i'm trying to open an url http://الاعلي-للاتصالات.قطر/ar/news-events/event/future-internet-privacy with the urllib2.urlopen but it reports always an error.
The similar occurs to http://الاعلي-للاتصالات.قطر/ar ... other pages (chinese ones) are opened ok.
Any ideas to point me to the right way to open those urls?
urllib2.urlopen("http://الاعلي-للاتصالات.قطر/ar/news-events/event/future-internet-privacy").read()
urllib2.urlopen('http://الاعلي-للاتصالات.قطر').read()
[Edited] the error is :
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.6/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib/python2.6/urllib2.py", line 391, in open
response = self._open(req, data)
File "/usr/lib/python2.6/urllib2.py", line 409, in _open
'_open', req)
File "/usr/lib/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/usr/lib/python2.6/urllib2.py", line 1170, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "/usr/lib/python2.6/urllib2.py", line 1142, in do_open
h.request(req.get_method(), req.get_selector(), req.data, headers)
File "/usr/lib/python2.6/httplib.py", line 914, in request
self._send_request(method, url, body, headers)
File "/usr/lib/python2.6/httplib.py", line 951, in _send_request
self.endheaders()
File "/usr/lib/python2.6/httplib.py", line 908, in endheaders
self._send_output()
File "/usr/lib/python2.6/httplib.py", line 780, in _send_output
self.send(msg)
File "/usr/lib/python2.6/httplib.py", line 759, in send
self.sock.sendall(str)
I also tried with the u'http://الاعلي-للاتصالات.قطر'.encode('utf-8') but the result url can't be opened too.
As @Donal says, the URL has to be punycoded. Luckily Python includes this already. Here is a sample Python code
domain = "الاعلي-للاتصالات.قطر"
domain_unicode = unicode(domain, "utf8")
domain_idna = domain_unicode.encode("idna")
urllib2.urlopen("http://" + domain_idna).read()
Hope this helps.
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