Question

我正在尝试进行JSON AsynTack服务调用以登录我的用户。不幸的是，它在openConnection上崩溃并显示错误消息：执行doInBackground（）时发生错误。我是一名试图学习Android的iOS开发人员，所以如果我遗漏了一些简单的东西，我会道歉。

 public class UserLoginTask extends AsyncTask<Void, Void, Boolean> {

        private final String mEmail;
        private final String mPassword;

        HttpURLConnection urlConnection = null;
        BufferedReader reader = null;

        UserLoginTask(String email, String password) {
            mEmail = email;
            mPassword = password;
        }


        @Override
        protected Boolean doInBackground(Void... params) {
            ConnectivityManager cm = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
            NetworkInfo info = cm.getActiveNetworkInfo();

            if(info==null || !info.isConnected()) {
                return false;
            }
                try {

                    URL url = new URL("http://urlForLoginHere");
                    urlConnection = (HttpsURLConnection) url.openConnection();
                    urlConnection.setReadTimeout(10000);
                    urlConnection.setConnectTimeout(15000);
                    urlConnection.setRequestMethod("POST");
                    urlConnection.setDoInput(true);
                    urlConnection.setDoOutput(true);

                    Uri.Builder builder = new Uri.Builder()
                            .appendQueryParameter("username", mEmail)
                            .appendQueryParameter("password", mPassword);
                    String query = builder.build().getEncodedQuery();

                    OutputStream os = urlConnection.getOutputStream();
                    BufferedWriter writer = new BufferedWriter(
                            new OutputStreamWriter(os, "UTF-8"));
                    writer.write(query);
                    writer.flush();
                    writer.close();
                    os.close();

                    urlConnection.connect();


                    // Read the input stream into a String
                    InputStream inputStream = urlConnection.getInputStream();
                    StringBuffer buffer = new StringBuffer();
                    if (inputStream == null) {
                        // Nothing to do.
                        return false;
                    }
                   // reader = new BufferedReader(new InputStreamReader(inputStream));

                    if (buffer.length() == 0) {
                        // Stream was empty.  No point in parsing.
                        return false;
                    }

                    String jsonString = buffer.toString();


                    urlConnection.setConnectTimeout(3000);
                    urlConnection.connect();
                    if (urlConnection.getResponseCode() == 200) {
//                        try {
//
//                        }
                        return true;
                    }
                } catch (MalformedURLException e1) {
                    // TODO Auto-generated catch block
                    e1.printStackTrace();
                } catch (IOException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            return  false;
        }

Answer 1

不确定这是否是原因，但您的urlConnection语句似乎不正确。请尝试使用以下代码：

    URL url = new URL("http://urlForLoginHere");
    HttpsURLConnection urlConnection = (HttpsURLConnection) url.openConnection();
    urlConnection.setReadTimeout(10000);
    urlConnection.setConnectTimeout(15000);
    urlConnection.setRequestMethod("POST");
    urlConnection.setDoInput(true);
    urlConnection.setDoOutput(true);

    Uri.Builder builder = new Uri.Builder()
            .appendQueryParameter("username", mEmail)
            .appendQueryParameter("password", mPassword);
    String query = builder.build().getEncodedQuery();

    OutputStream os = urlConnection.getOutputStream();
    BufferedWriter writer = new BufferedWriter(
            new OutputStreamWriter(os, "UTF-8"));
    writer.write(query);
    writer.flush();
    writer.close();
    os.close();

    // reader = new BufferedReader(new InputStreamReader(inputStream));

    String jsonString = buffer.toString();

    if (urlConnection.getResponseCode() == 200) {
        InputStream inputStream = urlConnection.getInputStream();
        StringBuffer buffer = new StringBuffer();
        // do stuff here
        return true;
    }
    else {
        InputStream inputStream = urlConnection.getErrorStream();
        // :(
    }

应用程序崩溃在HttpsURLConnection上openConnection（）应用程序可能在其主线程上做了太多工作

1 个答案: