请问如果在plain java里遇到connect exception,如何实现retry机制？以及如何将用户名密码隐藏起来？没有使用spring框架

代码如下，希望如果发送POST Request失败后可以重新发送，以及如何将用户密码等敏感信息隐藏起来，而不是直接码在代码里，没有使用spring框架。

 public void sendNotification() {
  try {
   URL url = new URL ("https://hello/api");
   String Username = "abc";
   String Password = "12345";//怎样可以将密码隐藏起来？
   HttpURLConnection con = (HttpURLConnection)url.openConnection();

   HashMap<String, String> in = new HashMap<>();
   input.put("short_description", "Hello");
   input.put("title", "Testing");

   String jsonStr = JSONValue.toJSONString(input);

   con.setRequestMethod("POST");
   con.setRequestProperty("Content-Type", "application/json; utf-8");
   con.setRequestProperty("Accept", "application/json");
   con.setDoOutput(true);

   String auth = Username + ":" + Password;
   byte[] encodedAuth = Base64.encodeBase64(auth.getBytes(StandardCharsets.UTF_8));
   String authHeaderValue = "Basic " + new String(encodedAuth);
   con.setRequestProperty("Authorization", authHeaderValue);

   try(OutputStream os = con.getOutputStream()) {
    byte[] inputStream = jsonStr.getBytes("utf-8");
    os.write(inputStream, 0, inputStream.length);
   }

   if (con.getResponseCode() == HttpURLConnection.HTTP_OK) {
    try(BufferedReader br = new BufferedReader(
      new InputStreamReader(con.getInputStream(), "utf-8"))) {
     StringBuilder response = new StringBuilder();
     String responseLine = null;
     while ((responseLine = br.readLine()) != null) {
      response.append(responseLine.trim());
     }
     LOG.info("Response:" + response.toString());
    }
   } else {
    LOG.error("Failed to send notification.");
    //should retry
   }
  } catch (Exception e) {
   LOG.error("Failed to send notification.");
   e.printStackTrace();
   //should retry
  }
 }

把下面的语句放在循环语句里面，用连接状态作为循环条件，或者设置循环次数

HttpURLConnection con = (HttpURLConnection)url.openConnection();