How can I unshorten a URL?

Question

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:

• Case 1: shortened --> unshortened
• Case 2: unshortened --> unshortened

e.g. bit.ly/silly in the input array should be google.com in the output array
e.g. google.com in the input array should be google.com in the output array

Answer 1

http://github.com/stef/urlclean

sudo pip install urlclean
urlclean.unshorten(url)

Answer 2

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

Answer 3

Unshorten.me has an api that lets you send a JSON or XML request and get the full URL returned.

Answer 4

Here a src code that takes into account almost of the useful corner cases:

• set a custom Timeout.
• set a custom User Agent.
• check whether we have to use an http or https connection.
• resolve recursively the input url and prevent ending within a loop.

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

Answer 5

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)

Answer 6

Using requests:

import requests

session = requests.Session()  # so connections are recycled
resp = session.head(url, allow_redirects=True)
print(resp.url)