18 
import urllib

print urllib.urlopen('http://www.reefgeek.com/equipment/Controllers_&_Monitors/Neptune_Systems_AquaController/Apex_Controller_&_Accessories/').read()

上面的脚本工作并返回预期的结果，同时：

import urllib2

print urllib2.urlopen('http://www.reefgeek.com/equipment/Controllers_&_Monitors/Neptune_Systems_AquaController/Apex_Controller_&_Accessories/').read()

引发以下错误：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.5/urllib2.py", line 124, in urlopen
    return _opener.open(url, data)
  File "/usr/lib/python2.5/urllib2.py", line 387, in open
    response = meth(req, response)
  File "/usr/lib/python2.5/urllib2.py", line 498, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.5/urllib2.py", line 425, in error
    return self._call_chain(*args)
  File "/usr/lib/python2.5/urllib2.py", line 360, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.5/urllib2.py", line 506, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: Not Found

有人知道为什么吗？我在没有代理设置的家庭网络上从笔记本电脑上运行它 - 只是直接从我的笔记本电脑到路由器，然后到 www。

4

1 回答 1

35

该 URL 确实会导致 404，但包含大量 HTML 内容。urllib2正在（正确地）将其作为错误条件进行处理。您可以像这样恢复该站点的 404 页面的内容：

import urllib2
try:
    print urllib2.urlopen('http://www.reefgeek.com/equipment/Controllers_&_Monitors/Neptune_Systems_AquaController/Apex_Controller_&_Accessories/').read()
except urllib2.HTTPError, e:
    print e.code
    print e.msg
    print e.headers
    print e.fp.read()
于 2009-12-22T15:50:59.593 回答