如何在django中获取用户IP地址?

问题描述 投票:257回答:9

如何在django中获取用户的IP?

我有这样的观点:

# 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())

但我得到这个错误:

KeyError at /mypage/
    'REMOTE_ADDR'
    Request Method: GET
    Request URL:    http://mywebsite.com/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
python django
9个回答
381
投票
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

确保正确配置了反向代理(如果有)(例如,为Apache安装了mod_rpaf)。

注意:以上使用X-Forwarded-For中的第一项,但您可能想要使用最后一项(例如,在Heroku的情况下:Get client's real IP address on Heroku

然后将请求作为参数传递给它;

get_client_ip(request)

0
投票

在django.VERSION(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()

如果你两次拨打上面的代码,你可能会得到

AttributeError(“'_ io.BytesIO'对象没有属性'stream'”,)

AttributeError(“'LimitedStream'对象没有属性'raw'”)


192
投票

您可以使用支持Python 2和3的django-ipware并处理IPv4和IPv6。

安装:

pip install django-ipware

简单用法:

获取客户端的IP地址。

# 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)

高级用法:

自定义标题 - 用于查看ipware的自定义请求标头

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'])

代理计数 - Django服务器支持固定数量的代理

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

受信任的代理 - Django服务器支持一个或多个已知和受信任的代理

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.'))

注意:请阅读此notice


74
投票

亚历山大的答案很棒,但缺乏有时在HTTP_X_FORWARDED_FOR标头中返回多个IP的代理的处理。

真正的IP通常位于列表的末尾,如下所述:http://en.wikipedia.org/wiki/X-Forwarded-For

解决方案是对Alexander代码的简单修改:

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

9
投票

我想建议改进yanchenko的答案。

我没有在X_FORWARDED_FOR列表中取第一个ip,而是采用第一个不是已知内部ip的ip,因为有些路由器不遵守协议,你可以看到内部ips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

我希望这可以帮助那些遇到同样问题的Google员工。


6
投票

最简单的解决方案(如果你使用的是fastcgi + nginx)是它的大猩猩评论:

谢谢你提出这个好问题。我的fastcgi没有传递REMOTE_ADDR元键。我在nginx.conf中添加了以下行并修复了问题:fastcgi_param REMOTE_ADDR $ remote_addr; - 伊戈里拉

Ps:我添加了这个答案只是为了让他的解决方案更加明显。


6
投票

这是一个简短的衬垫,以实现这一目标:

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

5
投票

在我的情况下,上述都不起作用,所以我必须检查uwsgi + django源代码并在nginx中传递静态参数,看看为什么/如何,以及下面是我找到的。

环境信息: python版本:2.7.5 Django版本:(1, 6, 6, 'final', 0) nginx版本:nginx/1.6.0 uwsgi:2.0.7

环境设定信息: nginx作为反向代理监听端口80 uwsgi作为上游unix socket,最终会响应请求

Django配置信息:

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

nginx配置:

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";

获取django app中的所有参数:

X-Forwarded-For :       10.10.10.10
HTTP_X_FORWARDED_FOR :  20.20.20.20

结论:

所以基本上,你必须在nginx中指定完全相同的字段/参数名称,并在django app中使用request.META[field/param]

现在您可以决定是否添加中间件(拦截器)或仅在某些视图中解析HTTP_X_FORWARDED_FOR


2
投票

最初从Django中删除功能的原因是标头最终不能被信任。原因是它很容易恶搞。例如,配置nginx反向代理的推荐方法是:

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

当你这样做时:

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

您在myhost.com中的nginx将向前发送:

X-Forwarded-For: 8.8.8.8, 192.168.1.2, 192.168.1.3

如果您盲目地遵循指示,X-Real-IP将是第一个先前代理的IP。

如果信任你的用户是谁是一个问题,你可以尝试像django-xffhttps://pypi.python.org/pypi/django-xff/


2
投票

在上面的回答中我也缺少代理。我使用了get_ip_address_from_requestdjango_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

这里是方法get_ip_address_from_request,IPv4和IPv6准备:

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
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