Question

如何在装饰器中的decorator中调用实例属性。以下是我的测试代码：

#!/usr/bin/env python
# coding=utf-8

class Foo(object):
    """"""
    def __init__(self):
        """"""
        self.a = 1

    def decorator(func):
        def wrapper(*args, **kw):
            """"""
            print self.a # or do something with self.a
            self.a += 1
        return wrapper

    @decorator
    def call_decorator(self):
        """"""
        pass

和

In [1]: import foo

In [2]: f = foo.Foo()

In [3]: f.call_decorator
Out[3]: <bound method Foo.wrapper of <foo.Foo object at 0x7fcc72c6a710>>

In [4]: f.call_decorator()
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-4-70d92b784c01> in <module>()
----> 1 f.call_decorator()

/home/fit/log/1228/foo.py in wrapper(*args, **kw)
     11         def wrapper(*args, **kw):
     12             """"""
---> 13             print self.a
     14         return wrapper
     15 

NameError: global name 'self' is not defined

In [5]:

如果我在self中添加def decorator(self, func):，并且@decorator中会出现错误，则需要2个参数。装饰师不容易理解。

所以，我只想在装饰器中使用实例a的属性f，我该怎么做？

Answer 1

wrapper在任何其他函数的参数中收到self的{{1}}，而不是decorator，所以定义应该是这样的：

def decorator(func):
    @functools.wraps(func)  # a good thing to do
    def wrapper(self, *args, **kw):
        print self.a # or do something with self.a
        self.a += 1
        # if you need to call it, otherwise what's the point of the decorator:
        return func(self, *args, **kw)  
    return wrapper

python，如何在decorator中调用实例的属性

1 个答案: