在类中定义python装饰器的正确方法是什么?

Question

我想要实现的是下面的代码输出以下内容:

• Here1
• Here2

我想知道我对__call__的使用是否在某种程度上破坏了functools.wraps;从某种程度上看，这些论点似乎是失败的。

我想要达到的是可能的吗?

from functools import wraps
class Decorator():
    def __init(self, something=None):
            self.something = something
    def __call__(self, func):
            print 'Here1'
            @wraps(func)
            def _wrapper(*args, **kwargs):
                return self.call(func, *args, **kwargs)
            return _wrapper
    def call(self, func, *args, **kwargs):
            print 'Here2'
            retsult = func(*args, **kwargs)
            return result

if __name__ == '__main__':
    decorator = Decorator()
    @decorator
    def do_the_thing(arg1='argOne', arg2='argTwo'):
            print 'Here3 {0} {1}'.format(arg1, arg2)
            return

Answer 1

似乎你只是有几个打字错误，实际上并没有调用函数do_the_thing。

改成这个，效果很好。

from functools import wraps

class Decorator():
    def __init__(self, something=None): # you are missing the __ on the right
            self.something = something
    def __call__(self, func):
            print 'Here1'
            @wraps(func)
            def _wrapper(*args, **kwargs):
                return self.call(func, *args, **kwargs)
            return _wrapper
    def call(self, func, *args, **kwargs):
            print 'Here2'
            result = func(*args, **kwargs) # result was misspelled
            return result

if __name__ == '__main__':
    @Decorator() # Just a bit cleaner
    def do_the_thing(arg1='argOne', arg2='argTwo'):
            print 'Here3 {0} {1}'.format(arg1, arg2)
    do_the_thing() # func was never called.