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

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。

其他回答

使用下面的函数获取ip地址:

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

在此之后,您可以从web应用程序http://www.iplocinfo.com/:获得用户位置数据和其他信息

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()

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
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

确保你有反向代理(如果有)配置正确(例如mod_rpaf安装Apache)。

注意:上面使用X-Forwarded-For中的第一项,但你可能想使用最后一项(例如,在Heroku的情况下:在Heroku上获取客户端的真实IP地址)

然后把请求作为参数传递给它;

get_client_ip(request)

Django文档HttpRequest。元

django。版本 (2,1,1, 'final', 0) 请求处理程序

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()

如果你调用以上代码两次,你可能会得到

AttributeError(“_io。BytesIO的对象没有属性“stream”

AttributeError("'LimitedStream'对象没有'raw'属性")

简单的添加

{{request.META。REMOTE_ADDR}}

在Django-Template中,你想让用户看到他们的IP地址。这是在您不感兴趣将其保存到DB的情况下。