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

Question

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())

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.

Answer 1

Add 'r' in starting of path:

path = r"D:\Folder\file.txt"

That works for me.

Answer 2

That is not a valid file path. You must either use a full path

open(r"C:\description_files\program_description.txt","r")

Or a relative path

open("program_description.txt","r")

Answer 3

I also ran into this fault when I used open(file_path). My reason for this fault was that my file_path had a special character like "?" or "<".

Answer 4

I received the same error when trying to print an absolutely enormous dictionary. When I attempted to print just the keys of the dictionary, all was well!

Answer 5

In my case, I was using an invalid string prefix.

Wrong:

path = f"D:\Folder\file.txt"

Right:

path = r"D:\Folder\file.txt"