raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) - python-2.7

Question

I try to read this url - http://malc0de.com/bl/BOOT in Python
import urllib2
threats = urllib2.urlopen("http://malc0de.com/bl/BOOT")
But I got this error:
Traceback (most recent call last):
File "C:\Android\android_workspace\pro2\test.py", line 2, in <module>
threats = urllib2.urlopen("http://malc0de.com/bl/BOOT")
File "C:\Python27\lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 437, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 550, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 475, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 558, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
What can I do to fix it?

Answer 1

This is a HTTP error unrelated to python or urllib. It says that, for some reason, you are not allowed to view this particular page.
It seems to me that the site owner filters access by bots/crawlers, because I can open it in Firefox, but not via urllib. It might filter based on user agent, which may be changed, see Changing user agent on urllib2.urlopen, although this might be bad etiquette.