open url from pythonanywhere

Question

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.codefor         


        

Answer 1

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.

Answer 2

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.

Answer 3

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?"