How to specify an authenticated proxy for a python

2019-01-03 09:03发布

What's the best way to specify a proxy with username and password for an http connection in python?

6条回答
趁早两清
2楼-- · 2019-01-03 09:29

This works for me:

import urllib2

proxy = urllib2.ProxyHandler({'http': 'http://
username:password@proxyurl:proxyport'})
auth = urllib2.HTTPBasicAuthHandler()
opener = urllib2.build_opener(proxy, auth, urllib2.HTTPHandler)
urllib2.install_opener(opener)

conn = urllib2.urlopen('http://python.org')
return_str = conn.read()
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▲ chillily
3楼-- · 2019-01-03 09:32

The best way of going through a proxy that requires authentication is using urllib2 to build a custom url opener, then using that to make all the requests you want to go through the proxy. Note in particular, you probably don't want to embed the proxy password in the url or the python source code (unless it's just a quick hack).

import urllib2

def get_proxy_opener(proxyurl, proxyuser, proxypass, proxyscheme="http"):
    password_mgr = urllib2.HTTPPasswordMgrWithDefaultRealm()
    password_mgr.add_password(None, proxyurl, proxyuser, proxypass)

    proxy_handler = urllib2.ProxyHandler({proxyscheme: proxyurl})
    proxy_auth_handler = urllib2.ProxyBasicAuthHandler(password_mgr)

    return urllib2.build_opener(proxy_handler, proxy_auth_handler)

if __name__ == "__main__":
    import sys
    if len(sys.argv) > 4:
        url_opener = get_proxy_opener(*sys.argv[1:4])
        for url in sys.argv[4:]:
            print url_opener.open(url).headers
    else:
        print "Usage:", sys.argv[0], "proxy user pass fetchurls..."

In a more complex program, you can seperate these components out as appropriate (for instance, only using one password manager for the lifetime of the application). The python documentation has more examples on how to do complex things with urllib2 that you might also find useful.

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地球回转人心会变
4楼-- · 2019-01-03 09:34

Use this:

import requests

proxies = {"http":"http://username:password@proxy_ip:proxy_port"}

r = requests.get("http://www.example.com/", proxies=proxies)

print(r.content)

I think it's much simpler than using urllib. I don't understand why people love using urllib so much.

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smile是对你的礼貌
5楼-- · 2019-01-03 09:40

Setting an environment var named http_proxy like this: http://username:password@proxy_url:port

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forever°为你锁心
6楼-- · 2019-01-03 09:52

Here is the method use urllib

import urllib.request

# set up authentication info
authinfo = urllib.request.HTTPBasicAuthHandler()
proxy_support = urllib.request.ProxyHandler({"http" : "http://ahad-haam:3128"})

# build a new opener that adds authentication and caching FTP handlers
opener = urllib.request.build_opener(proxy_support, authinfo,
                                     urllib.request.CacheFTPHandler)

# install it
urllib.request.install_opener(opener)

f = urllib.request.urlopen('http://www.python.org/')
"""
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劳资没心,怎么记你
7楼-- · 2019-01-03 09:54

Or if you want to install it, so that it is always used with urllib2.urlopen (so you don't need to keep a reference to the opener around):

import urllib2
url = 'www.proxyurl.com'
username = 'user'
password = 'pass'
password_mgr = urllib2.HTTPPasswordMgrWithDefaultRealm()
# None, with the "WithDefaultRealm" password manager means
# that the user/pass will be used for any realm (where
# there isn't a more specific match).
password_mgr.add_password(None, url, username, password)
auth_handler = urllib2.HTTPBasicAuthHandler(password_mgr)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)
print urllib2.urlopen("http://www.example.com/folder/page.html").read()
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