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
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
最初从Django中删除该功能的原因是头文件最终不能被信任。原因是它很容易被恶搞。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/
你的Nginx在myhost。示例将发送:
X-Forwarded-For: 8.8.8.8, 192.168.1.2, 192.168.1.3
如果盲目地按照说明操作,X-Real-IP将是前面第一个代理的IP。
如果信任你的用户是一个问题,你可以尝试django-xff: https://pypi.python.org/pypi/django-xff/
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")
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。