Avoiding TypeError: Incorrect padding with Python's base64 encoding

base64_padding.md

Avoiding padding errors with Python's base64 encoding

>>> import base64
>>> data = '{"u": "test"}'
>>> code = base64.b64encode(data)
>>> code
'eyJ1IjogInRlc3QifQ=='

Note the trailing == to make len a multiple of 4. This decodes properly

>>> len(code)
20
>>> base64.b64decode(code)
'{"u": "test"}'
>>> base64.b64decode(code) == data
True

without the == padding (this is how many things are encoded for e.g. access tokens)

>>> base64.b64decode(code[0:18]) == data
...
TypeError: Incorrect padding 

However, you can add back the padding

>>> base64.b64decode(code + "==") == data
True

Or add an arbitrary amount of padding (it will ignore extraneous padding)

>>> base64.b64decode(code + "========") == data
True

This last property of python's base64 decoding ensures that the following code adding 3 padding = will never succumb to the TypeError and will always produce the same result.

>>> base64.b64decode(code + "===") == data
True

It's clumsy but effective method to deal with strings from different implementations of base64 encoders

That is genuinely a horrible but beautiful thing. It saves so much extra and padding code!

Very thank you!

Very helpful!

Didn't feel quite right to pad more that needed so I ended up doing this

code = code.ljust((int)(math.ceil(len(code) / 4)) * 4, '=')
data = base64.b64decode(code)

A similar way to add the necessary padding:

code_with_padding = f"{code}{'=' * (len(code) % 4)}"

You saved my bacon buddy. Thank you.

A similar way to add the necessary padding:

code_with_padding = f"{code}{'=' * (len(code) % 4)}"

That is not quite right. Modulo alone isn't going to give you the number of padding characters you need. This works:

code_with_padding = f"{code}{'=' * ((4 - len(code) % 4) % 4)}"

Thank you!

I don't understand this theme

I don't know what black magic is this, but it is working. Thanks

Works like a charm