Python调用日志记录。获取基类名称 我有一个Python源代码： 它打印： 我想：

可能吗？

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我在几个小时内找不到答案。

您这样做的原因是可以讨论的，但我认为最好是手动指定类名

class base_class(object):  
    def base_method(self):  
      logging.debug("  %s:%s() {" % ( base_class.__name__, (inspect.stack()[0][3])))
      # . . .
      logging.debug("  } // %s:%s()" % (base_class.__name__, (inspect.stack()[0][3])))

class derived_class(base_class):
  def derived_method(self):

      logging.debug("%s:%s()." % ( derived_class.__name__, (inspect.stack()[0][3])))
      self.base_method()
      logging.debug("} // %s:%s()" % ( derived_class.__name__, (inspect.stack()[0][3])))

DEBUG:root:derived_class:derived_method().
DEBUG:root:  derived_class:base_method() {
DEBUG:root:  } // derived_class:base_method()
DEBUG:root:} // derived_class:derived_method()

DEBUG:root:derived_class:derived_method().
DEBUG:root:  base_class:base_method() {
DEBUG:root:  } // base_class:base_method()
DEBUG:root:} // derived_class:derived_method()

class base_class(object):  
    def base_method(self):  
      logging.debug("  %s:%s() {" % ( base_class.__name__, (inspect.stack()[0][3])))
      # . . .
      logging.debug("  } // %s:%s()" % (base_class.__name__, (inspect.stack()[0][3])))

class derived_class(base_class):
  def derived_method(self):

      logging.debug("%s:%s()." % ( derived_class.__name__, (inspect.stack()[0][3])))
      self.base_method()
      logging.debug("} // %s:%s()" % ( derived_class.__name__, (inspect.stack()[0][3])))