如何从 python 中受密码保护的 URL 下载文件？

Question

from urllib.request import Request, urlopen   
import base64  
import logging  
import urllib  
import urllib.request  
import requests  

def run(file_path,password):  
    if file_path:  
            try:  
                request = Request(file_path)  
                base64string = base64.b64encode('{}:{}'.format('', password).encode())  
                #input_file = tempfile.NamedTemporaryFile()  
                request.add_header("Authorization", "Basic %s" % base64string)  
                result = urlopen(request)  
                print (result)  
                ur = result.geturl()  
                a = ur.split("/")[:-1]   
                print (a)  
                ', '.join(a[0:])  
                url = '/'.join(a)  
                print (url)  
                b = ur + "/download"  
                filename = "myfile"   
                input_file = requests.get(b,filename)  
            except Exception as e:  
                print(e)  

l = run("https://cloud.abc.co.uk/s/523aX3O8B5uOWlP","test123")
print (l)

This is the URL https://cloud.abc.co.uk/s/523aX3O8B5uOWlP which is password protected. I have tried using urllib and base64 for authentication. This gives mehttps://cloud.abc.co.uk/s/523aX3O8B5uOWlP/authenticate. This URL redirects to https://cloud.abc.co.uk/s/523aX3O8B5uOWlP/download from where the file can be downloaded.

request = Request(file_path)  
        base64string = base64.b64encode('{}{}'.format('',password).encode())   
        request.add_header("Authorization", "Basic %s" % base64string)  
        result = urlopen(request)  

I want to download the file bypassing authentication credentials.

Answer 1

I had a same problem several weeks ago with JSON file. Tried something like this. Hope it works for you as well.

import urllib2
url = "url"
password_mgr = HTTPPasswordMgrWithDefaultRealm()
password_mgr.add_password(None, url, 'username', 'password')
handler = HTTPBasicAuthHandler(password_mgr)
opener = build_opener(handler)
response = opener.open(url)