code: public InputStream open(String fileName) {
          try {
              URL urlToFile = new URL(url, fileName);
              return urlToFile.openStream();
          } catch (MalformedURLException e) {
              throw new RuntimeException(String.format("error while forming new URL from URL %s and filename %s", url.getPath(), fileName), e);
          } catch (IOException e) {
              return null;
          }
      }
docs: Tries to open the given file.
  A filename that starts with '/' is understood as an absolute path,
  i.e. the URLFileOpener forgets about its path.
  
  @param fileName the filename
  @return InputStream of the resource found at the given URL