Get Url Parameters In Django

Posted by picomon on Stack Overflow See other posts from Stack Overflow or by picomon
Published on 2013-11-12T09:29:46Z Indexed on 2013/11/12 9:54 UTC
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I want to get current transaction id in url. it should be like this

www.example.com/final_result/53432e1dd34b3

.

I wrote the below codes, but after successful payment, I'm redirected to Page 404.

(www.example.com/final_result//)

Views.py

@csrf_exempt
def pay_notif(request, v_transaction_id):
    if request.method=='POST':
        v_transaction_id=request.POST.get('transaction_id')
        endpoint='https://testpay.com/?v_transaction_id={0}&type=json'
        req=endpoint.format(v_transaction_id)
        last_result=urlopen(req).read()
        if 'Approved' in last_result:
            session=Pay.objects.filter(session=request.session.session_key).latest('id')
        else:
            return HttpResponse(status=204)
    return render_to_response('final.html',{'session':session},context_instance=RequestContext(request))

Urls.py

url(r'^final_result/(?P<v_transaction_id>[-A-Za-z0-9_]+)/$', 'digiapp.views.pay_notif', name="pay_notif"),

Template:

<input type='hidden' name='v_merchant_id' value='{{newpayy.v_merchant_id}}' />
<input type='hidden' name='item_1' value='{{ newpayy.esell.up_name }}' />
<input type='hidden' name='description_1' value='{{ newpayy.esell.up_description }}' />
<input type='hidden' name='price_1' value='{{ newpayy.esell.up_price }}' />

#page to be redirected to after successful payment
 <input type='hidden' name='success_url'  value='http://127.0.0.1:8000/final_result/{{newpayy.v_transaction_id}}/' />

How can I go about this?

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