Python3.4 urllib.request错误(http 403)
Python3.4 urllib.request错误(http 403)
提问于 2015-02-08 23:57:27
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。然而，我想转到python3.4，在那里我得到了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, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden它适用于其他不使用https的URL。
url = 'http://www.stopforumspam.com/ipcheck/212.91.188.166'没问题。
回答 1
Stack Overflow用户
发布于 2016-04-11 01:18:57
以下是我在学习python-3时在urllib上收集的一些笔记：
我保存着它们，以防它们会派上用场，或者帮助别人。
如何导入urllib.request和urllib.parse
import urllib.request as urlRequest
import urllib.parse as urlParse如何发出GET请求：
url = "http://www.example.net"
# open the url
x = urlRequest.urlopen(url)
# get the source code
sourceCode = x.read()如何发出POST请求：
url = "https://www.example.com"
values = {"q": "python if"}
# encode values for the url
values = urlParse.urlencode(values)
# encode the values in UTF-8 format
values = values.encode("UTF-8")
# create the url
targetUrl = urlRequest.Request(url, values)
# open the url
x = urlRequest.urlopen(targetUrl)
# get the source code
sourceCode = x.read()如何发出POST请求(403 forbidden响应)：
url = "https://www.example.com"
values = {"q": "python urllib"}
# pretend to be a chrome 47 browser on a windows 10 machine
headers = {"User-Agent": "Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.106 Safari/537.36"}
# encode values for the url
values = urlParse.urlencode(values)
# encode the values in UTF-8 format
values = values.encode("UTF-8")
# create the url
targetUrl = urlRequest.Request(url = url, data = values, headers = headers)
# open the url
x = urlRequest.urlopen(targetUrl)
# get the source code
sourceCode = x.read()如何发出GET请求(403 forbidden响应)：
url = "https://www.example.com"
# pretend to be a chrome 47 browser on a windows 10 machine
headers = {"User-Agent": "Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.106 Safari/537.36"}
req = urlRequest.Request(url, headers = headers)
# open the url
x = urlRequest.urlopen(req)
# get the source code
sourceCode = x.read()页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/28396036
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