CODEWITHSUNDEEP
pythondjangodjango-viewsdjango-urls
Kevin
Kevin

Reputation: 2277

Where does Django request object originated in `class View`?

Here is the code segment I am referring to

@classonlymethod
def as_view(cls, **initkwargs):
    """
    Main entry point for a request-response process.
    """
    for key in initkwargs:
        if key in cls.http_method_names:
            raise TypeError("You tried to pass in the %s method name as a "
                            "keyword argument to %s(). Don't do that."
                            % (key, cls.__name__))
        if not hasattr(cls, key):
            raise TypeError("%s() received an invalid keyword %r. as_view "
                            "only accepts arguments that are already "
                            "attributes of the class." % (cls.__name__, key))

    def view(request, *args, **kwargs):
        self = cls(**initkwargs)
        if hasattr(self, 'get') and not hasattr(self, 'head'):
            self.head = self.get
        self.request = request
        self.args = args
        self.kwargs = kwargs
        return self.dispatch(request, *args, **kwargs)
    view.view_class = cls
    view.view_initkwargs = initkwargs

    # take name and docstring from class
    update_wrapper(view, cls, updated=())

    # and possible attributes set by decorators
    # like csrf_exempt from dispatch
    update_wrapper(view, cls.dispatch, assigned=())
    return view

I am looking for the code that the request object is passed in to.

The common place where as_view is used is in url

However I couldn't reference to request object in

def url(regex, view, kwargs=None, name=None, prefix=''):
    if isinstance(view, (list, tuple)):
        # For include(...) processing.
        urlconf_module, app_name, namespace = view
        return RegexURLResolver(regex, urlconf_module, kwargs, app_name=app_name, namespace=namespace)
    else:
        if isinstance(view, six.string_types):
            warnings.warn(
                'Support for string view arguments to url() is deprecated and '
                'will be removed in Django 1.10 (got %s). Pass the callable '
               'instead.' % view,
            RemovedInDjango110Warning, stacklevel=2
            )
            if not view:
                raise ImproperlyConfigured('Empty URL pattern view name not permitted (for pattern %r)' % regex)
            if prefix:
                view = prefix + '.' + view
        return RegexURLPattern(regex, view, kwargs, name)

Can someone point me a direction?

Upvotes: 1

Views: 521

Answers (3)

Alasdair
Alasdair

Reputation: 309039

Note that the request is never passed to as_view().

The as_view() method is called when the url config is loaded, before any requests are handled. It defines a method view, and returns it.

def view(request, *args, **kwargs):
    ...
return view

This view method takes an argument request, positional and keyword arguments. The view method is then passed to the url instance. Note that url simply requires a callable that takes a request argument. This could be a callable returned by calling as_view() for a class based view, or a regular function based view, it makes no difference to how the request is passed to the view.

def function_view(request, *args, **kwargs):
    return HttpResponse("I'm a function based view")

url(r'^cbv/$', MyView.as_view()),
url(r'^fv/$', function_view),

Then, when a request is handled, the url is resolved into this view, and BaseHandler.get_response calls the view with the request, and args and kwargs captured from the url.

Upvotes: 3

grantmcconnaughey
grantmcconnaughey

Reputation: 10719

The request is created from the WSGIHandler class.

James Bennett talks about this in Django In Depth at around 2 hours and 14 minutes. Slides can be found here.

Upvotes: 1

Blender
Blender

Reputation: 298432

The request is processed by BaseHandler.get_response:

wrapped_callback = self.make_view_atomic(callback)
try:
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
    ...

Upvotes: 1

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