open url from pythonanywhere

This code works well on my local machine, but when I upload and run it on pythonanywhere.com it gives me this error.

My Code:

url = "http://www.codeforces.com/api/contest.list?gym=false"
hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
                'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
                'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
                'Accept-Encoding': 'none',
                'Accept-Language': 'en-US,en;q=0.8',
                'Connection': 'keep-alive'}
         req = urllib2.Request(url, headers=hdr)
         opener = urllib2.build_opener()
         openedReq = opener.open(req, timeout=300)

The error:

Traceback (most recent call last):
File "/home/GehadAbdallah/main.py", line 135, in openApi
    openedReq = opener.open(req, timeout=300)
  File "/usr/lib/python2.7/urllib2.py", line 410, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 523, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 448, in error
    return self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 531, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden

P.S. i’m working on python 2.7

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Free accounts on PythonAnywhere are restricted to a whitelist of sites, http/https only, and access goes via a proxy. There’s more info here:

PythonAnywhere wiki: “why do I get a 403 forbidden error when opening a url?”

Method 2

I recently used urllib2 with a flask project on pythonanywhere using their free account to access an api at donorschoose.org

This might be helpful,

@app.route('/funding')
def fundingByState():
    urllib2.install_opener(urllib2.build_opener(urllib2.ProxyHandler({'http': 'proxy.server:3128'})))
    donors_choose_url = "http://api.donorschoose.org/common/json_feed.html?historical=true&APIKey=DONORSCHOOSE"
    response = urllib2.urlopen(donors_choose_url)
    json_response = json.load(response)
    return json.dumps(json_response)

This does work.

Method 3

If you’re using paid account but still get this error message
try this pythonanywhere_forums

To me, I have to delete the console then restart a new one.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0