OSError [Errno 22] invalid argument when use open() in Python

def choose_option(self):
        if self.option_picker.currentRow() == 0:
            description = open(":/description_files/program_description.txt","r")
        elif self.option_picker.currentRow() == 1:
            requirements = open(":/description_files/requirements_for_client_data.txt", "r")
        elif self.option_picker.currentRow() == 2:
            menus = open(":/description_files/menus.txt", "r")

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.

Leave a Comment