在Python中，urllib2給出錯誤

Question

我嘗試運行這個

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

但這給了這樣的錯誤，有人可以告訴我解決方案嗎？

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

雙重檢查域是否可訪問。

我現在在域tycho.usno.navy.mil收到504網關超時錯誤。

看起來該網站已關閉，也downforeveryoneorjustme.com表示

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

這就是為什么getaddrinfo失敗

Answer 2

包裝try..except可以使它保持整潔：

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