所以我试图在python3中获取页面的URL ...
如果我执行以下操作,
from urllib.request import urlopen
html = urlopen("http://google.com/")
html.read()
我根据需要获得了HTML。 但是,如果我要选择不同的URL,如下所示,
from urllib.request import urlopen
html = urlopen("http://www.stackoverflow.com/")
html.read()
第二行行后出现以下错误:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 153, in urlopen
return opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 461, in open
response = meth(req, response)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 574, in http_response
'http', request, response, code, msg, hdrs)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 499, in error
return self._call_chain(*args)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 433, in _call_chain
result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 582, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden
为什么会发生这种情况以及如何解决这个问题?
答案 0 :(得分:2)
如果您仔细查看错误消息,您会发现它是一个HTTP错误和一个特殊错误:
HTTP Error 403: Forbidden
所以你和服务器谈过并得到你的回复,但你不知道为什么你被拒绝了。
您可以在服务器返回的HTML中获得更详细的消息,如下所示:
from urllib.request import urlopen
from urllib.error import HTTPError
try:
html = urlopen("http://www.stackoverflow.com/")
except HTTPError as e:
print(e.read().decode('utf-8'))
html.read()
对我而言,它说:
<h2 data-translate="what_happened">What happened?</h2>
<p>The owner of this website (www.stackoverflow.com) has banned your access based on your browser's signature (213702c58d2116a6-ua48).</p>
您可以将HTTPError
视为文件对象(https://docs.python.org/3/library/urllib.error.html#urllib.error.HTTPError):
虽然是异常(URLError的子类),但HTTPError可以 也可以作为一个非特殊的文件类返回值(相同 urlopen()返回的东西)。处理异国情调时这很有用 HTTP错误,例如身份验证请求。