Get name of current class?
How do I get the na开发者_StackOverflowme of the class I am currently in?
Example:
def get_input(class_name):
[do things]
return class_name_result
class foo():
input = get_input([class name goes here])
Due to the nature of the program I am interfacing with (vistrails), I cannot use __init__()
to initialize input
.
obj.__class__.__name__
will get you any objects name, so you can do this:
class Clazz():
def getName(self):
return self.__class__.__name__
Usage:
>>> c = Clazz()
>>> c.getName()
'Clazz'
Within the body of a class, the class name isn't defined yet, so it is not available. Can you not simply type the name of the class? Maybe you need to say more about the problem so we can find a solution for you.
I would create a metaclass to do this work for you. It's invoked at class creation time (conceptually at the very end of the class: block), and can manipulate the class being created. I haven't tested this:
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'
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
You can access it by the class' private attributes:
cls_name = self.__class__.__name__
EDIT:
As said by Ned Batchelder
, this wouldn't work in the class body, but it would in a method.
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'
>>>
@Yuval Adam answer using @property
class Foo():
@property
def name(self):
return self.__class__.__name__
f = Foo()
f.name # will give 'Foo'
I think, it should be like this:
class foo():
input = get_input(__qualname__)
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()
I'm using python3.8 and below is example to get your current class name.
class MyObject():
@classmethod
def print_class_name(self):
print(self.__name__)
MyObject.print_class_name()
Or without @classmethod you can use
class ClassA():
def sayhello(self):
print(self.getName())
def getName(self):
return self.__class__.__name__
ClassA().sayhello()
Hope that helps others !!!
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