小编典典

urllib.error.URLError: ?

python

因此,我的代码只有4行。我正在尝试连接到一个网站,但此后我想做的事无关紧要,因为没有其他代码就出现了错误。

import urllib.request
from bs4 import BeautifulSoup

html=urllib.request.urlopen('http://python-data.dr-chuck.net/known_by_Fikret.html').read()
soup=BeautifulSoup(html,'html.parser')

和错误(简要总结一个):

for res in _socket.getaddrinfo(host, port, family, type, proto, flags):
socket.gaierror: [Errno 11002] getaddrinfo failed
During handling of the above exception, another exception occurred:
urllib.error.URLError: <urlopen error [Errno 11002] getaddrinfo failed>

这是我尝试过的。

  1. 我在Google上搜索了返回“ urlopen错误[Errno 11002]的错误”,尤其是在stackoverflow上,没有返回任何有用的信息(实际上,对此错误11002的询问不多)。
  2. 所以,然后我尝试用另一个网站“ http://www.pythonlearn.com/code替换urlopen函数中的网站参数(即“ http://python-data.dr-chuck.net/known_by_Fikret.html ”)/urllinks.py “。而且效果很好。没有错误出现。
  3. 因此,我想这个错误与该特定网站本身有关。该网站是动态的,我的意思是网站的内容将发生变化,并变成另一种完全不同的事物。但是我不仅仅了解描述我所见的知识。

以及错误的较长和完整版本:

Traceback (most recent call last):
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1240, in do_open
h.request(req.get_method(), req.selector, req.data, headers)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1083, in request
self._send_request(method, url, body, headers)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1128, in _send_request
self.endheaders(body)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1079, in endheaders
self._send_output(message_body)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 911, in _send_output
self.send(msg)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 854, in send
self.connect()
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 826, in connect
(self.host,self.port), self.timeout, self.source_address)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\socket.py", line 693, in create_connection
for res in getaddrinfo(host, port, 0, SOCK_STREAM):
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\socket.py", line 732, in getaddrinfo
for res in _socket.getaddrinfo(host, port, family, type, proto, flags):
socket.gaierror: [Errno 11002] getaddrinfo failed

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
File "D:/baiduyundownload/Tempo/Active/Python/Python Examples/Fileanalysis11111.py", line 4, in <module>
html=urllib.request.urlopen('http://python-data.dr-chuck.net/known_by_Fikret.html').read()
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 162, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 465, in open
response = self._open(req, data)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 483, in _open
'_open', req)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 443, in _call_chain
result = func(*args)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1268, in http_open
return self.do_open(http.client.HTTPConnection, req)
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1242, in do_open
raise URLError(err)
urllib.error.URLError: <urlopen error [Errno 11002] getaddrinfo failed>

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2021-01-20

共1个答案

小编典典

这意味着您的DNS系统无法正常工作,或者您必须在网络上使用代理并且未正确定义代理。

如果您需要使用代理,请将环境变量HTTP_PROXY(以及,可选HTTPS_PROXY)设置为网络的正确配置。格式为http://proxy.example.com:80;
如果您的代理服务器需要用户名和密码,则应将其传递,例如:http://username:password@proxy.example.com:80

对于DNS问题,请尝试从命令行查找域。打开命令提示符并键入,nslookup python-data.dr- chuck.net然后查看它是否返回给您IP地址。

2021-01-20