python - __getattr__ 的对应部分

Question

我正在尝试找到一种方法来设置封装到 class 中的 dict 值，例如使用 __getattr__ 我可以返回内部 dict 值，但是即使属性存在，也会调用 __setattr__，这让我的实现变得丑陋。下面的示例简化了我的实际类继承自 Subject class(观察者模式的主题部分)

我正在努力实现这样的目标:

obj = Example()
obj.username = 'spidername' # all OK username is a key in the internal dict
# but company is not a key in the internal dict so
obj.company = 'ABC' # will raise AttributeError

我想问是否有比我在下面做的方式更好的方法:

class Example(object):
    def __init__(self, table=None):
        self._fields = {}
        self._table = table

    def _set_fields(self):
        """
        this method will be implemented by 
        subclasses and used to set fields names and values
        i.e.
        self._field['username'] = Field(default='unknown', is_primary=False)
        """
        raise NotImplementedError

    def __getattr__(self, name):
        """
        great this method is only called when "name"
        is not an attribute of this class
        """
        if name in self._fields:
            return self._fields[name].value
        return None

    def __setattr__(self, name, value):
        """
        not so great, this method is called even for
        attributes that exists in this class

        is there a better way to do the following?             
        this can be in __init__, but its still ugly
        """
        attribs = ['_fields', '_table'] 
        if name in attribs:
            super(Example, self).__setattr__(name, value)
        else:
            if name in self._fields:
                self._fields[name].value = value
            else:
                raise AttributeError

编辑:调整了代码中的注释，添加了遗漏的引号

Answer 1

问题是属性在第一次赋值时并不存在。在 __init__ 中，当您第一次将字典分配给 _fields 时，_fields 不是属性。它只有在分配后才成为现有属性。如果您事先知道属性是什么，则可以使用 __slots__，但我猜您不知道。所以我的建议是手动将它们插入到实例字典中:

class Example(object):
    def __init__(self, table=None):
        self.__dict__['_fields'] = {}
        self.__dict__['_table'] = table

    ...

    def __setattr__(self, name, value):
        if name in self._fields:
            self._fields[name].value = value
        else:
            raise AttributeError

但是，对于此实现，您以后可以添加或更改实例属性的唯一方法是通过 __dict__。但我认为这不太可能。