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我是新来的python,并试图提取页面的内容。当我做urlopen('http://www.google.com')
时,出现以下错误:urllib2.urlopen('你')给出错误
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)
对此的任何解决方案?
请正确缩进您的错误追溯。另外,请包含您的实际代码。 – 2011-06-03 10:15:40
错误信息是什么? – 2011-06-03 10:28:39
错误追溯开始时间短,单间隔,易于阅读的事情。为什么它在这里双倍间距? – 2011-06-03 11:04:25