BASE64 as URL parameter

Problem
In a REST API, I wanted to pass a URL as a BASE64-encoded string, e.g. “http://host/api/v2/url/aHR0cHM6...“. It worked well for a while but I got an error for a URL. As it turned out, a BASE64 string can contain the “/” sign, and it caused the problem.

Solution
Replace the “+” and “/” signs with “-” and “_“, respectively. Fortunately, Python has functions for that (see here).

Here are my modified, URL-safe functions:

def base64_to_str(b64):
    return base64.urlsafe_b64decode(b64.encode()).decode()

def str_to_base64(s):
    data = base64.urlsafe_b64encode(s.encode())
    return data.decode()

You can also quote and unquote a URL instead of using BASE64:

>>> url = "https://www.youtube.com/watch?v=V6w24Lg3zTI"
>>>
>>> import urllib.parse
>>>
>>> new = urllib.parse.quote(url)
>>>
>>> new
>>> 'https%3A//www.youtube.com/watch%3Fv%3DV6w24Lg3zTI'    # notice the "/" signs!
>>>
>>> urllib.parse.quote(url, safe='')
>>> 'https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DV6w24Lg3zTI'    # no "/" signs!
>>>
>>> new = urllib.parse.quote(url, safe='')
>>>
>>> urllib.parse.unquote(new)
>>> 'https://www.youtube.com/watch?v=V6w24Lg3zTI'

Simplicity is the final achievement.

“Simplicity is the final achievement. After one has played a vast quantity of notes and more notes, it is simplicity that emerges as the crowning reward of art.”

Frederic Chopin

urllib.quote, urllib.unquote

>>> import urllib
>>> urllib.quote('hello world')
'hello%20world'
>>> urllib.unquote('hello%20world')
'hello world'

Inspiration

“I hear and I forget. I see and I remember. I do and I understand.” – Confucius

Ref.: https://beginnertomaster.wordpress.com/2011/03/30/inspirational-quotes/.