写自己的装饰和添加一些秘密的头你的要求。的https://$c$c.djangoproject.com/browser/django/trunk/django/views/decorators/csrf.py

 高清csrf_exempt（view_func）：
        
        标记视图功能是免征CSRF视图保护。
        
        ＃我们可以只是view_func.csrf_exempt = TRUE，但装饰
        ＃是，如果他们没有副作用更好，所以我们返回一个新的
        ＃ 功能。
        高清wrapped_view（要求*的args，** kwargs）：
            返回view_func（要求*的args，** kwargs）
            如果request.META.has_key（'HTTP_X_SKIP_CSRF'）：
                wrapped_view.csrf_exempt = TRUE
        返回包装（view_func，分配= available_attrs（view_func））（wrapped_view）
 

I would like my android app to be able to send some information to my django server. So I made the android app send a post request to the mysite/upload page and django's view for this page would do work based on the post data. The problem is the response the server gives for the post request complains about csrf verication failed. Looking in to the problem it seems I might have to get a csrf token from the server first then do the post with that token But I am unsure how I do this. Edit: I have found out that I can knock off the crsf verification for this view using a view decorator @csrf_exempt but I am not sure if this is the best solution. My android code:

// Create a new HttpClient and Post Header
                    HttpClient httpclient = new DefaultHttpClient();
                    HttpPost httppost = new HttpPost(URL);

                    // Add your data
                    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
                    nameValuePairs.add(new BasicNameValuePair("scoreone", scoreone));
                    nameValuePairs.add(new BasicNameValuePair("scoretwo", scoretwo));
                    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                    System.out.println("huzahhhhhhh");
                    // Execute HTTP Post Request
                    HttpResponse response = httpclient.execute(httppost);
                    BufferedReader in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
                    StringBuffer sb = new StringBuffer("");
                    String line = "";
                    String NL = System.getProperty("line.separator");
                    while ((line = in.readLine()) != null) {
                        sb.append(line + NL);
                    }
                    in.close();
                    String result = sb.toString();
                    System.out.println("Result: "+result);

and my views code for handling the upload:

# uploads a players match
def upload(request):
    if request.method == 'POST':
        scoreone = int(request.POST['scoreone'])
        scoretwo = int(request.POST['scoretwo'])
        m = Match.objects.create()
        MatchParticipant.objects.create(player = Player.objects.get(pk=1), match = m, score = scoreone)
        MatchParticipant.objects.create(player = Player.objects.get(pk=2), match = m, score = scoretwo)
    return HttpResponse("Match uploaded" )

enter code here

Write own decorator and add some "secret" header to your request. https://code.djangoproject.com/browser/django/trunk/django/views/decorators/csrf.py

def csrf_exempt(view_func):
        """
        Marks a view function as being exempt from the CSRF view protection.
        """
        # We could just do view_func.csrf_exempt = True, but decorators
        # are nicer if they don't have side-effects, so we return a new
        # function.
        def wrapped_view(request,*args, **kwargs):
            return view_func(request, *args, **kwargs)
            if request.META.has_key('HTTP_X_SKIP_CSRF'):
                wrapped_view.csrf_exempt = True
        return wraps(view_func, assigned=available_attrs(view_func))(wrapped_view)