我试图通过php将android连接到mysql。我的代码中没有错误，但我想，是出了什么问题。我试图运行应用程序，但对话框的标题只显示。我认为 result 可能有问题吗？

public class BackgroundWorker extends AsyncTask<String,Void,String> {

    Context context;
    AlertDialog alertDialog;
    BackgroundWorker (Context ctx){
        context = ctx;
    }

    @Override
    protected String doInBackground(String[] voids) {
        String type = "login";

        String login_url = "http://192.168.254.109/ITSP/login.php";
        if(type.equals("login")){
            try {
                String username = voids[1];
                String password =voids[2];
                URL url = new URL(login_url);
                HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
                httpURLConnection.setRequestMethod("POST");
                httpURLConnection.setDoOutput(true);
                httpURLConnection.setDoInput(true);
                OutputStream outputStream = httpURLConnection.getOutputStream();
                BufferedWriter bufferedWriter =  new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
                String post_data = URLEncoder.encode("username", "UTF-8")+"="+URLEncoder.encode(username, "UTF-8")+"&"
                        +URLEncoder.encode("password", "UTF-8")+"="+URLEncoder.encode(password, "UTF-8");
                bufferedWriter.write(post_data);
                bufferedWriter.flush();
                bufferedWriter.close();
                outputStream.close();
                InputStream inputStream = httpURLConnection.getInputStream();
                BufferedReader bufferedReader =  new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
                String result = "";
                String line;
                while((line = bufferedReader.readLine()) != null){
                    result += line;
                }
                bufferedReader.close();
                inputStream.close();
                httpURLConnection.disconnect();
                return result;
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return null;
    }

    @Override
    protected void onPreExecute() {
        alertDialog = new AlertDialog.Builder(context).create();
        alertDialog.setTitle("Login Status");
    }

    @Override
    protected void onPostExecute(String result) {
        alertDialog.setMessage(result);
        alertDialog.show();
    }

    @Override
    protected void onProgressUpdate(Void... values) {
        super.onProgressUpdate(values);
    }
}

从这里得到价值：

public void OnLogin(View view){
    String username = UsernameEt.getText().toString();
    String password = PasswordEt.getText().toString();
    String type = "login";
    BackgroundWorker backgroundWorker = new BackgroundWorker(this);
    backgroundWorker.execute(type, username, password);
}

我刚刚使用了php连接的简单代码。 (conn.php) 当我按下 Log In 按钮，只有 alertDialog = "Login Status" 将显示。