在類中定義python裝飾器的正確方法是什么？

Question

我想要實現的是，下面的代碼放出了以下內容：

• 這里1
• 這里2
• Here3 argOne argTwo

我想知道我使用__call__是否以某種方式破壞了functools.wraps; 在某些時候似乎也失去了論點。

我要實現的目標有可能嗎？

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

似乎您只是有一些錯別字，實際上並沒有調用函數do_the_thing 。

將其更改為此並可以正常工作。

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.