如何在Django中获取用户的IP ?

我的观点是这样的:

# 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())

但是我得到了这个错误:

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

当前回答

下面是一个简短的一行代码来完成这个任务:

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

其他回答

我想对yanchenko的回答提出一个改进。

而不是在X_FORWARDED_FOR列表中取第一个ip,我取第一个不知道内部ip,因为一些路由器不尊重协议,你可以看到内部ip作为列表的第一个值。

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

我希望这能帮助那些有同样问题的谷歌同事。

下面是一个简短的一行代码来完成这个任务:

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

在获得ip地址后,您需要找到位置

# 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

Alexander的回答很好,但是缺少对代理的处理,代理有时会在HTTP_X_FORWARDED_FOR报头中返回多个IP。

真实IP通常在列表的末尾,如下所示:http://en.wikipedia.org/wiki/X-Forwarded-For

解决方案是对Alexander的代码进行简单的修改:

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

最简单的解决方案(如果你正在使用fastcgi+nignx)是itgorilla评论的:

谢谢你这个好问题。我的fastcgi没有传递REMOTE_ADDR元键。我在nginx.conf中添加了下面的行,并修复了这个问题:——itgorilla

Ps:我添加这个答案只是为了让他的解决方案更明显。