how to get derived class name from base class

Question

I have a base class Person and derived classes Manager and Employee. Now, what I would like to know is the object created is Manager or the Employee.

The person is given as belows:

from Project.CMFCore.utils import getToolByName
schema = getattr(Person, 'schema', Schema(())).copy() + Schema((TextField('FirstName', required = True, widget = StringWidget(label='First Name', i18n_domain='project')), TextField('Last Name', required = True, widget = StringWidget(label='Last Name', i18n_domain='i5', label_msgid='label_pub_city'))
class Manager(BaseContent):
  def get_name(self):
    catalog = getToolByName(self, "portal_catalog")
      people = catalog(portal_type='Person')
      person={}
      for object in people:
        fname = object.firstName
        lname = object.lastName
        person['name'] = fname+' '+ lname
        # if the derived class is Employee then i would like go to the method title of employee and if its a Manager then go to the title method of Manager
        person['post'] = Employee/Manager.title()
      return person

For Manager and employees they are like (employee is also similar but some different methods)

from Project.Person import Person
class Manager(Person):
    def title(self):
      return "Manager"

For Employee the title is 'Employee'. When I create a Person it is either Manager or the Employee. When I get the person object the class is Person but I would like to know whether it is from the derived class 'Manager' or 'Employee'.

Solution 1

You can use x.__class__.__name__ to retrieve the class name as a string, e.g.

class Person:
    pass

class Manager(Person):
    pass

class Employee(Person):
    pass

def get_class_name(instance):
    return instance.__class__.__name__

>>> m = Manager()
>>> print get_class_name(m)
Manager
>>> print get_class_name(Employee())
Employee

Or, you could use isinstance to check for different types:

>>> print isinstance(m, Person)
True
>>> print isinstance(m, Manager)
True
>>> print isinstance(m, Employee)
False

So you could do something like this:

def handle_person(person):
    if isinstance(person, Manager):
        person.read_paper()     # method of Manager class only
    elif isinstance(person, Employee):
        person.work_hard()      # method of Employee class only
    elif isinstance(person, Person):
        person.blah()           # method of the base class
    else:
        print "Not a person"

Solution 2

I don't know if this is what you want, and the way you'd like it implemented, but here's a try:

>>> class Person(object):
    def _type(self):
        return self.__class__.__name__


>>> p = Person()
>>> p._type()
'Person'
>>> class Manager(Person):
    pass

>>> m = Manager()
>>> m._type()
'Manager'
>>> 

Pros: only one definition of the _type method.

Solution 3

Python objects provide a __class__ attribute which stores the type used to make that object. This in turns provides a __name__ attribute which can be used to get the name of the type as a string. So, in the simple case:

class A(object):
    pass
class B(A):
    pass

b = B()
print b.__class__.__name__

Would give:

'B'

So, if I follow your question correctly you would do:

m = Manager()
print m.__class__.__name__
'Manager'

Solution 4

Would you be looking for something like this?

>>> class Employee:
...     pass
... 
>>> class Manager(Employee):
...     pass
... 
>>> e = Employee()
>>> m = Manager()
>>> print e.__class__.__name__
Employee
>>> print m.__class__.__name__
Manager
>>> e.__class__.__name__ == 'Manager'
False
>>> e.__class__.__name__ == 'Employee'
True

Solution 5

The best way to "do this" is to not do it. Instead, create methods on Person that are overridden on Manager or Employee, or give the subclasses their own methods that extend the base class.

class Person(object):
    def doYourStuff(self):
        print "I'm just a person, I don't have anything to do"

class Manager(object):
    def doYourStuff(self):
        print "I hereby manage you"

class Employee(object):
    def doYourStuff(self):
        print "I work long hours"

If you find yourself needing to know in the base class which subclass is being instantiated, your program probably has a design error. What will you do if someone else later extends Person to add a new subclass called Contractor? What will Person do when the subclass isn't any of the hard-coded alternatives it knows about?

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Comments

[Tom Russell]

Succinct. Nice use of interactive output.

[martineau]

Doubtful this is the "best way" or that the OP has a broken design. Simpler to implement as shown in this answer in which the base class has a method that all the subclasses can use (for whatever purpose they choose).

[Saurav Kumar]

Many Thanks.. This is what I was looking for!

[Chandan]

What if I have made 3 types of User classes and I want to know about the class of the object I get from request.user. request.user returns an object of User class.

[Emmanuel]

I'm sorry @Chandan bu t I don't understand your question. Can you be more precise?

[Chandan]

I have a User class "class User(AbstractBaseUser)", "class Manager(User)" and "class Employee(User)". Django creates tables for Manager, Employee as well as for User. If I type in the shell: u1 = User.objects.all()[0], now I want to know if u1 is an Employee or a Manager. How can I do that? Or am I doing anything wrong here that is why a table for User is being created separately?

[Emmanuel]

Ok, you are in the Django context, that was not clear at all... Well, I don't know how Django manages these classes, but I'm pretty sure you can know. What about creating a dedicated post in SO for that?

[Shlomo Gottlieb]

What if I want _type to be static? (decorated by @staticmethod)

[Emmanuel]

@ShlomoGottlieb: this completely changes the solution, since this one is based on the object instance via self... I guess you have to define a static method in every derived class, which is not scalable. But you can use @classmethod similarly: def type2(cls): return cls.__name_

[Shlomo Gottlieb]

@Emmanuel Very nice! the classmethod solution works as expected.