Question

我正在尝试打开网址http：//الاعلي-للاتصالات.قطر/ ar / news-events / event / future-internet-privacy 使用urllib2.urlopen但它总是报告错误。

类似的情况发生在http：//الاعلي-للاتصالات.قطر/ ar ...其他页面（中文页面）打开确定。

有什么想法指出我打开这些网址的正确方法吗？

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

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

将帖子错误是：

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)

我也尝试过使用u'http：//الاعلي-للاتصالات.قطر'.coding（'utf-8'），但结果网址也无法打开。

Answer 1

正如@Donal所说，网址必须是punycoded。幸运的是Python已经包含了这个。这是一个示例Python代码

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

希望这有帮助。

使用urllib2打开persian url域

1 个答案: