How do I get the IP address from a http request using the requests library?

Question:

I am making HTTP requests using the requests library in python, but I need the IP address from the server that responded to the HTTP request and I’m trying to avoid making two calls (and possibly having a different IP address from the one that responded to the request).

Is that possible? Does any python HTTP library allow me to do that?

PS: I also need to make HTTPS requests and use an authenticated proxy.

Update 1:

Example:

import requests

proxies = {
  "http": "http://user:[email protected]:3128",
  "https": "http://user:[email protected]:1080",
}

response = requests.get("http://example.org", proxies=proxies)
response.ip # This doesn't exist, this is just an what I would like to do

Then, I would like to know to which IP address requests are connected from a method or property in the response. In other libraries, I was able to do that by finding the sock object and using the getpeername() function.

Asked By: gawry

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Answers:

It turns out that it’s rather involved.

Here’s a monkey-patch while using requests version 1.2.3:

Wrapping the _make_request method on HTTPConnectionPool to store the response from socket.getpeername() on the HTTPResponse instance.

For me on python 2.7.3, this instance was available on response.raw._original_response.

from requests.packages.urllib3.connectionpool import HTTPConnectionPool

def _make_request(self,conn,method,url,**kwargs):
    response = self._old_make_request(conn,method,url,**kwargs)
    sock = getattr(conn,'sock',False)
    if sock:
        setattr(response,'peer',sock.getpeername())
    else:
        setattr(response,'peer',None)
    return response

HTTPConnectionPool._old_make_request = HTTPConnectionPool._make_request
HTTPConnectionPool._make_request = _make_request

import requests

r = requests.get('http://www.google.com')
print r.raw._original_response.peer

Yields:

('2a00:1450:4009:809::1017', 80, 0, 0)

Ah, if there’s a proxy involved or the response is chunked, the HTTPConnectionPool._make_request isn’t called.

So here’s a new version patching httplib.getresponse instead:

import httplib

def getresponse(self,*args,**kwargs):
    response = self._old_getresponse(*args,**kwargs)
    if self.sock:
        response.peer = self.sock.getpeername()
    else:
        response.peer = None
    return response


httplib.HTTPConnection._old_getresponse = httplib.HTTPConnection.getresponse
httplib.HTTPConnection.getresponse = getresponse

import requests

def check_peer(resp):
    orig_resp = resp.raw._original_response
    if hasattr(orig_resp,'peer'):
        return getattr(orig_resp,'peer')

Running:

>>> r1 = requests.get('http://www.google.com')
>>> check_peer(r1)
('2a00:1450:4009:808::101f', 80, 0, 0)
>>> r2 = requests.get('https://www.google.com')
>>> check_peer(r2)
('2a00:1450:4009:808::101f', 443, 0, 0)
>>> r3 = requests.get('http://wheezyweb.readthedocs.org/en/latest/tutorial.html#what-you-ll-build')
>>> check_peer(r3)
('162.209.99.68', 80)

Also checked running with proxies set; proxy address is returned.


Update 2016/01/19

est offers an alternative that doesn’t need the monkey-patch:

rsp = requests.get('http://google.com', stream=True)
# grab the IP while you can, before you consume the body!!!!!!!!
print rsp.raw._fp.fp._sock.getpeername()
# consume the body, which calls the read(), after that fileno is no longer available.
print rsp.content  

Update 2016/05/19

From the comments, copying here for visibility, Richard Kenneth Niescior offers the following that is confirmed working with requests 2.10.0 and Python 3.

rsp=requests.get(..., stream=True)
rsp.raw._connection.sock.getpeername()

Update 2019/02/22

Python3 with requests version 2.19.1.

resp=requests.get(..., stream=True)
resp.raw._connection.sock.socket.getsockname()

Update 2020/01/31

Python3.8 with requests 2.22.0

resp = requests.get('https://www.google.com', stream=True)
resp.raw._connection.sock.getsockname()
Answered By: MattH

Try:

import requests

proxies = {
  "http": "http://user:[email protected]:3128",
  "https": "http://user:[email protected]:1080",
}

response = requests.get('http://jsonip.com', proxies=proxies)
ip = response.json()['ip']
print('Your public IP is:', ip)
Answered By: Sergey Bushmanov