我试过运行这个,
>>> urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl')但是它给出了这样的错误,谁能告诉我一个解决方案?
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>发布于 2010-05-12 19:05:23
仔细检查域是否可访问。
目前,我在domain - tycho.usno.navy.mil这里收到504网关超时错误。
看起来网站宕机了,downforeveryoneorjustme.com也说
不仅仅是你!http://tycho.usno.navy.mil从这里往下看。
这就是getaddrinfo失败的原因
发布于 2010-05-12 19:48:11
包装在try..except中可以帮助保持整洁:
try:
urllib2.urlopen('http://tycho.usno.navy.mil/cgi-bin/timer.pl')
except URLError:
print "Error opening URL"https://stackoverflow.com/questions/2818098
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