Question

我有一个类似下面的课程

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)

如上面的评论所示，self._count = <expression>应该获得超类中的count属性。如果是方法，我们可以像super(CachingPaginator,self).count() AFAIK一样调用它。我在SO中提到了许多问题，但没有一个对我有帮助。任何人都可以帮助我。

Answer 1

属性只是类属性。要获取父类的class属性，可以在父类（Paginator.count）或super()调用上使用直接查找。现在，在这种情况下，如果在父类上使用直接查找，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)

如果您想要直接父类查找，请替换：

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

与

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

如何从子类python中的超类获取属性名称

1 个答案: