如何在 Django 中获取用户 IP 地址？

Question

How do I get user's IP in Django?

I have a view like this:

# Create your views
from django.contrib.gis.utils import GeoIP
from django.template import  RequestContext
from django.shortcuts import render_to_response

def home(request):
  g = GeoIP()
  client_ip = request.META['REMOTE_ADDR']
  lat,long = g.lat_lon(client_ip)
  return render_to_response('home_page_tmp.html',locals())

But I get this error:

KeyError at /mypage/
    'REMOTE_ADDR'
    Request Method: GET
    Request URL:    http://mywebsite.example/mypage/
    Django Version: 1.2.4
    Exception Type: KeyError
    Exception Value:
    'REMOTE_ADDR'
    Exception Location: /mysite/homepage/views.py in home, line 9
    Python Executable:  /usr/bin/python
    Python Version: 2.6.6
    Python Path:    ['/mysite', '/usr/local/lib/python2.6/dist-packages/flup-1.0.2-py2.6.egg', '/usr/lib/python2.6', '/usr/lib/python2.6/plat-linux2', '/usr/lib/python2.6/lib-tk', '/usr/lib/python2.6/lib-old', '/usr/lib/python2.6/lib-dynload', '/usr/local/lib/python2.6/dist-packages', '/usr/lib/python2.6/dist-packages', '/usr/lib/pymodules/python2.6']
    Server time:    Sun, 2 Jan 2011 20:42:50 -0600

Answer 1

def get_client_ip(request):
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR')
    if x_forwarded_for:
        ip = x_forwarded_for.split(',')[0]
    else:
        ip = request.META.get('REMOTE_ADDR')
    return ip

Make sure you have reverse proxy (if any) configured correctly (eg mod_rpaf installed for Apache).

Note: the above uses the first item in X-Forwarded-For , but you might want to use the last item (eg, in the case of Heroku: Get client's real IP address on Heroku )

And then just pass the request as argument to it;

get_client_ip(request)

Django documentation for HttpRequest.META

Answer 2

You can use django-ipware which supports Python 2 & 3 and handles IPv4 & IPv6 .

Install:

pip install django-ipware

Simple Usage:

# In a view or a middleware where the `request` object is available

from ipware import get_client_ip
ip, is_routable = get_client_ip(request)
if ip is None:
    # Unable to get the client's IP address
else:
    # We got the client's IP address
    if is_routable:
        # The client's IP address is publicly routable on the Internet
    else:
        # The client's IP address is private

# Order of precedence is (Public, Private, Loopback, None)

Advanced Usage:

• Custom Header - Custom request header for ipware to look at:

   i, r = get_client_ip(request, request_header_order=['X_FORWARDED_FOR']) i, r = get_client_ip(request, request_header_order=['X_FORWARDED_FOR', 'REMOTE_ADDR'])

• Proxy Count - Django server is behind a fixed number of proxies:

   i, r = get_client_ip(request, proxy_count=1)

• Trusted Proxies - Django server is behind one or more known & trusted proxies:

   i, r = get_client_ip(request, proxy_trusted_ips=('177.2.2.2')) # For multiple proxies, simply add them to the list i, r = get_client_ip(request, proxy_trusted_ips=('177.2.2.2', '177.3.3.3')) # For proxies with fixed sub-domain and dynamic IP addresses, use partial pattern i, r = get_client_ip(request, proxy_trusted_ips=('177.2.', '177.3.'))

Note: read this notice .

Answer 3

Alexander's answer is great, but lacks the handling of proxies that sometimes return multiple IP's in the HTTP_X_FORWARDED_FOR header.

The real IP is usually at the end of the list, as explained here: http://en.wikipedia.org/wiki/X-Forwarded-For

The solution is a simple modification of Alexander's code:

def get_client_ip(request):
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR')
    if x_forwarded_for:
        ip = x_forwarded_for.split(',')[-1].strip()
    else:
        ip = request.META.get('REMOTE_ADDR')
    return ip

Answer 4

No More confusion In the recent versions of Django it is mentioned clearly that the Ip address of the client is available at

request.META.get("REMOTE_ADDR")

for more info check the Django Docs

Answer 5

I would like to suggest an improvement to yanchenko's answer.

Instead of taking the first ip in the X_FORWARDED_FOR list, I take the first one which in not a known internal ip, as some routers don't respect the protocol, and you can see internal ips as the first value of the list.

PRIVATE_IPS_PREFIX = ('10.', '172.', '192.', )

def get_client_ip(request):
    """get the client ip from the request
    """
    remote_address = request.META.get('REMOTE_ADDR')
    # set the default value of the ip to be the REMOTE_ADDR if available
    # else None
    ip = remote_address
    # try to get the first non-proxy ip (not a private ip) from the
    # HTTP_X_FORWARDED_FOR
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR')
    if x_forwarded_for:
        proxies = x_forwarded_for.split(',')
        # remove the private ips from the beginning
        while (len(proxies) > 0 and
                proxies[0].startswith(PRIVATE_IPS_PREFIX)):
            proxies.pop(0)
        # take the first ip which is not a private one (of a proxy)
        if len(proxies) > 0:
            ip = proxies[0]

    return ip

I hope this helps fellow Googlers who have the same problem.

Answer 6

这是一个简短的班轮来实现这一点：

request.META.get('HTTP_X_FORWARDED_FOR', request.META.get('REMOTE_ADDR', '')).split(',')[0].strip()

Answer 7

In my case none of above works, so I have to check uwsgi + django source code and pass static param in nginx and see why/how, and below is what I have found.

Env info:
python version: 2.7.5
Django version: (1, 6, 6, 'final', 0)
nginx version: nginx/1.6.0
uwsgi: 2.0.7

Env setting info:
nginx as reverse proxy listening at port 80 uwsgi as upstream unix socket, will response to the request eventually

Django config info:

USE_X_FORWARDED_HOST = True # with or without this line does not matter

nginx config:

uwsgi_param      X-Real-IP              $remote_addr;
// uwsgi_param   X-Forwarded-For        $proxy_add_x_forwarded_for;
// uwsgi_param   HTTP_X_FORWARDED_FOR   $proxy_add_x_forwarded_for;

// hardcode for testing
uwsgi_param      X-Forwarded-For        "10.10.10.10";
uwsgi_param      HTTP_X_FORWARDED_FOR   "20.20.20.20";

getting all the params in django app:

X-Forwarded-For :       10.10.10.10
HTTP_X_FORWARDED_FOR :  20.20.20.20

Conclusion:

So basically, you have to specify exactly the same field/param name in nginx, and use request.META[field/param] in django app.

And now you can decide whether to add a middleware (interceptor) or just parse HTTP_X_FORWARDED_FOR in certain views.

Answer 8

The simpliest solution (in case you are using fastcgi+nignx) is what itgorilla commented:

Thank you for this great question. My fastcgi was not passing the REMOTE_ADDR meta key. I added the line below in the nginx.conf and fixed the problem: fastcgi_param REMOTE_ADDR $remote_addr; – itgorilla

Ps: I added this answer just to make his solution more visible.

Answer 9

The reason the functionality was removed from Django originally was that the header cannot ultimately be trusted. The reason is that it is easy to spoof. For example the recommended way to configure an Nginx reverse proxy is to:

add_header X-Forwarded-For $proxy_add_x_forwarded_for;
add_header X-Real-Ip       $remote_addr;

When you do:

curl -H 'X-Forwarded-For: 8.8.8.8, 192.168.1.2' http://192.168.1.3/

Your Nginx in myhost.example will send onwards:

X-Forwarded-For: 8.8.8.8, 192.168.1.2, 192.168.1.3

The X-Real-IP will be the IP of the first previous proxy if you follow the instructions blindly.

