Django Rest Framework for function based views

Question

I'm having trouble getting the JSON for function based views in django. I have the below code. I basically would like the function to return either json or an html page based on the user request.

@api_view(['GET'])
@renderer_classes((JSONRenderer,TemplateHTMLRenderer,BrowsableAPIRenderer))
def meld_live_orders(request):
    if request.method =='GET':
        current_orders = Meld_Sales.objects.values_list('TicketNo',flat=True).distinct()
        prev_orders = Meld_Order.objects.values_list('TicketNo',flat =True).distinct()

        live_orders = live_order_generator(current_orders,prev_orders)

        return render(request,'live_orders.html',{'live_orders':live_orders})

When i go to the url - http://localhost:8000/live-orders.json

I'm getting an error which states the below -meld_live_orders() got an unexpected keyword argument 'format'

Is this because i need to include the serializer class somewhere the same way in CBVs? Doesnt the @API_VIEW serialize the response?

i tried including format = '' in the function argument. but the problem is that it still renders html when i want it to render json.

Rahul Gupta · Accepted Answer · 2015-08-30 10:27:25Z

You need to make some changes to your code.

Firstly, you need to use format_suffix_patterns in your urls if you have not defined it. This will allow us to use filename extensions on URLs thereby providing an endpoint for a given media type.

from rest_framework.urlpatterns import format_suffix_patterns

urlpatterns = [
     ...
]

urlpatterns = format_suffix_patterns(urlpatterns, allowed=['json', 'html']) # allow us to use '.json' and '.html' at the end of the url

Secondly. your view does not have a format parameter in the definition.

When using format_suffix_patterns, you must make sure to add the 'format' keyword argument to the corresponding views.

@api_view(['GET'])
@renderer_classes((JSONRenderer,TemplateHTMLRenderer,BrowsableAPIRenderer))
def meld_live_orders(request, format=None): # add a 'format' parameter
    ...

Thirdly, you need to return a DRF response and not a Django response which you are returning at the end of the view.

hsfzxjy · Accepted Answer · 2015-08-30 08:34:41Z

0

You must have match a format parameter in the url pattern, but in the view function there is not an argument named format. Change the view definition into:

def meld_live_orders(request, format = ""):

answered Aug 30, 2015 at 8:34

hsfzxjy

1,3224 gold badges14 silver badges22 bronze badges

6 Comments

Jayaram Over a year ago

I actually tried that .. sorry for skipping it in the question .. it still renders in html when i want it to render a json

hsfzxjy Over a year ago

Would you please present your url pattern?

hsfzxjy Over a year ago

Well i know, in the last line you have rendered a HTML page as the response, which should be a rest_framework.response.Response object.

Jayaram Over a year ago

is there a way i could have both? i.e the user gets json data only by including .json after the url? my url pattern - url(r'^live-orders$',meld_live_orders,name='live-orders')

hsfzxjy Over a year ago

Please lookup the document carefully. The correct way is to pass data of the serializer to the Response object. For JSONRenderer, the data will be converted into JSON format; for TemplateHTMLRenderer, the data will be pass to the template as a context variable.

|

Collectives™ on Stack Overflow

Django Rest Framework for function based views

2 Answers 2

Comments

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related