相关疑难解决方法(0)

我正在尝试打开并解析一个html页面.在python 2.7.8中我没有问题:

import urllib
url = "https://ipdb.at/ip/66.196.116.112"
html = urllib.urlopen(url).read()

一切都很好.但是我想转移到python 3.4并在那里得到HTTP错误403(禁止).我的代码:

import urllib.request
html = urllib.request.urlopen(url) # same URL as before

File "C:\Python34\lib\urllib\request.py", line 153, in urlopen
return opener.open(url, data, timeout)
File "C:\Python34\lib\urllib\request.py", line 461, in open
response = meth(req, response)
File "C:\Python34\lib\urllib\request.py", line 574, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python34\lib\urllib\request.py", line 499, in error
return self._call_chain(*args)
File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain
result = func(*args)
File "C:\Python34\lib\urllib\request.py", line 582, in http_error_default
raise HTTPError(req.full_url, code, …

我正在使用urllib.request.urlretrieve将文件下载到本地.

urllib.request.urlretrieve(url_string,file_name)

它抛出错误:

ssl.CertificateError了未处理由用户代码消息:主机名"foo.net"不匹配的任一"a248.e.akamai.net"," .akamaihd.net"," .akamaihd-staging.net",' . akamaized.net','. akamaized-staging.net'

如果您将网址复制到Chrome中,它会向您显示通知,您需要说"继续关注网址".