问题
I'm learning Django and I found class-based views and I wonder how to implement Ajax on those views.
I searched github for a django project and I found some using class-based views but not ajax.
So... Anybody knows an open source project that use both things? It easier to learn that way.
Thank you :)
回答1:
An ajax view isn't much different to a normal view except that you usually want to return a different format then when processing a normal request. This format is usually JSON.
The documentation has an example of a mixin that can be used to return JSON, so this is a good starting point:
https://docs.djangoproject.com/en/dev/topics/class-based-views/mixins/#more-than-just-html
Do you want your view to reply to normal requests or only deal with AJAX requests? If the former, the only trick would be to write in a small check in the render_to_response method to reject any normal GET requests. If the latter, the above link goes on to discuss a situation where you can create a view that will deal with ajax requests and with normal requests.
回答2:
without using the popular dajaxic and dajax packages, its still a straightforward affair.
It would help to write a decorator that just wraps around django's is_ajax() function for request objects like so:
def ajax_request(function):
def wrapper(request, *args, **kwargs):
if not request.is_ajax():
return render_to_response('error/ajax_required.html', {},
context_instance=RequestContext(request))
else:
return function(request, *args, **kwargs)
return wrapper
assuming there is a template called ajax_required to handle this particular failure. Something like this prevents a user from entering your ajax specific url in the browser if thats what you don't want.
Because it makes for a shorter example, the following is a class based ajax view that renders a template.
from django.views.generic.base import TemplateView
class AjaxGeneral(TemplateView):
template_name= None
def get(self, request):
data={}
return render_to_response(self.template_name, data,
context_instance=RequestContext(request))
@method_decorator(ajax_request)
def dispatch(self, *args, **kwargs):
return super(AjaxGeneral, self).dispatch(*args, **kwargs)
now for everything ajax that just needs to render an html snippet you can define short class based views like:
class ShowSomeTable(AjaxGeneral):
template_name="some_table.html"
Assuming some_table.html has some html snippet in it.
Now your urls.py entry for this view will look like:
url(r'showtable/$', ShowSomeTable.as_view()),
and you can call it in the js as normal like:
$(#dynamic-content).load('/showtable');
回答3:
If you want to support both AJAX and traditional views, you can add something like this to your CreateView:
# at top of file
from django.http import JSONResponse
# inside CreateView class
def render_to_response(self, context, **response_kwargs):
""" Allow AJAX requests to be handled more gracefully """
if self.request.is_ajax():
return JSONResponse('Success',safe=False, **response_kwargs)
else:
return super(CreateView,self).render_to_response(context, **response_kwargs)
This handles regular views normally (with redirect etc.) but for AJAX, it return a JSONResponse instead. Where it returns 'Success', you may want to add more sophisticated logic but this is the basic idea.
来源:https://stackoverflow.com/questions/8059160/django-apps-using-class-based-views-and-ajax