Django Form using Ajax and jQuery! Can't work out how to pass control from Ajax to Python in URL -

how use python file (project/controller/functions/email/sendemail.py) process sending of email (or other task) ajax using jquery? or going wrong?

i don't want redirect page django tuts teach, want refresh form , show success message only. (i have messages , refresh of form working fine not handling of url in ajax).

i have searched high , low on , have come trumps without useful. advice or links examples appreciated.

my files similar below, except using jquery validate bit different same principal , form laid out using bootstrap 3 on version.

index.html

<form method="post" action="sendemail" id="sendemail">   <input name="name" type="text" />   <input name="email" type="email" />   <input name="submit" type="submit" value="send" /> </form> 

main.js

$('form#sendemail').on('submit', function() {   $.ajax({     url: $(this).attr('action'), #(targeting action form. working can console.log correct action sendemail),     type: type,     data: data   }); }); 

i tried redirecting urls target file use url ajax target file below.

urls.py

from django.conf.urls import include, url django.contrib import admin . import views .controller.functions.email import sendemail  urlpatterns = [   url(r'^$', views.home, name='home'),   url(r'^sendemail', sendemail, name='sendemail'),   url(r'^admin/', include(admin.site.urls)), ]     

i 500 server error in console this.

sendemail.py

from django.core.mail import send_mail  def sendemail(request):   if (send_mail('subject', 'here message.', 'from@example.com', ['to@example.com'], fail_silently=false)):     print(1) #success   else:     print(99) #fail 

views.py

from django.shortcuts import render django.views.decorators.csrf import csrf_protect .forms import contactform  @csrf_protect def home(request):   tpl = 'main/index.html'   contactnotify = ""   contactform = request.post     if request.method == 'post':     contactform = contactform(request.post)     if contactform.is_valid():       return render(request, tpl, context)    else:     contactform = contactform()  context = {     'contactform'       : contactform }  return render(request, tpl, context) 

in php use echo return jquery presume print it's equivalent in python return value rather return.

on send following log in console: post http://localhost:8888/sendemail 500 (internal server error)

django works differently

<form method="post" action="{% 'sendemail' %}" id="sendemail"> 

urls.py

from . import views .controller.functions.email import sendemail  urlpatterns = [   url(r'^$', views.home, name='home'),   url(r'^sendemail$/', views.sendemail_view, name='sendemail'),   url(r'^admin/', include(admin.site.urls)), ]   

views.py

from .helpers import sendemail  def sendemail_view(request):     # here sendemail()     # , return httpresponse 

and remember,

1. after urlconf match, django calls function views. how django designed, views == controllers frameworks
2. django views function names should lowercase always.

i dont know why listening post requests in home views.. think there more things fix work, please try suggestions