Question

由于某些原因，这不起作用。我确实得到了doInBackground的响应，并将其保存在结果中。我能够在doInBackground之后看到结果，并且收到了正确的结果，只是onPostExecute没有运行。

private class AttemptLogin extends AsyncTask<String, Void, Void>
    {
        String result;
        TextView error =(TextView)findViewById(R.id.errorlogin);
        @Override
        protected void onPreExecute() {}
        @Override
        protected Void doInBackground(String... params) {


            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(url);

            try {

                nameValuePairs.add(new BasicNameValuePair("username", params[0]));
                nameValuePairs.add(new BasicNameValuePair("password", params[1]));

                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                System.out.print(nameValuePairs);

                result = httpclient.execute(httppost, new BasicResponseHandler());
                System.out.print(result);


                //System.out.println("check");
                nameValuePairs.clear();




            } catch (ClientProtocolException e) {

                System.out.println("check1");


            } catch (IOException e) {

                System.out.println(e.getMessage());
            }

            return null;
        }
        @Override
        protected void onProgressUpdate(Void... values) {}
        @Override
        protected void onPostExecute(Void voids) {
            super.onPostExecute(voids);
            if (result.equals("Success"))
            {validate();
            System.out.print(result);}

            else if (result.equals("Failure")){
                System.out.print(result);
                error.setText("Don't match");
            }
            else{
                System.out.print("no response");
            }

        }






    }

Answer 1

我认为你必须将结果作为参数传递，就像它应该是：

检查此样本：

class AttemptLogin extends AsyncTask<String, Void, String>
{
    TextView error =(TextView)findViewById(R.id.errorlogin);

    @Override
    protected void onPreExecute() {}

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

        if (params[0].equals(""))
            return null;

        String result = null;
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(url);

        try {

            nameValuePairs.add(new BasicNameValuePair("username", params[0]));
            nameValuePairs.add(new BasicNameValuePair("password", params[1]));

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            System.out.print(nameValuePairs);

            result = httpclient.execute(httppost, new BasicResponseHandler());
            System.out.print(result);

            //System.out.println("check");
            nameValuePairs.clear();

            return result.trim();

        } catch (ClientProtocolException e) {

            System.out.println("check1");
            return null;

        } catch (IOException e) {

            System.out.println(e.getMessage());
            return null;
        }
    }
    @Override
    protected void onProgressUpdate(Void... values) {}

    @Override
    protected void onPostExecute(String result) {

        if (result.equals("Success"))
        {
            validate();
            System.out.print(result);}
        }
        else if (result.equals("Failure")){
            System.out.print(result);
            error.setText("Don't match");
        }
        else{
            System.out.print("no response");
        }
    }
}

在AsyncTask中没有调用onPostExecute

1 个答案: