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

PythonHttpUrllib2

Python Problem Overview


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 Files\Wing IDE 4.0\src\debug\tserver\_sandbox.py", line 1, in <module>
    # Used internally for debug sandbox under external interpreter
  File "c:\Python26\Lib\urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "c:\Python26\Lib\urllib2.py", line 397, in open
    response = meth(req, response)
  File "c:\Python26\Lib\urllib2.py", line 510, in http_response
    'http', request, response, code, msg, hdrs)
  File "c:\Python26\Lib\urllib2.py", line 435, in error
    return self._call_chain(*args)
  File "c:\Python26\Lib\urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "c:\Python26\Lib\urllib2.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?

Python Solutions


Solution 1 - Python

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())

Solution 2 - Python

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. "

Solution 3 - Python

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

Solution 4 - Python

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?

Solution 5 - Python

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

Solution 6 - Python

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

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRam RachumView Question on Stackoverflow
Solution 1 - PythonJochen RitzelView Answer on Stackoverflow
Solution 2 - PythonS.LottView Answer on Stackoverflow
Solution 3 - PythonEliView Answer on Stackoverflow
Solution 4 - PythonChris FosterView Answer on Stackoverflow
Solution 5 - PythonHello WorldView Answer on Stackoverflow
Solution 6 - PythonPhilView Answer on Stackoverflow