Python 3.4 urllib.request错误（http 403）

Question

我正试图打开并解析一个html页面。在Python2.7.8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'

没关系。

Answer 1

以下是我在学习python-3时收集到的一些笔记：
我留着它们，以防它们能派上用场或帮助别人。

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()

Answer 2

网站似乎不喜欢Python3.x的用户代理

指定User-Agent将解决您的问题：

import urllib.request
req = urllib.request.Request(url, headers={'User-Agent': 'Mozilla/5.0'})
html = urllib.request.urlopen(req).read()

注意Python 2.x urllib版本也接收403状态，但与Python 2.x urllib2和Python 3.x urllib不同，它不会引发异常。

您可以通过以下代码确认：

print(urllib.urlopen(url).getcode())  # => 403