-1

我是 Android 应用程序开发的新手。请找到我的 AsyncTask 代码，用于在用户单击按钮时连接 URL。

    package in.mosto.geeglobiab2bmobile;

    import java.io.IOException;
    import java.net.HttpURLConnection;
    import java.util.ArrayList;
    import java.util.List;

    import org.apache.http.HttpResponse;
    import org.apache.http.NameValuePair;
    import org.apache.http.StatusLine;
    import org.apache.http.client.ClientProtocolException;
    import org.apache.http.client.HttpClient;
    import org.apache.http.client.entity.UrlEncodedFormEntity;
    import org.apache.http.client.methods.HttpPost;
    import org.apache.http.impl.client.DefaultHttpClient;
    import org.apache.http.message.BasicNameValuePair;
    import org.apache.http.util.EntityUtils;

    import android.os.AsyncTask;

    public class OnbuttonclickHttpPost extends AsyncTask<String, String, String> {
        @Override
        protected String doInBackground(String... params) {
            byte[] result = null;
            String str = "";
           // Create a new HttpClient and Post Header
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://URL_HERE/login.php");

            try {
                    // Add your data
                    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
                    nameValuePairs.add(new BasicNameValuePair("mobile", params[0]));
                    nameValuePairs.add(new BasicNameValuePair("password", params[1]));
                    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

                    // Execute HTTP Post Request
                    HttpResponse response = httpclient.execute(httppost);
                    StatusLine statusLine = response.getStatusLine();
                    if(statusLine.getStatusCode() == HttpURLConnection.HTTP_OK){
                    result = EntityUtils.toByteArray(response.getEntity());
                    str = new String(result, "UTF-8");
                }
              } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
            } catch (IOException e) {
                // TODO Auto-generated catch block
            }  
            return str;
        }

        /**
         * on getting result
         */
        @Override
        protected void onPostExecute(String result) {
            // something with data retrieved from server in doInBackground
            //Log.v("Response ", result);

            MainActivity main = new MainActivity();

            if(result.equals("NO_USER_ERROR")){
                main.showNewDialog("There is no user existed with the given details.");
            }else if(result.equals("FIELDS_ERR")){
                main.showNewDialog("Please enter username and password.");
            }else{
                main.startRecharge(result);
            }
        }
    }

请参阅我的 MainActivity 方法：

    OnbuttonclickHttpPost t = new OnbuttonclickHttpPost();
    t.execute(mobile.getText().toString(),pass.getText().toString());

public void startRecharge(String param){
    Intent inte = new Intent(MainActivity.this.getBaseContext(), StartRecharge.class);
    startActivity(inte);
}

public void showNewDialog(String alert){
    new AlertDialog.Builder(this)
        .setTitle("Please correct following error")
        .setMessage(alert)
        .setPositiveButton("Okay", new DialogInterface.OnClickListener() {

            @Override
            public void onClick(DialogInterface dialog, int which) {
                // TODO Auto-generated method stub
                dialog.dismiss();
            }
        })
        .show();
}

在这里我得到一个错误。我不知道我的代码有什么问题。谁能帮帮我吗 ？

4

3 回答 3

4

这是错误的。您不应该创建活动类的实例。

 MainActivity main = new MainActivity();

活动由 启动startActivity。

您可以使您的 asynctask 成为您的活动类的内部类并在onPostExecute

或者使用接口

如何从 AsyncTask 返回布尔值？

编辑

   new OnbuttonclickHttpPost(ActivityName.this).execute(params);

在你的 asyctask

  TheInterface listener;

在建设者

  public OnbuttonclickHttpPost(Context context)
{

    listener = (TheInterface) context;
    c= context;

}

界面

  public interface TheInterface {

    public void theMethod(String result);

     }

在 onPostExecute

    if (listener != null) 
  {
      listener.theMethod(result);
      }

在您的活动类中实现接口

     implements OnbuttonclickHttpPos.TheInterface

实现方法

     @Override
     public void theMethod(STring result) {
   // update ui using result
     }
于 2013-09-10T15:42:55.103 回答
2

这是错误的做法。最好的方法是，如果你AsyncTask在其他任何地方都不需要它，那么让它成为你的Activity. 然后它将可以访问您的Activity和函数的成员变量，以便它可以轻松地更新您的onPostExecute().

如果要将其保存为单独的文件，则可以使用 aninterface来监听回调。

这个答案在答案底部有一个界面示例。

于 2013-09-10T15:43:28.960 回答
0

只需更改您的 onPostExecute()

@Override
protected void onPostExecute(String result) {
    parseResponse(result);      
}

然后你有

private void parseResponse(String result) {
    if(result.equals("NO_USER_ERROR")){
        showNewDialog("There is no user existed with the given details.");
    }else if(result.equals("FIELDS_ERR")){
        showNewDialog("Please enter username and password.");
    }else{
        startRecharge(result);
    }
}

这不是什么都不做

MainActivity main = new MainActivity();

于 2013-09-10T15:46:54.877 回答