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Home/ Questions/Q 7512233
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T23:48:06+00:00 2026-05-29T23:48:06+00:00

I would like my android app to be able to send some information to

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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
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T23:48:07+00:00Added an answer on May 29, 2026 at 11:48 pm

    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)
    
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