如何按名称访问对象的属性？

Question

class a(type): 
    def __str__(self): 
     return 'aaa' 
    def __new__(cls, name, bases, attrs): 
     attrs['cool']='cool!!!!' 
     new_class = super(a,cls).__new__(cls, name, bases, attrs) 
       #if 'media' not in attrs: 
        #new_class.media ='media' 
     return new_class 

class b(object): 
    __metaclass__=a 
    def __str__(self): 
     return 'bbb' 

print b 
print b()['cool']#how can i print 'cool!!!!' 

Answer 1

print b().cool 

attrs你的__new__方法变成了对象的字典。 Python对象的属性以.语法引用。

Answer 2

print "cool!!!" 

还是我错过了什么？