In case trusting who your users are is an issue, you could try something like django-xff : https://pypi.python.org/pypi/django-xff/

Answer 10

I was also missing proxy in above answer. I used get_ip_address_from_request from django_easy_timezones .

from easy_timezones.utils import get_ip_address_from_request, is_valid_ip, is_local_ip
ip = get_ip_address_from_request(request)
try:
    if is_valid_ip(ip):
        geoip_record = IpRange.objects.by_ip(ip)
except IpRange.DoesNotExist:
    return None

And here is method get_ip_address_from_request , IPv4 and IPv6 ready:

def get_ip_address_from_request(request):
    """ Makes the best attempt to get the client's real IP or return the loopback """
    PRIVATE_IPS_PREFIX = ('10.', '172.', '192.', '127.')
    ip_address = ''
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR', '')
    if x_forwarded_for and ',' not in x_forwarded_for:
        if not x_forwarded_for.startswith(PRIVATE_IPS_PREFIX) and is_valid_ip(x_forwarded_for):
            ip_address = x_forwarded_for.strip()
    else:
        ips = [ip.strip() for ip in x_forwarded_for.split(',')]
        for ip in ips:
            if ip.startswith(PRIVATE_IPS_PREFIX):
                continue
            elif not is_valid_ip(ip):
                continue
            else:
                ip_address = ip
                break
    if not ip_address:
        x_real_ip = request.META.get('HTTP_X_REAL_IP', '')
        if x_real_ip:
            if not x_real_ip.startswith(PRIVATE_IPS_PREFIX) and is_valid_ip(x_real_ip):
                ip_address = x_real_ip.strip()
    if not ip_address:
        remote_addr = request.META.get('REMOTE_ADDR', '')
        if remote_addr:
            if not remote_addr.startswith(PRIVATE_IPS_PREFIX) and is_valid_ip(remote_addr):
                ip_address = remote_addr.strip()
    if not ip_address:
        ip_address = '127.0.0.1'
    return ip_address

Answer 11

In django.VERSION (2, 1, 1, 'final', 0) request handler

sock=request._stream.stream.raw._sock
#<socket.socket fd=1236, family=AddressFamily.AF_INET, type=SocketKind.SOCK_STREAM, proto=0, laddr=('192.168.1.111', 8000), raddr=('192.168.1.111', 64725)>
client_ip,port=sock.getpeername()

if you call above code twice,you may got

AttributeError("'_io.BytesIO' object has no attribute 'stream'",)

AttributeError("'LimitedStream' object has no attribute 'raw'")

Answer 12

Simply add

{{ request.META.REMOTE_ADDR }}

In Django-Template where you want the user to see their IP address. That is if you are not interested in saving this to the DB.

Answer 13

After getting ip address you need to find location

# pip install geocoder

import geocoder

def get_client_ip(request):
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR')
    if x_forwarded_for:
        ip = x_forwarded_for.split(',')[0]
        ip_location = geocoder.ip(f"{ip}")
        ip_location = geocoder.ip("me")
        print(ip_location.city)
        # you can get city such as "New York"
    else:
        ip = request.META.get('REMOTE_ADDR')
    return ip

Answer 14

Get the ip address with this function:

def get_ip_address(request):
    x_forwarded_for = request.META.get('HTTP_X_FORWARDED_FOR')
    if x_forwarded_for:
        ip = x_forwarded_for.split(',')[0]
    else:
        ip = request.META.get('REMOTE_ADDR')
    return ip

after that you can get the user location data and other info from that web app http://www.iplocinfo.com/ :

import requests
def get_ip_data(request):
    ip_address = get_ip_address(request)
    api_key = "your api key"
    endPoint = f'https://www.iplocinfo.com/api/v1/{ip_address}?apiKey={api_key}'
    data = requests.get(endPoint)
    return data.json()