Question

我无法使用urllib2打开一个特定网址。同样的方法适用于其他网站，例如“http://www.google.com”，但不适用于此网站（在浏览器中也能正常显示）。

我的简单代码：

from BeautifulSoup import BeautifulSoup
import urllib2

url="http://www.experts.scival.com/einstein/"
response=urllib2.urlopen(url)
html=response.read()
soup=BeautifulSoup(html)
print soup

任何人都可以帮助我让它发挥作用吗？

这是我得到的错误：

Traceback (most recent call last):
  File "/Users/jontaotao/Documents/workspace/MedicalSchoolInfo/src/AlbertEinsteinCollegeOfMedicine_SciValExperts/getlink.py", line 12, in <module>
    response=urllib2.urlopen(url);
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
    response = meth(req, response)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 432, in error
    result = self._call_chain(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 619, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
    response = meth(req, response)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 438, in error
    return self._call_chain(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 521, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: Not Found

谢谢

Answer 1

我刚尝试了这个并收到了404代码和页面。

猜测它正在进行用户代理检测，无论是偶然还是故意都不会向python urllib提供内容。

澄清，urllib，我收到urlopen返回的响应对象，其中包含404代码和HTML内容。 urllib2.urlopen urllib2.HTTPError引发了{{1}}例外。

我建议你尝试将用户代理设置为看起来像浏览器的东西。这里有一个问题：Changing user agent on urllib2.urlopen

Answer 2

您可以使用try except捕获错误

try:
    u = urllib2.urlopen(req)
except urllib2.HTTPError, e:
    print e.code
    print e.msg
    return

Answer 3

嗯...你确定这个URL有效吗？尝试“http://www.google.com”我有类似的代码，urllib没有问题。或者您可以使用try - except语句查看错误的详细信息。当然，MattH的答案非常类似于真相:)

对于在浏览器中显示正常的网站，urllib2返回404

3 个答案: