Python / Cryptography / Blowfish — Using Blowfish and b64decode to store encrypted data


I used this just the other day, with reference from its use in Satchmo payment processing to build a unique encrypted link for our affiliate clients.

I wanted them to have an easy, one click link that they can use to visit their information instead of having to remember a password (FAILS, unless service is important enough).

I used Blowfish in the Python Cryptography Toolkit.

import bas64from Crypto.Ciphers import Blowfish

encryption_object ="my-key-of-some-kind") # keep this key.. I used the django settings.SECRET_KEY

# requires multiple of 8 character input, so we must check if our string to encrypt is a multiple of 8 and add extra characters when it's short 
padding = ''
if (len(STRING_TO_ENCRYPT) % 8  ) <> 0    :  
    padding = 'X' * (8-(len(STRING_TO_ENCRYPT) % 8))  
    # len(STRING_TO_ENCRYPT) % 8 returns the remainder of division
    # subtracting this from 8 gives you the number needed to reach the multiple of 8.

encrypted_string = base64.b64encode(encryption_object.encrypt(STIRNG_TO_ENCRYPT+padding))
# b64encode to create a "more" url-safe string.


Now I can encrypt something like a slug and use it as a unique link that is only available to the person who I send the link to.

Another way I’ve done that is to use a 1 way hash: store a hash on the model of the slug and look up the DB object by hash.

I just didn’t want to add a new field and deal with updating DBs to add this feature.


One thought on “Python / Cryptography / Blowfish — Using Blowfish and b64decode to store encrypted data

  1. Instead of
    8 -(x % 8)
    to calculate how much to add to x to make it a multiple of 8, you can just use
    -x % 8

    Loose proof:

    Let 8*n +r = x, where 0 <= r < 8, n being integer.

    -x % 8 = (-8*n -r) % 8

    adding or subtracting a multiple of 8 from a number does not change the remainder, so

    -x % 8 = -r % 8
    = (8 -r) % 8 (doesn't simplify to 8-r due to case r=0)
    = 0 for r=0 OR 8-r for 0<r<8.

    x +(-x %8) =
    x when r=0
    8n +r +8-r = 8(n+1) for 0<r<8 {smallest multiple of 8 larger than x)

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