On 28/09/10 01:50, Brian Jones wrote:
On Mon, Sep 27, 2010 at 11:43 AM, Brian Jones mailto:bkjo...@gmail.com>> wrote:
How about this:
d = [digits, punctuation, ascii_uppercase, ascii_lowercase]
s = 'asdf1234A'
for c in d:
if not [x for x in s if x in c]:
print x, ' not in ', c
Ju
On 28/09/10 01:46, Jerry Hill wrote:
The way you've written it obviously works fine. That being said, I'd
probably do something like this:
from string import ascii_lowercase, ascii_uppercase, digits, punctuation
def complex_password(password):
'''Checks to make sure a password is complex'
set does seem to have what you want: isdisjoint() could do the trick.
Eg:
if set(punctuation).isdisjoint(password) or
set(digits).isdisjoint(password) or set(ascii_uppercase).isdisjoint(password)
or set(ascii_lowercase).isdisjoint(password):
return False
return True
You co
I've got a small function that I'm using to check whether a password is
of a certain length and contains mixed case, numbers and punctuation.
Originally I was using multiple "if re.search" for the patterns but it
looked terrible so I've read up on list comprehensions and it's slightly
improved
