I don't see why wouldn't your model's .save() work, maybe we'd need to see the model code, but I know you probably want to read this:
https://docs.djangoproject.com/en/dev/topics/forms/modelforms/
once you've implemented a ModelForm, you'd pretty much would just do
form = RoomForm(request.POST)
if form.is_valid():
form.save()
On Wed, Jul 4, 2012 at 3:58 PM, rafiee.nima <rafiee.nima@gmail.com> wrote:
> Hi Im some how new to django . I write a view to handle ajax request
> but I find out that save() method dose not save model instance in to the
> database
> here is my code
>
> def add_room(request):
> context={}
> status=''
> if request.is_ajax:
> if request.POST:
> hotel_instance=Hotel.objects.get(id=request.POST['hotel'])
> room_id=int(request.POST['id'])
> if room_id > 0 :
> room=HotelRoom.objects.get(id=request.POST['id'])
> room.hotel=hotel_instance
> room.number=request.POST['number']
> room.bed_count=request.POST['bed_count']
> room.ground_sleep=request.POST['ground_sleep']
> room.view=request.POST['view']
> room.reserved=request.POST['reserved']
> room.tv=request.POST['tv'].capitalize()
> room.phone=request.POST['phone']
> room.refrigerator=request.POST['refrigerator']
> room.air_condition=request.POST['air_condition']
> room.toilet=request.POST['toilet']
> room.creator=request.user
> room.save()
> room_json=model_to_dict(room)
> status="room successfully updated"
> else:
> room = HotelRoom(
> hotel=hotel_instance,
> number=request.POST['number'],
> bed_count=request.POST['bed_count'],
> ground_sleep=request.POST['ground_sleep'],
> view=request.POST['view'],
> reserved=request.POST['reserved'],
> tv=request.POST['tv'],
> phone=request.POST['phone'],
> refrigerator=request.POST['refrigerator'],
> air_condition=request.POST['air_condition'],
> toilet=request.POST['toilet'],
> creator=request.user )
> room.save()
> status="new room successfully added "
> raw_data={'status':status,'data':room_json}
> data=simplejson.dumps(raw_data)
> return HttpResponse(data, mimetype="application/json")
>
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