현재 수업 이름을 얻으시겠습니까?

현재 수업 이름을 얻으시겠습니까?

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현재 수업 이름을 어떻게 알 수 있습니까?

예:

def get_input(class_name):
    [do things]
    return class_name_result

class foo():
    input = get_input([class name goes here])

내가 (vistrails) 인터페이스하고있는 프로그램의 특성으로 인해을 (를) __init__()초기화 하는 데 사용할 수 없습니다 input.

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obj.__class__.__name__ 개체 이름을 가져 오므로 다음과 같이 할 수 있습니다.

class Clazz():
    def getName(self):
        return self.__class__.__name__

용법:

>>> c = Clazz()
>>> c.getName()
'Clazz'

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클래스 본문 내에서 클래스 이름이 아직 정의되지 않았으므로 사용할 수 없습니다. 단순히 수업 이름을 입력 할 수 없습니까? 해결책을 찾을 수 있도록 문제에 대해 더 많이 말해야 할 수도 있습니다.

이 작업을 수행하기 위해 메타 클래스를 만들 것입니다. 클래스 생성시 (개념적으로 class : block의 맨 끝에서) 호출되며 생성되는 클래스를 조작 할 수 있습니다. 나는 이것을 테스트하지 않았습니다.

class InputAssigningMetaclass(type):
    def __new__(cls, name, bases, attrs):
        cls.input = get_input(name)
        return super(MyType, cls).__new__(cls, name, bases, newattrs)

class MyBaseFoo(object):
    __metaclass__ = InputAssigningMetaclass

class foo(MyBaseFoo):
    # etc, no need to create 'input'

class foo2(MyBaseFoo):
    # etc, no need to create 'input'

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클래스의 개인 속성으로 액세스 할 수 있습니다.

cls_name = self.__class__.__name__

편집하다:

As said by Ned Batcheler, this wouldn't work in the class body, but it would in a method.

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EDIT: Yes, you can; but you have to cheat: The currently running class name is present on the call stack, and the traceback module allows you to access the stack.

>>> import traceback
>>> def get_input(class_name):
...     return class_name.encode('rot13')
... 
>>> class foo(object):
...      _name = traceback.extract_stack()[-1][2]
...     input = get_input(_name)
... 
>>> 
>>> foo.input
'sbb'

However, I wouldn't do this; My original answer is still my own preference as a solution. Original answer:

probably the very simplest solution is to use a decorator, which is similar to Ned's answer involving metaclasses, but less powerful (decorators are capable of black magic, but metaclasses are capable of ancient, occult black magic)

>>> def get_input(class_name):
...     return class_name.encode('rot13')
... 
>>> def inputize(cls):
...     cls.input = get_input(cls.__name__)
...     return cls
... 
>>> @inputize
... class foo(object):
...     pass
... 
>>> foo.input
'sbb'
>>> 

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PEP 3155 introduced __qualname__, which was implemented in Python 3.3.

For top-level functions and classes, the __qualname__ attribute is equal to the __name__ attribute. For nested classes, methods, and nested functions, the __qualname__ attribute contains a dotted path leading to the object from the module top-level.

It is accessible from within the very definition of a class or a function, so for instance:

class Foo:
    print(__qualname__)

will effectively print Foo. You'll get the fully qualified name (excluding the module's name), so you might want to split it on the . character.

However, there is no way to get an actual handle on the class being defined.

>>> class Foo:
...     print('Foo' in globals())
... 
False

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import sys

def class_meta(frame):
    class_context = '__module__' in frame.f_locals
    assert class_context, 'Frame is not a class context'

    module_name = frame.f_locals['__module__']
    class_name = frame.f_code.co_name
    return module_name, class_name

def print_class_path():
    print('%s.%s' % class_meta(sys._getframe(1)))

class MyClass(object):
    print_class_path()

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I think, it should be like this:

    class foo():
        input = get_input(__qualname__)

참고URL : https://stackoverflow.com/questions/6943182/get-name-of-current-class