java，如果我们在用户名和密码中键入错误的字段。它显示登录失败警报，但它正在调用下一个活动。要修这个吗？

Question

此代码用于我的登录活动

我使用了一个意图来调用onPostExecute()方法中的下一个页面，但它无法正常工作
如果我们在用户名中键入了错误的值&；通过，它将显示登录失败警报，但会调用下一个活动

如何解决这个问题

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

    Context context;
    AlertDialog alertDialog;    

    BackgroundWorker(Context ctx) {
        context = ctx;
    }

    @Override
    protected String doInBackground(String... params) {

        String type = params[0];    
        String login_url = "http://techblog.96.lt/login.php";

        if (type.equals("Login")) {

            try {
                String username = params[1];
                String pass = params[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("pass", "UTF-8") + "=" + URLEncoder.encode(pass, "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 Success");
    }

    @Override
    protected void onPostExecute(String result) {

            Intent i = new Intent(context, NewActivity.class);
            context.startActivity(i);

            alertDialog.setMessage(result);
            alertDialog.show();
        }

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

Answer 1

您应该使用onPostExecute方法。您需要检查doInBackground方法中的响应代码，并根据结果返回适当的String。然后在onPostExecute内检查来自doInBackground的响应并采取适当的措施

Answer 2

onPostExecute将始终启动下一个活动，而不管String result的值如何

您需要检查该值以确定是否成功登录。还要检查它是否为null，因为它也由doInBackground返回