在Python中，urllib2给出错误

Question

我尝试运行这个

>>> urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl')

但这给了这样的错误，有人可以告诉我解决方案吗？

Traceback (most recent call last):
  File "<pyshell#11>", line 1, in <module>
    urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl')
  File "C:\Python26\lib\urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python26\lib\urllib2.py", line 391, in open
    response = self._open(req, data)
  File "C:\Python26\lib\urllib2.py", line 409, in _open
    '_open', req)
  File "C:\Python26\lib\urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "C:\Python26\lib\urllib2.py", line 1161, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "C:\Python26\lib\urllib2.py", line 1136, in do_open
    raise URLError(err)
URLError: <urlopen error [Errno 11001] getaddrinfo failed>

Answer 1

双重检查域是否可访问。

我现在在域tycho.usno.navy.mil收到504网关超时错误。

看起来该网站已关闭，也downforeveryoneorjustme.com表示

不只是你！ http://tycho.usno.navy.mil从这里向下看。

这就是为什么getaddrinfo失败

Answer 2

包装try..except可以使它保持整洁：

try:
    urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl')
except URLError:
    print "Error opening URL"