将方法参数传递给它的父类构造函数（Flask - ModelView.as_view（））

Question

这是一个比Flask问题更普遍的Python问题。

这段代码来自https://github.com/mitsuhiko/flask/blob/master/flask/views.py#L18。

@classmethod 
def as_view(cls, name, *class_args, **class_kwargs): 
    """Converts the class into an actual view function that can be used 
    with the routing system. Internally this generates a function on the 
    fly which will instantiate the :class:`View` on each request and call 
    the :meth:`dispatch_request` method on it. 

    The arguments passed to :meth:`as_view` are forwarded to the 
    constructor of the class. 
    """ 
    def view(*args, **kwargs): 
     self = view.view_class(*class_args, **class_kwargs) 
     return self.dispatch_request(*args, **kwargs) 

    if cls.decorators: 
     view.__name__ = name 
     view.__module__ = cls.__module__ 
     for decorator in cls.decorators: 
      view = decorator(view) 

    # we attach the view class to the view function for two reasons: 
    # first of all it allows us to easily figure out what class-based 
    # view this thing came from, secondly it's also used for instantiating 
    # the view class so you can actually replace it with something else 
    # for testing purposes and debugging. 
    view.view_class = cls 
    view.__name__ = name 
    view.__doc__ = cls.__doc__ 
    view.__module__ = cls.__module__ 
    view.methods = cls.methods 
    return view 

有人可以给我解释一下这个as_view()功能到底如何，因为它说，它的方法评论它的参数转发到View类的构造函数？

如果不是一个直接的解释，也许是在正确的方向上推进具体什么我需要学习Python的明智，以更好地了解发生了什么事情。

感谢

Answer 1

这条线路是最关键的一步：

self = view.view_class(*class_args, **class_kwargs) 

它正在传递到as_view的参数和使用他们创造的view使用这些参数的实例。