How can I open a website with urllib via proxy in Python?

Question

I have this program that check a website, and I want to know how can I check it via proxy in Python…

this is the code, just for example

while True:
    try:
        h = urllib.urlopen(website)
        break
    except:
        print '['+time.strftime('%Y/%m/%d %H:%M:%S')+'] '+'ERROR. Trying again in a few seconds...'
        time.sleep(5)

Asked By: Bruno 'Shady'

||

Source

Answer 1

By default, urlopen uses the environment variable http_proxy to determine which HTTP proxy to use:

$ export http_proxy='http://myproxy.example.com:1234'
$ python myscript.py  # Using http://myproxy.example.com:1234 as a proxy

If you instead want to specify a proxy inside your application, you can give a proxies argument to urlopen:

proxies = {'http': 'http://myproxy.example.com:1234'}
print("Using HTTP proxy %s" % proxies['http'])
urllib.urlopen("http://www.google.com", proxies=proxies)

Edit: If I understand your comments correctly, you want to try several proxies and print each proxy as you try it. How about something like this?

candidate_proxies = ['http://proxy1.example.com:1234',
                     'http://proxy2.example.com:1234',
                     'http://proxy3.example.com:1234']
for proxy in candidate_proxies:
    print("Trying HTTP proxy %s" % proxy)
    try:
        result = urllib.urlopen("http://www.google.com", proxies={'http': proxy})
        print("Got URL using proxy %s" % proxy)
        break
    except:
        print("Trying next proxy in 5 seconds")
        time.sleep(5)

Answered By: Pär Wieslander

Answer 2

Here example code guide how to use urllib to connect via proxy:

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.google.com/')
"""

Answered By: daz

Answer 3

Python 3 is slightly different here. It will try to auto detect proxy settings but if you need specific or manual proxy settings, think about this kind of code:

#!/usr/bin/env python3
import urllib.request

proxy_support = urllib.request.ProxyHandler({'http' : 'http://user:pass@server:port', 
                                             'https': 'https://...'})
opener = urllib.request.build_opener(proxy_support)
urllib.request.install_opener(opener)

with urllib.request.urlopen(url) as response:
    # ... implement things such as 'html = response.read()'

Refer also to the relevant section in the Python 3 docs

Answered By: DomTomCat

Answer 4

For http and https use:

proxies = {'http':'http://proxy-source-ip:proxy-port',
           'https':'https://proxy-source-ip:proxy-port'}

more proxies can be added similarly

proxies = {'http':'http://proxy1-source-ip:proxy-port',
           'http':'http://proxy2-source-ip:proxy-port'
           ...
          }

usage

filehandle = urllib.urlopen( external_url , proxies=proxies)

Don’t use any proxies (in case of links within network)

filehandle = urllib.urlopen(external_url, proxies={})

Use proxies authentication via username and password

proxies = {'http':'http://username:password@proxy-source-ip:proxy-port',
           'https':'https://username:password@proxy-source-ip:proxy-port'}

Note: avoid using special characters such as :,@ in username and passwords

Answered By: mayure098

How can I open a website with urllib via proxy in Python?

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