Python urllib2 URLError 异常?

本文介绍了Python urllib2 URLError 异常?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我之前在 Windows XP 机器上安装了 Python 2.6.2 并运行以下代码:

I installed Python 2.6.2 earlier on a Windows XP machine and run the following code:

import urllib2 import urllib page = urllib2.Request('www.python/fish.html') urllib2.urlopen( page )

我收到以下错误.

Traceback (most recent call last):<br> File "C:\Python26\test3.py", line 6, in <module><br> urllib2.urlopen( page )<br> File "C:\Python26\lib\urllib2.py", line 124, in urlopen<br> return _opener.open(url, data, timeout)<br> File "C:\Python26\lib\urllib2.py", line 383, in open<br> response = self._open(req, data)<br> File "C:\Python26\lib\urllib2.py", line 401, in _open<br> '_open', req)<br> File "C:\Python26\lib\urllib2.py", line 361, in _call_chain<br> result = func(*args)<br> File "C:\Python26\lib\urllib2.py", line 1130, in http_open<br> return self.do_open(httplib.HTTPConnection, req)<br> File "C:\Python26\lib\urllib2.py", line 1105, in do_open<br> raise URLError(err)<br> URLError: <urlopen error [Errno 11001] getaddrinfo failed><br><br><br>

推荐答案

import urllib2 response = urllib2.urlopen('www.python/fish.html') html = response.read()

你做错了.