如何從子類蟒蛇

Question

超類獲得的屬性名稱我有一個像下面

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) 

如上評論指出一類，self._count = <expression>應該得到超一流的計數屬性。如果是方法我們可以這樣稱呼它012AYAFAIK。我提到過很多問題，但沒有一個能幫助我。任何人都可以幫助我。

Answer 1

屬性只是類屬性。要獲得父級的屬性，可以使用直接查找父類（Paginator.count）或super()調用。現在，在這種情況下，如果你在父類中使用直接查找，你必須手動調用描述符的協議，這是一個有點冗長，因此使用super()是最簡單的解決方案：

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) 

如果你想有一個直接父類的查找，替換：

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

與

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