Python's `urllib2`: Why do I get error 403 when I `urlopen` a Wikipedia page?

Question

I have a strange bug when trying to urlopen a certain page from Wikipedia. This is the page:

http://en.wikipedia.org/wiki/OpenCola_(drink)

This is the shell session:

>>> f = urllib2.urlopen('http://en.wikipedia.org/wiki/OpenCola_(drink)')
Traceback (most recent call last):
  File "C:Program FilesWing IDE 4.0srcdebugtserver_sandbox.py", line 1, in <module>
    # Used internally for debug sandbox under external interpreter
  File "c:Python26Liburllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "c:Python26Liburllib2.py", line 397, in open
    response = meth(req, response)
  File "c:Python26Liburllib2.py", line 510, in http_response
    'http', request, response, code, msg, hdrs)
  File "c:Python26Liburllib2.py", line 435, in error
    return self._call_chain(*args)
  File "c:Python26Liburllib2.py", line 369, in _call_chain
    result = func(*args)
  File "c:Python26Liburllib2.py", line 518, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden

This happened to me on two different systems in different continents. Does anyone have an idea why this happens?

Asked By: Ram Rachum

||

Source

Answer 1

Some websites will block access from scripts to avoid ‘unnecessary’ usage of their servers by reading the headers urllib sends. I don’t know and can’t imagine why wikipedia does/would do this, but have you tried spoofing your headers?

Answered By: Chris Foster

Answer 2

Often times websites will filter access by checking if they are being accessed by a recognised user agent. Wikipedia is just treating your script as a bot and rejecting it. Try spoofing as a browser. The following link takes to you an article to show you how.

http://wolfprojects.altervista.org/changeua.php

Answered By: Eli

Answer 3

To debug this, you’ll need to trap that exception.

try:
    f = urllib2.urlopen('http://en.wikipedia.org/wiki/OpenCola_(drink)')
except urllib2.HTTPError, e:
    print e.fp.read()

When I print the resulting message, it includes the following

“English

Our servers are currently experiencing
a technical problem. This is probably
temporary and should be fixed soon.
Please try again in a few minutes. “

Answered By: S.Lott

Answer 4

Wikipedias stance is:

Data retrieval: Bots may not be used
to retrieve bulk content for any use
not directly related to an approved
bot task. This includes dynamically
loading pages from another website,
which may result in the website being
blacklisted and permanently denied
access. If you would like to download
bulk content or mirror a project,
please do so by downloading or hosting
your own copy of our database.

That is why Python is blocked. You’re supposed to download data dumps.

Anyways, you can read pages like this in Python 2:

req = urllib2.Request(url, headers={'User-Agent' : "Magic Browser"}) 
con = urllib2.urlopen( req )
print con.read()

Or in Python 3:

import urllib
req = urllib.request.Request(url, headers={'User-Agent' : "Magic Browser"}) 
con = urllib.request.urlopen( req )
print(con.read())

Answered By: Jochen Ritzel

Answer 5

As Jochen Ritzel mentioned, Wikipedia blocks bots.

However, bots will not get blocked if they use the PHP api.
To get the Wikipedia page titled “love”:

http://en.wikipedia.org/w/api.php?format=json&action=query&titles=love&prop=revisions&rvprop=content

Answered By: Hello World

Answer 6

I made a workaround for this using php which is not blocked by the site I needed.

it can be accessed like this:

path='http://phillippowers.com/redirects/get.php? 
file=http://website_you_need_to_load.com'
req = urllib2.Request(path)
response = urllib2.urlopen(req)
vdata = response.read()

This will return the html code to you

Answered By: Phil

Python's `urllib2`: Why do I get error 403 when I `urlopen` a Wikipedia page?

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