python - 将方法参数传递给其父类构造函数(Flask -- ModelView.as_view())

Question

这更像是一个一般的 Python 问题，而不是 Flask 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

谁能给我解释一下这个 as_view() 函数是如何将它的参数转发给 View 类的构造函数的，就像它在方法注释中所说的那样？

如果不是直接的解释，也许是朝着正确的方向推进，特别是我需要学习 Python 方面的知识，以便更好地理解正在发生的事情。

谢谢

Answer 1

这一行是关键:

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

它采用传递给 as_view 的参数，并使用它们创建带有这些参数的 view 实例。