Find where a t.co link goes to [closed]

Question

Given a "t.co" link, how can I find out what the link resolves to? For example, if I have "t.co/foo", I want a function or process that returns "domain.com/bar".

Answer 1

I would stay away from external APIs over which you have no control. That will simply introduce a dependency into your application that is a potential point of failure, and could cost you money to use.

CURL can do this quite nicely. Here's how I did it in PHP:

function unshorten_url($url) {
  $ch = curl_init($url);
  curl_setopt_array($ch, array(
    CURLOPT_FOLLOWLOCATION => TRUE,  // the magic sauce
    CURLOPT_RETURNTRANSFER => TRUE,
    CURLOPT_SSL_VERIFYHOST => FALSE, // suppress certain SSL errors
    CURLOPT_SSL_VERIFYPEER => FALSE, 
  ));
  curl_exec($ch); 
  return curl_getinfo($ch, CURLINFO_EFFECTIVE_URL);
}

I'm sure this could be adapted to other languages or even scripted with the curl command on UNIXy systems.

http://jonathonhill.net/2012-05-18/unshorten-urls-with-php-and-curl/

Answer 2

If you want to do it from the command line, curl's verbose option comes to the rescue:

curl -v <url>

gives you the HTTP reply. For t.co it seems to give you an HTTP/301 reply (permanently moved). Then, there's a Location field, which points to the URL behind the shortened one.

Answer 3

curl -s -o /dev/null --head -w "%{url_effective}\n" -L "https://t.co/6e7LFNBv"

• --head or -I only downloads HTTP headers
• -w or --write-out prints the specified string after the output
• -L or --location follows location headers

Answer 4

Here is a Python solution.

import urllib2

class HeadRequest(urllib2.Request):
    def get_method(self): return "HEAD"

def get_real(url):
    res = urllib2.urlopen(HeadRequest(url))
    return res.geturl()

Tested with an actual twitter t.co link:

url = "http://t.co/yla4TZys"
expanded = get_real(url)

expanded = http://twitter.com/shanselman/status/276958062156320768/photo/1

Wrap it up with a try-except and you are good to go.

Answer 5

Another Python solution, this time relying on the requests module instead of urllib2 (and all the rest of those libraries):

#!/usr/bin/env python

import requests

shorturl = raw_input("Enter the shortened URL in its entirety: ")
r = requests.get(shorturl)

print("""
The shortened URL forwards to:

    %s
""" % r.url)

Answer 6

Here is an R solution, ported from other answers in this thread, and from example() code of the RCurl Package:

unshorten_url <- function(uri){
        require(RCurl)
        if(RCurl::url.exists(uri)){
                # listCurlOptions()
                opts <- list(
                        followlocation = TRUE,  # resolve redirects
                        ssl.verifyhost = FALSE, # suppress certain SSL errors
                        ssl.verifypeer = FALSE, 
                        nobody = TRUE, # perform HEAD request
                        verbose = FALSE
                );
                curlhandle = getCurlHandle(.opts = opts)
                getURL(uri, curl = curlhandle)
                info <- getCurlInfo(curlhandle)
                rm(curlhandle)  # release the curlhandle!
                info$effective.url
        } else {
                # just return the url as-is
                uri
        }
}

Answer 7

Twitter expands the URL. Assume you have a single tweet using twitter API encoded as json file.

import json
urlInfo=[]

tweet=json.loads(tweet)
keyList=tweet.keys() # list of all posssible keys
tweet['entities'] # gives us values linked to entities 

You can observe that there is a value called 'urls' tweet['entities']['urls'] # gives values mapped to key urls

urlInfo=tweet['entities']['expanded_url'] # move it to a list
# iterating over the list.. gives shortened URL
# and expanded URL
for item in urlInfo:
  if "url" and "expanded_url" in urlInfo.keys():
    print(item["url"] + " "+item["expanded_url"])