我有一列我正在使用urllib
来检查的网址。它的工作很好,直到它遇到阻止请求的网站。在这种情况下,我只想跳过它并继续到列表中的下一个URL。任何想法如何做到这一点?如何跳过一个在Python 3中给出HTTP 403错误代码的网站?
以下是完整的错误:
Traceback (most recent call last):
File "C:/Users/Goris/Desktop/ssser/link.py", line 51, in <module>
x = urllib.request.urlopen(req)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 223, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 532, in open
response = meth(req, response)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 570, in error
return self._call_chain(*args)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 504, in _call_chain
result = func(*args)
File "C:\Users\Goris\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 650, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden
什么是错误? – Ryan
urllib.error.HTTP错误:HTTP错误403:禁止 – Goriss
好的,请将这些信息编辑到您的问题中。你可能会想用'try' /'except'来捕获这个错误。如果您不知道如何将其应用到您的脚本,那么您也可能想要将相关代码编辑到您的问题中。 – Ryan