Protecting against XSS

Aug 14, 2011 23:49

I wanted a module to strip out potential XSS injections.

I looked at the set of allowed HTML on the LJ post and was came up with this idea.

use BeautifulSoup to parse the submitted html, remove all tags that aren't in a safe html whitelist. And then for img & a tags process the url and require they start with an allowed set of protocols. The main downside is that
wont work. you have to list the http:

This seems like a good method for sanitizing user input while allowing some html -- but how can you really know you're protecting against all the possible ways to inject a hostile payload. There's some really funky techniques for tricking the browser at http://ha.ckers.org/xss.html



from BeautifulSoup import BeautifulSoup, Tag
import urlparse

SAFE_HTML = set(['a','img',
'b','big','blockquote','br',
'center','cite','code',
'dd','div','dl','dt',
'em','font','h1','h2','h3','hr',
'i','input',
'li','nobr','ol','option','p','pre',
's','small','span','strike','strong','sub','sup',
'table','td','th','tr','tt','u','ul'])
SAFE_URL_SCHEME = set(['ftp', 'gopher', 'http', 'https', 'mailto',
'svn', 'svn+ssh'])

def pasturize_html(body):
soup = BeautifulSoup(body)
pasturize_soup_contents(soup)
return unicode(soup)

def pasturize_soup_contents(contents):
for element in contents:
if isinstance(element, Tag):
if element.next is not None:
pasturize_soup_contents(element)
name = element.name.lower()
if name == 'a':
href = pasturize_url(element.get('href'))
if href is not None:
element['href'] = href
elif name == 'img':
src = pasturize_url(element.get('src'))
if src is not None:
element['src'] = src
elif name not in SAFE_HTML:
element.replaceWith('')

def pasturize_url(url):
if url is None:
return None

parts = urlparse.urlparse(url)
if parts.scheme.lower() not in SAFE_URL_SCHEME:
return ''
return urlparse.urlunparse(parts)

body = pasturize_html('''
">''')
assert body == '''
">'''

body = pasturize_html('''
''')
assert body == '''
">'''

python

Previous post Next post
Up