I am having trouble finding solid examples of how to build a simple script in urllib3 which opens a url (via a proxy), then reads it and finally prints it. The proxy requires a user/pass to authenticate however it's not clear to me how you do this? Any help would be appreciated.
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I don't know the library, but is it reasonably similar to [Proxy with urllib2](http://stackoverflow.com/q/1450132) ? – Nathan Tuggy Jul 01 '15 at 03:12
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1Not really, I have done this will the urllib2 library already. I am interested to see how this is done with urllib3. – Tom Jul 01 '15 at 03:21
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use `requests`? http://stackoverflow.com/questions/8287628/proxies-with-python-requests-module – maxymoo Jul 01 '15 at 07:15
2 Answers
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urllib3 has a ProxyManager
component which you can use. You'll need to build headers for the Basic Auth component, you can either do that manually or use the make_headers
helper in urllib3.
All together, it would look something like this:
from urllib3 import ProxyManager, make_headers
default_headers = make_headers(proxy_basic_auth='myusername:mypassword')
http = ProxyManager("https://myproxy.com:8080/", proxy_headers=default_headers)
# Now you can use `http` as you would a normal PoolManager
r = http.request('GET', 'https://stackoverflow.com/')
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Is the "headers" the login for the proxy, or can it be used to log into a website as well? – SPYBUG96 Dec 05 '17 at 16:12
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1@SPYBUG96 It's for the proxy. You can make separate headers for the website itself. – shazow Dec 05 '17 at 20:11
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This only worked for me when I changed "headers=default_headers" to "proxy_headers=default_headers". See https://stackoverflow.com/questions/58131573/urllib3-proxymanager-is-giving-authentication-error – Kyle Apr 03 '20 at 01:54
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I believe the correct answer to this should be
from urllib3 import ProxyManager, make_headers
default_headers = make_headers(proxy_basic_auth='myusername:mypassword')
http = ProxyManager("https://myproxy.com:8080/", headers=default_headers)
# Now you can use `http` as you would a normal PoolManager
r = http.request('GET', 'https://stackoverflow.com/')
(note: proxy_basic_auth, not basic_auth)
I was trying this with basic_auth in my environment without any luck. shazow you committed this comment to git which pointed me in the right direction

Jean-François Fabre
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