Template Literals Not Being read properly in Python and returning: Errno 22, Invalid Argument

I am writing a very simple function in Python where the user calls the function with one paramater: website, and the function simply makes an empty markdown file with the website name, for example www.example.com.md.

For this purpose I am trying to use a template literal using f'', however Python returns invalid argument when trying to use a template literal. When I hardcode the website 'www.example.com' it works fine.

So how do you get template strings to work in Python without returning an error?

    def fileCreator(website):
        outputFile = open(f'{website}.md', 'w')
    
        print(outputFile)

 fileCreator('https://www.example.com/')

Files in windows can't have a forward slash. You can use the re library in python to remove characters like spaces and slashes to make it a valid file name.

Edit: modified to store the original link as another var while also outputting the final modified file name.

import re

def fileCreator(website):
    original_url = website
    website = re.sub(r'[^\w\s-]', '-', website).strip().lower()
    website = re.sub(r'[-\s]+', '-', website)
    outputFile = open(f'{website}.md', 'w')
    print(outputFile)
    return original_url

original_url = fileCreator('https://www.example.com/')
print(original_url)

Output:

<_io.TextIOWrapper name='https-www-example-com-.md' mode='w' encoding='cp1252'>

The problem simply is that the file name contains forward slash ("/")
So either pass the website name without such character like that:

def fileCreator(website):
    outputFile = open(f'{website}.md', 'w')

    print(outputFile)

fileCreator('www.example.com')

Or remove them in the "fileCreator" function like that:

def fileCreator(website):
    website = website.replace("https://","")
    website = website.replace("/","")
    outputFile = open(f'{website}.md', 'w')

    print(outputFile)

fileCreator('https://www.example.com/')

Use the package urllib.parse from the standard library to encode and decode a url into a valid file format. It also provide a way to decode it back to a valid url.

import urllib.parse

url = "https://docs.python.org/3/library/urllib.parse.html"

# as valid path
url2path = urllib.parse.quote_plus(url)
print(url2path)
#https%3A%2F%2Fdocs.python.org%2F3%2Flibrary%2Furllib.parse.html

# as url
path2url = urllib.parse.unquote(url2path)
print(path2url)
#'https://docs.python.org/3/library/urllib.parse.html'

An url is made of different components, scheme, hostname, ... which are easily accessible as a named tuple of an instance of urllib.parse.ParseResult.

r = urllib.parse.urlparse(url) # ParseResult object
print(r)
#ParseResult(scheme='https', netloc='docs.python.org', path='/3/library/urllib.parse.html', params='', query='', fragment='')

print(r.hostname)
#docs.python.org

So if you care only about the "website name" use urlparse.hostname and I strongly discourage to use solutions based on direct string manipulation such as regex. Try first to use what the standard library offers for that task!