Question

import urllib

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

以上脚本可以工作并返回预期结果：

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

有谁知道这是为什么？我在我的家庭网络上使用笔记本电脑运行它，没有代理设置 - 只需从我的笔记本电脑直接到路由器再到www。

Answer 1

该URL确实会产生404，但包含大量HTML内容。 urllib2正在将其（正确地）作为错误条件处理。您可以像这样恢复该网站的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()

urllib2.urlopen（）vs urllib.urlopen（） - urllib2在urllib工作时抛出404！为什么？

1 个答案: