I want to be able to take a shortened or non-shortened URL and return its un-shortened form. How can I make a python program to do this?
Additional Clarification:
http://github.com/stef/urlclean
sudo pip install urlclean
urlclean.unshorten(url)
Here a src code that takes into account almost of the useful corner cases:
The src code is on github @ https://github.com/amirkrifa/UnShortenUrl
comments are welcome ...
import logging
logging.basicConfig(level=logging.DEBUG)
TIMEOUT = 10
class UnShortenUrl:
def process(self, url, previous_url=None):
logging.info('Init url: %s'%url)
import urlparse
import httplib
try:
parsed = urlparse.urlparse(url)
if parsed.scheme == 'https':
h = httplib.HTTPSConnection(parsed.netloc, timeout=TIMEOUT)
else:
h = httplib.HTTPConnection(parsed.netloc, timeout=TIMEOUT)
resource = parsed.path
if parsed.query != "":
resource += "?" + parsed.query
try:
h.request('HEAD',
resource,
headers={'User-Agent': 'curl/7.38.0'}
}
)
response = h.getresponse()
except:
import traceback
traceback.print_exec()
return url
logging.info('Response status: %d'%response.status)
if response.status/100 == 3 and response.getheader('Location'):
red_url = response.getheader('Location')
logging.info('Red, previous: %s, %s'%(red_url, previous_url))
if red_url == previous_url:
return red_url
return self.process(red_url, previous_url=url)
else:
return url
except:
import traceback
traceback.print_exc()
return None
Send an HTTP HEAD request to the URL and look at the response code. If the code is 30x, look at the Location
header to get the unshortened URL. Otherwise, if the code is 20x, then the URL is not redirected; you probably also want to handle error codes (4xx and 5xx) in some fashion. For example:
# This is for Py2k. For Py3k, use http.client and urllib.parse instead, and
# use // instead of / for the division
import httplib
import urlparse
def unshorten_url(url):
parsed = urlparse.urlparse(url)
h = httplib.HTTPConnection(parsed.netloc)
h.request('HEAD', parsed.path)
response = h.getresponse()
if response.status/100 == 3 and response.getheader('Location'):
return response.getheader('Location')
else:
return url
Open the url and see what it resolves to:
>>> import urllib2
>>> a = urllib2.urlopen('http://bit.ly/cXEInp')
>>> print a.url
http://www.flickr.com/photos/26432908@N00/346615997/sizes/l/
>>> a = urllib2.urlopen('http://google.com')
>>> print a.url
http://www.google.com/
Using requests:
import requests
session = requests.Session() # so connections are recycled
resp = session.head(url, allow_redirects=True)
print(resp.url)
To unshort, you can use requests. This is a simple solution that works for me.
import requests
url = "http://foo.com"
site = requests.get(url)
print(site.url)