Python:将类名作为参数传递给函数？

Question

class TestSpeedRetrieval(webapp.RequestHandler):
  """
  Test retrieval times of various important records in the BigTable database 
  """
  def get(self):
      commandValidated = True 
      beginTime = time()
      itemList = Subscriber.all().fetch(1000) 

      for item in itemList: 
          pass 
      endTime = time()
      self.response.out.write("<br/>Subscribers count=" + str(len(itemList)) + 
           " Duration=" + duration(beginTime,endTime)) 

如何将上述内容转换为我传递类名称的函数？在上面的示例中,Subscriber(在Subscriber.all(..fetch语句中)是一个类名,这是您使用Python在Google BigTable中定义数据表的方式.

我想做这样的事情:

       TestRetrievalOfClass(Subscriber)  
or     TestRetrievalOfClass("Subscriber")  

谢谢,Neal Walters

Answer 1

class TestSpeedRetrieval(webapp.RequestHandler):
  """
  Test retrieval times of various important records in the BigTable database 
  """
  def __init__(self, cls):
      self.cls = cls

  def get(self):
      commandValidated = True 
      beginTime = time()
      itemList = self.cls.all().fetch(1000) 

      for item in itemList: 
          pass 
      endTime = time()
      self.response.out.write("<br/>%s count=%d Duration=%s" % (self.cls.__name__, len(itemList), duration(beginTime,endTime))

TestRetrievalOfClass(Subscriber)  

Answer 2

如果直接传递类对象,就像在"喜欢这个"和"或"之间的代码中那样,可以将其名称作为__name__属性.

从名称开始(如在"或" 之后的代码中)使得检索类对象非常困难(并且不明确),除非您有关于类对象可能包含的位置的指示 - 所以为什么不传递类对象代替？!