python之Django 请求对象起源于 `class View` 在哪里

下面是我的代码段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 

我正在寻找请求对象传入的代码。

as_view常用的地方在url

但是我无法在

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) 

谁能给我指明方向？

请您参考如下方法:

请注意，请求永远不会传递给 as_view()。

在加载 url 配置时调用 as_view() 方法，然后再处理任何请求。它定义了一个方法view，并返回它。

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

此view 方法采用参数request、位置参数和关键字参数。 view 方法随后被传递给 url 实例。请注意，url 只需要一个带有 request 参数的可调用对象。这可以是通过为基于类的 View 或基于常规函数的 View 调用 as_view() 返回的可调用对象，它对请求传递到 View 的方式没有影响。

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),  

然后，当请求被处理时，url 被解析到这个 View ，BaseHandler.get_response 调用带有请求的 View ，并从中捕获 args 和 kwargs网址。