Validate (X)HTML in Python

I think the most elegant way it to invoke the W3C Validation Service at

http://validator.w3.org/

programmatically. Few people know that you do not have to screen-scrape the results in order to get the results, because the service returns non-standard HTTP header paramaters

X-W3C-Validator-Recursion: 1
X-W3C-Validator-Status: Invalid (or Valid)
X-W3C-Validator-Errors: 6
X-W3C-Validator-Warnings: 0

for indicating the validity and the number of errors and warnings.

For instance, the command line

curl -I "http://validator.w3.org/check?uri=http%3A%2F%2Fwww.stalsoft.com"

returns

HTTP/1.1 200 OK
Date: Wed, 09 May 2012 15:23:58 GMT
Server: Apache/2.2.9 (Debian) mod_python/3.3.1 Python/2.5.2
Content-Language: en
X-W3C-Validator-Recursion: 1
X-W3C-Validator-Status: Invalid
X-W3C-Validator-Errors: 6
X-W3C-Validator-Warnings: 0
Content-Type: text/html; charset=UTF-8
Vary: Accept-Encoding
Connection: close

Thus, you can elegantly invoke the W3C Validation Service and extract the results from the HTTP header:

# Programmatic XHTML Validations in Python
# Martin Hepp and Alex Stolz
# mhepp@computer.org / alex.stolz@ebusiness-unibw.org

import urllib
import urllib2

URL = "http://validator.w3.org/check?uri=%s"
SITE_URL = "http://www.heppnetz.de"

# pattern for HEAD request taken from 
# http://stackoverflow.com/questions/4421170/python-head-request-with-urllib2

request = urllib2.Request(URL % urllib.quote(SITE_URL))
request.get_method = lambda : 'HEAD'
response = urllib2.urlopen(request)

valid = response.info().getheader('X-W3C-Validator-Status')
if valid == "Valid":
    valid = True
else:
    valid = False
errors = int(response.info().getheader('X-W3C-Validator-Errors'))
warnings = int(response.info().getheader('X-W3C-Validator-Warnings'))

print "Valid markup: %s (Errors: %i,