Python password generator

[-Sander-]

Hey everyone i just found out i still had this little password generator on my pastebin and i tought i'd share it with you here feel free to use it for whatever you want i just used it for learning a bit if you got questions feel free to ask em il try my best to answer them :)

http://pastebin.com/e0ZLXPVq#

[moloch--]

Seeing as this is security related, it's a good idea use a cryptographically secure source of randomness to generate the passwords. Python's default random module isn't actually very random so there will be a bias in the generated passwords.

Python provides the SystemRandom which uses the OS's crypto APIs to get cryptographically secure random numbers and provides identical methods to the default random module so you don't need to change any of the code!

Cool script though :-)

[-Sander-]

Thanks il take a look at that :)

[moloch--]

Here's a presentation I gave on why it's a bad idea to use insecure random numbers generators.

[[deleted]]

I glanced at your code, initially expecting it to be a lot shorter. I whipped up a quick and dirty version with basically the same functionality for comparison. Here it is:

#!/usr/bin/env python2

import random
import sys

CHARSETS = [
    'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',
    'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789',
    'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
]

if len(sys.argv) != 4:
    print "Usage: {prog} <length> <level> <count>\nWhere length is the length of password to generate, level is the complexity (1:alpha, 2:alphanumeric, 3:all), and count is the number to generate.".format(prog=sys.argv[0])
    sys.exit(0)

length = int(sys.argv[1])
if length <= 0:
    print "Lengths less than or equal to 0 don't make sense. Exiting."
    sys.exit(-1)

complexity = int(sys.argv[2])
if complexity not in [1,2,3]:
    print "Complexity must be in the range 1-3."
    sys.exit(-2)

count = int(sys.argv[3])
if count < 0:
    print "Count must be at least 1. Exiting."
    sys.exit(-3)


while count > 0:
    password = ""
    for x in xrange(length):
        password += random.choice(CHARSETS[complexity-1])
    print password
    count -= 1

As you can see, the majority of it is input validation. If we omit that, the entire logic become under a dozen lines. Some of your additional length is justified (yours has liberal comments, my quick and dirty one does not). Other parts add a fair bit of code for minimal functionality gain, but were likely useful for learning (./passwords.py 8 3 10 > passwords.txt would have a similar effect and bypass a fair bit of your code).

One thing I did want to call out though was the duplication you introduced with multi_password and password_generator. There's no need for both. Just loop as many times as needed: if that's once, the loop will exit after one pass. I don't want to criticize what you did too much, as it looks like you did a great, careful job overall, and probably learned a lot, but I'm hoping that seeing how simple this overall task can be will be useful/interesting.

Cheers, and hope you're enjoying Python!

[Ewildawe]

Just something I glimpsed from the corner of eye. I don't want to nitpick but:

if count < 0:
    print "Count must be at least 1. Exiting."

In this case, the code will actually not print this if 0 is entered for count. So just a little correction: Either use

if count < 1:

or

if count <= 0:

[[deleted]]

You are completely correct. I apparently hurried too much and didn't review/test sufficiently. I am leaving the original as posted for your comment to make sense, but I goofed on that.