Android如何将数据从类传递到活动？

我在将数据（值）从java类传递到活动并在textView中显示时遇到问题。SharedReferences不起作用。 这是我课堂上的片段：

else if (type.equals("getUser")) {
            try {
                String user_name = params[1];
//                String password = params[2];
                URL url = new URL(getUser_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("user_name", "UTF-8") + "=" + URLEncoder.encode(user_name, "UTF-8") + "&"
//                        + URLEncoder.encode("password", "UTF-8") + "=" + URLEncoder.encode(password, "UTF-8");
                String post_data = URLEncoder.encode("user_name", "UTF-8") + "=" + URLEncoder.encode(user_name, "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();
                System.out.println(result);
                return result;
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return null;
@Override
    protected void onPostExecute(String result) {
        String temp = "Login success. Welcome!";

        if (result.equals(temp)) {
            Intent intent = new Intent(context, ProjectsActivity.class);
            context.startActivity(intent);
        } else if (result.contains("[{")) {

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

我想在活动中将我的结果传递给textview。此活动（UserAreaActivity）

非常感谢你的帮助

 SharedPreferences myprefs = getSharedPreferences("user", MODE_WORLD_READABLE);
    String username = myprefs.getString("username", null);
    String password = myprefs.getString("password", null); //Typo, you had used "username" here

在调用此函数之前，您没有将数据保存在SharedReferences中，因此您的用户名和密码都为null，并且您没有看到任何内容

您需要先保存数据，然后才能随时访问它们

 //Code for writing into shared preferences

  SharedPreferences myprefs = getSharedPreferences("user", MODE_WORLD_READABLE);
   myprefs.edit().putString("username","Name to be saved").commit();
   myprefs.edit().putString("password","Password to be saved").commit();

---

用于在不同的应用程序执行之间保存数据（当您关闭和打开应用程序时，您将能够访问数据，前提是您先保存数据）。

这是一个打字错误，用户名和密码都是相同的键吗？问了一个类似的问题。但我想问一个结果。它是来自mysql的json。不是密码和用户名。在json中有一些信息，比如名字、surrname、年龄等。我想获取这个json并进行解析，然后用这些信息填充TextView。但是我想获取一个结果。它是来自mysql的json。不是密码和用户名。在json中有诸如名称、surrname、年龄等信息。我想获取这个json并进行解析，然后用这些信息填充TextView。在这种情况下，您根本不需要SharedReference…因为您有一个后端数据库来保留信息。您需要从数据库中读取数据

 //Code for writing into shared preferences

  SharedPreferences myprefs = getSharedPreferences("user", MODE_WORLD_READABLE);
   myprefs.edit().putString("username","Name to be saved").commit();
   myprefs.edit().putString("password","Password to be saved").commit();