title: OSError [Errno 22] invalid argument when use open() in Python
body: <pre><code>def choose_option(self):
          if self.option_picker.currentRow() == 0:
              description = open(":/description_files/program_description.txt","r")
              self.information_shower.setText(description.read())
          elif self.option_picker.currentRow() == 1:
              requirements = open(":/description_files/requirements_for_client_data.txt", "r")
              self.information_shower.setText(requirements.read())
          elif self.option_picker.currentRow() == 2:
              menus = open(":/description_files/menus.txt", "r")
              self.information_shower.setText(menus.read())
  </code></pre>
  
  <p>I am using resource files and something is going wrong when i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine.</p>