Python urllib库不适用于http

如何解决Python urllib库不适用于http

我认为我在以下链接中找到了答案； Python urllib doesn't open any http URL

但是该页面上的答案并未清楚说明要解决此问题需要做什么。

我正在尝试通过Jupyter Notebook在Python 3.7.4版本中运行此代码；

from urllib.request import urlopen

html = urlopen('http://pythonscraping.com/pages/page1.html')
print(html.read())

然后我收到此错误；

ValueError                                Traceback (most recent call last)
C:\Anaconda3\lib\http\client.py in _get_hostport(self,host,port)
    886                 try:
--> 887                     port = int(host[i+1:])
    888                 except ValueError:

ValueError: invalid literal for int() with base 10: 'port'

During handling of the above exception,another exception occurred:

InvalidURL                                Traceback (most recent call last)
<ipython-input-2-b7a1a86203a2> in <module>
      1 from urllib.request import urlopen
      2 
----> 3 html = urlopen('http://pythonscraping.com/pages/page1.html')
      4 print(html.read())

C:\Anaconda3\lib\urllib\request.py in urlopen(url,data,timeout,cafile,capath,cadefault,context)
    220     else:
    221         opener = _opener
--> 222     return opener.open(url,timeout)
    223 
    224 def install_opener(opener):

C:\Anaconda3\lib\urllib\request.py in open(self,fullurl,timeout)
    523             req = meth(req)
    524 
--> 525         response = self._open(req,data)
    526 
    527         # post-process response

C:\Anaconda3\lib\urllib\request.py in _open(self,req,data)
    541         protocol = req.type
    542         result = self._call_chain(self.handle_open,protocol,protocol +
--> 543                                   '_open',req)
    544         if result:
    545             return result

C:\Anaconda3\lib\urllib\request.py in _call_chain(self,chain,kind,meth_name,*args)
    501         for handler in handlers:
    502             func = getattr(handler,meth_name)
--> 503             result = func(*args)
    504             if result is not None:
    505                 return result

C:\Anaconda3\lib\urllib\request.py in http_open(self,req)
   1343 
   1344     def http_open(self,req):
-> 1345         return self.do_open(http.client.HTTPConnection,req)
   1346 
   1347     http_request = AbstractHTTPHandler.do_request_

C:\Anaconda3\lib\urllib\request.py in do_open(self,http_class,**http_conn_args)
   1283 
   1284         # will parse host:port
-> 1285         h = http_class(host,timeout=req.timeout,**http_conn_args)
   1286         h.set_debuglevel(self._debuglevel)
   1287 

C:\Anaconda3\lib\http\client.py in __init__(self,port,source_address,blocksize)
    849         self._tunnel_headers = {}
    850 
--> 851         (self.host,self.port) = self._get_hostport(host,port)
    852 
    853         # This is stored as an instance variable to allow unit

C:\Anaconda3\lib\http\client.py in _get_hostport(self,port)
    890                         port = self.default_port
    891                     else:
--> 892                         raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
    893                 host = host[:i]
    894             else:

InvalidURL: nonnumeric port: 'port'

如果我尝试使用链接的 https 版本使用相同的代码，

from urllib.request import urlopen

html = urlopen('https://pythonscraping.com/pages/page1.html')
print(html.read())

然后它起作用了

b'<html>\n<head>\n<title>A Useful Page</title>\n</head>\n<body>\n<h1>An Interesting Title</h1>\n<div>\nLorem ipsum dolor sit amet,consectetur adipisicing elit,sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident,sunt in culpa qui officia deserunt mollit anim id est laborum.\n</div>\n</body>\n</html>\n'

我该怎么做才能使其也适用于 http ？

谢谢。

编辑：我也没有使用Anaconda尝试了此操作，并得到了相同的错误。然后，我在Anaconda中创建了另一个环境，并安装了最新版本的Python（当前为3.8.5），并再次遇到相同的错误。当我在在线解释器上尝试时，它可以工作。我想这与本地设置有关，但我无法弄清楚。

Python urllib库不适用于http

如何解决Python urllib库不适用于http

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