如何在python子类中从超类中获取属性名

Question

我有下面这样的课

class Paginator(object):
    @cached_property
    def count(self):
        some_implementation

class CachingPaginator(Paginator):
    def _get_count(self):
        if self._count is None:
            try:
                key = "admin:{0}:count".format(hash(self.object_list.query.__str__()))
                self._count = cache.get(key, -1)
                if self._count == -1:
                    self._count = self.count # Here, I want to get count property in the super-class, this is giving me -1 which is wrong
                    cache.set(key, self._count, 3600)
            except:
                self._count = len(self.object_list)
    count = property(_get_count)

如上面的注释所示，self._count = <expression>应该在超类中获得count属性。如果是方法，我们可以这样叫它super(CachingPaginator,self).count()AFAIK。我在报告中提到了许多问题，但都没有帮助我。有人能帮我吗。你知道吗

Answer 1

属性只是类属性。要获取父类的class属性，可以对父类使用直接查找（Paginator.count）或super()调用。在本例中，如果对父类使用直接查找，则必须手动调用描述符协议，这有点冗长，因此使用super()是最简单的解决方案：

class Paginator(object):
    @property
    def count(self):
        print "in Paginator.count"
        return 42

class CachingPaginator(Paginator):
    def __init__(self):
        self._count = None

    def _get_count(self):
        if self._count is None:
            self._count = super(CachingPaginator, self).count 
        # Here, I want to get count property in the super-class, this is giving me -1 which is wrong
        return self._count
    count = property(_get_count)

如果要直接查找父类，请替换：

self._count = super(CachingPaginator, self).count 

与

self._count = Paginator.count.__get__(self, type(self))