我是python新手,正在尝试提取页面的内容。当我执行urlopen('http://www.google.com')
操作时,我得到以下错误:
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib/python2.7/urllib2.py", line 391, in open
response = self._open(req, data)
File "/usr/lib/python2.7/urllib2.py", line 409, in _open
'_open', req)
File "/usr/lib/python2.7/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 1185, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "/usr/lib/python2.7/urllib2.py", line 1160, in do_open
raise URLError(err)
有什么解决办法吗?在
如果您的网络处于脱机状态,则会出现该错误消息
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