在Django中基于类视图的参数的函数装饰器 [英] Function decorators with parameters on a class based view in Django

问题描述

官方文档解释如何装饰基于类的视图，但是我找不到有关如何向装饰器提供参数的任何信息。

The official documentation explains how to decorate a class based view, however I could not find any information on how to provide parameters to the decorator.

I想要实现类似于

class MyView(View):
    @method_decorator(mydecorator, some_parameters)
    def dispatch(self, *args, **kwargs):
        return super(MyView, self).dispatch(*args, **kwargs)

应该等于

@mydecorator(some_parameters)
def my_view(request):
    ....

如何处理这样的情况？

推荐答案

@method_decorator 将一个函数作为参数。如果要传递带参数的装饰器，则只需要：

@method_decorator takes a function as parameter. If you want to pass a decorator with parameters, you only need to:

• 评估decorator-creator函数中的参数。 >
  
• 将评估值传递给 @method_decorator 。

• Evaluate the parameters in the decorator-creator function.
• Pass the evaluated value to @method_decorator.

在显式的Python代码中，这将是：

In explicit Python code this would be:

decorator = mydecorator(arg1, arg2, arg...)
method_dec = method_decorator(decorator)

class MyClass(View):
    @method_dec
    def my_view(request):
        ...

所以，完全使用语法糖：

So, using the syntactic sugar completely:

class MyClass(View):
    @method_decorator(mydecorator(arg1, arg2, arg...))
    def my_view(request):
        ...

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