {
        Log.d("ANDRO_ASYNC_ERROR", "Error is "+e.toString());
    }
    Log.d("ANDRO_ASYNC_ERROR", responseString);
     httpClient.getConnectionManager().shutdown();
     return responseString;

    }


    @Override
    protected void onPostExecute(String response) {
         super.onPostExecute(response); 
            if(pDialog!=null) 
            pDialog.dismiss();
    }
}

所以请建议我必须进行哪些更改才能使 processDialog 正确显示在设备的中心

Answer 1

//在进度条中添加样式

 protected void onPreExecute() {
            super.onPreExecute();
pDialog = new ProgressDialog(context);
pDialog.setProgressStyle(ProgressDialog.STYLE_SPINNER);
        pDialog.setMessage("Authenticating user..."); 

if (pDialog != null && !pDialog.isShowing()) {
                        pDialog.show();
                    }
}

Answer 2

AsyncTask return value only after using get() method

从上面的链接中绘图

调用AsyncTask的get()方法会阻塞主线程，等待返回结果。这有效地使使用 AsyncTask 成为同步操作，在这种情况下使用 AsyncTask 毫无意义。

我能想到使用 get() 方法的唯一原因是来自主 (UI) 线程以外的线程，尽管我想不出很多这样做的理由。

点击按钮

       RequestClient reqClient = new RequestClient(LoginActivity.this,new TheInterface() {
             @Override
             public void theMethod(String result) {
                 Log.i("Result  =",result); 
            }  
        });
       reqClient.execute(url); // no get(). pass url to doInBackground()      

在你的活动课上

    public interface TheInterface {  
    public void theMethod(String result);

       }
     }

异步任务

public class RequestClient extends AsyncTask<String, Void, String>{
    ProgressDialog pDialog;
    Context context;
    TheInterface listener; 
    public RequestClient(Context c,TheInterface listen) {
        context = c;
        listener = listen;   
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(context);
        pDialog.setMessage("Authenticating user...");
        pDialog.show();

    }

    @Override
    protected String doInBackground(String... aurl){
    String responseString="";
     HttpClient client; 
    try {
         HttpClient client = new DefaultHttpClient();  
         HttpGet get = new HttpGet(aurl[0]); // url 
         HttpResponse responseGet = client.execute(get);  
         HttpEntity resEntityGet = responseGet.getEntity();  
         if (resEntityGet != null) {  
             responseString = EntityUtils.toString(resEntityGet);
             Log.i("GET RESPONSE", responseString);
         }
    } catch (Exception e) {
        Log.d("ANDRO_ASYNC_ERROR", "Error is "+e.toString());
    }
    Log.d("ANDRO_ASYNC_ERROR", responseString);
     client.getConnectionManager().shutdown();
     return responseString;

    }


    @Override
    protected void onPostExecute(String response) {
         super.onPostExecute(response); 
              pDialog.dismiss();
              if (listener != null) 
              {
                listener.theMethod(result);
              }

    }
}

Answer 3

您的按钮代码似乎不正确，因为它是异步的，但您正在尝试将其用作标准同步代码。

尝试将此代码移动到 onPostExecute：

String status = ValidateLoginStatus.checkLoginStatus(response);
            Log.d("Status recived", status);

            if(status.equals("200")){
                saveInformation(userId,pass);
                startingActivity(HOST_URL);
            }else{
                error.setText("Incorrect UserName or Password");
            }

并让这个按钮点击代码：

public  void postHttpRequest(String userId,String pass,TextView error){
        RequestClient reqClient = new RequestClient(LoginActivity.this);
        String AppResponse = null;
        try {
            url = "myurl";
            Log.d("URL", url);
            reqClient.execute();
        } catch (Exception e) {
            Log.e("Exception Occured", "Exception is "+e.getMessage());
        }
    }