获取实例的类名？

class A:
  pass

a = A()
str(a.__class__)

上面的示例代码(当在交互式解释器中输入时)将产生'__main__.A',而不是在调用属性时'A'产生的代码__name__.只需将结果传递A.__class__给str构造函数,就可以为您处理解析.但是,如果您想要更明确的内容,也可以使用以下代码.

"{0}.{1}".format(a.__class__.__module__,a.__class__.__name__)

如果您在单独的模块中定义了具有相同名称的类,则此行为更可取.

上面提供的示例代码在Python 2.7.5中进行了测试.

type(instance).__name__ != instance.__class__.__name  #if class A is defined like
class A():
   ...

type(instance) == instance.__class__                  #if class A is defined like
class A(object):
  ...

例:

>>> class aclass(object):
...   pass
...
>>> a = aclass()
>>> type(a)

>>> a.__class__

>>>
>>> type(a).__name__
'aclass'
>>>
>>> a.__class__.__name__
'aclass'
>>>


>>> class bclass():
...   pass
...
>>> b = bclass()
>>>
>>> type(b)

>>> b.__class__

>>> type(b).__name__
'instance'
>>>
>>> b.__class__.__name__
'bclass'
>>>

好问题.

这是一个基于GHZ的简单示例,它可以帮助某些人:

>>> class person(object):
        def init(self,name):
            self.name=name
        def info(self)
            print "My name is {0}, I am a {1}".format(self.name,self.__class__.__name__)
>>> bob = person(name='Robert')
>>> bob.info()
My name is Robert, I am a person

或者你可以使用classmethod装饰器:

class A:
    @classmethod
    def get_classname(cls):
        return cls.__name__

    def use_classname(self):
        return self.get_classname()

>>> A.get_classname()
'A'
>>> a = A()
>>> a.get_classname()
'A'
>>> a.use_classname()
'A'

class Spam:
    def meth(self):
        pass
    class Bar:
        pass

>>> s = Spam()
>>> type(s).__name__ 
'Spam'
>>> type(s).__qualname__
'Spam'
>>> type(s).Bar.__name__       # type not needed here
'Bar'
>>> type(s).Bar.__qualname__   # type not needed here 
'Spam.Bar'
>>> type(s).meth.__name__
'meth'
>>> type(s).meth.__qualname__
'Spam.meth'