使用urllib2打开波斯网址域名

Question

我正在尝试用 urllib2.urlopen 打开一个网址：http://الاعلي-للاتصالات.قطر/ar/news-events/event/future-internet-privacy，但总是报错。

类似的情况也发生在 http://الاعلي-للاتصالات.قطر/ar 这个网址上……其他一些页面（比如中文的）可以正常打开。

有没有什么建议可以帮我找到打开这些网址的正确方法？

urllib2.urlopen("http://الاعلي-للاتصالات.قطر/ar/news-events/event/future-internet-privacy").read()

urllib2.urlopen('http://الاعلي-للاتصالات.قطر').read()

[编辑]

错误信息是：

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)

我也试过用 u'http://الاعلي-للاتصالات.قطر'.encode('utf-8')，但是这个网址也打不开。

Answer 1

正如@Donal所说，网址需要用一种叫做Punycode的方式进行编码。幸运的是，Python已经包含了这个功能。下面是一段示例的Python代码：

domain = "الاعلي-للاتصالات.قطر"
domain_unicode = unicode(domain, "utf8")
domain_idna = domain_unicode.encode("idna")
urllib2.urlopen("http://" + domain_idna).read()

希望这对你有帮助。