Python 在Django中使用SimpleXMLRPCDispatcher时如何获取IP

其代码源于：

如何在“post_log”定义中获取客户端的IP地址？ 我见过，但不能把它应用到我的案例中

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谢谢。

好的，我可以。。。有一些妙招

首先，我创建了自己的SimpleXMLRPCDispatcher副本，它继承了它的所有内容，并覆盖了2个方法：

class MySimpleXMLRPCDispatcher (SimpleXMLRPCDispatcher) :
    def _marshaled_dispatch(self, data, dispatch_method = None, request = None):
        # copy and paste from /usr/lib/python2.6/SimpleXMLRPCServer.py except
        response = self._dispatch(method, params)
        # which becomes
        response = self._dispatch(method, params, request)

    def _dispatch(self, method, params, request = None):
        # copy and paste from /usr/lib/python2.6/SimpleXMLRPCServer.py except
        return func(*params)
        # which becomes
        return func(request, *params)

那么在我的代码中，要做的就是：

# ...
if len(request.POST):
    response = HttpResponse(mimetype="application/xml")
    response.write(dispatcher._marshaled_dispatch(request.raw_post_data, request = request))
# ...
def post_log(request, message = "", tags = []):
    """ Code called via RPC. Want to know here the remote IP (or hostname). """
    ip = request.META["REMOTE_ADDR"]
    hostname = socket.gethostbyaddr(ip)[0]

就这样。 我知道它不是很干净。。。欢迎对更清洁的解决方案提出任何建议

# ...
if len(request.POST):
    response = HttpResponse(mimetype="application/xml")
    response.write(dispatcher._marshaled_dispatch(request.raw_post_data, request = request))
# ...
def post_log(request, message = "", tags = []):
    """ Code called via RPC. Want to know here the remote IP (or hostname). """
    ip = request.META["REMOTE_ADDR"]
    hostname = socket.gethostbyaddr(ip)[0]