Django model - interface design to avoid need of passing object in CHILD class calling method defined in PARENT class

Django model - interface design to avoid need of passing object in CHILD class calling method defined in PARENT class



I am using django 2.0.8 and Python 3.5. I have written a base class which encapsulates behavior in a base class.



When using the interface in the child class, I find that I have to pass the object of the child class to the parent - which is not only ugly, is error prone.



I do not want to use composition (instead of an interface), because AFAIK fields in django models are saved to the DB - that aside, I prefer the sub classing approach, since all the functionality can remain in the base class.



Is there any way I can (in the parent class), find/obtain the instance (or at least the name of the class and it's id) that invoked the method call?



Here is my code:


class Likeable(models.Model):
likes = GenericRelation(Like)


def action_is_permissible(self, actionable_object, actor):
ct = ContentType.objects.get_for_model(actionable_object)
object_id = actionable_object.id
found_objects = Like.objects.filter(content_type=ct, object_id=object_id, liker=actor)

return ((len(found_objects) == 0), ct, object_id, found_objects)


def add_like(self, actionable_object, actor):

can_add, ct, object_id, found_objects = self.action_is_permissible(actionable_object, actor)

if can_add:
like = self.likes.create(content_type=ct, object_id=object_id, liker=actor)
else:
# do nothing
return


class Meta:
abstract = True


class Foo(Likeable):
name = models.CharField(max_length=255,default='')
objects = models.Manager()


foo = Foo.objects.get(id=1)
p = User.objects.get(id=1)

foo.add_like(foo, p) # <- nasty API calling convention




1 Answer
1



You can access it using self.
self refers to the object which is calling.


self


self



Ref: What is the purpose of self?



EDIT (code changes):


class Likeable(models.Model):
def add_like(self, actor):
# update `actionable_object` to `self`

foo.add_like(p)





Look at the usage example. I can't invoke foo.add_like(self, p) in code - as self would be unidentified. Look at the signature of the add_like method in the base class.
– Homunculus Reticulli
Sep 5 '18 at 17:27


foo.add_like(self, p)


self


add_like





You dont have to pass self yourself. Python does that for you.
– Sachin Kukreja
Sep 5 '18 at 17:28


self





Sigh ... you don't seem to grasp what the problem is here. The method is implemented in the subclass (i.e. parent class of Foo): Likeable. At the point where the method is called, self points to an instance of the Likeable class. The problem is that I don't know how (or indeed if it's possible) to obtain an instance of the CHILD (in this case Foo) who invoked the method at the point that the code is being executed from within the Likeable class. Put simply, I'm trying to "reach out" from the Likeable class to see which class invoked the method.
– Homunculus Reticulli
Sep 5 '18 at 17:34



self


Likeable


Likeable


Likeable





"The method is implemented in the subclass". I think you meant that the method is implemented in the parent class. And, I tried this code. self does refer to foo and self.id prints out the id of the foo object.
– Sachin Kukreja
Sep 5 '18 at 17:44


The method is implemented in the subclass


self


foo


self.id


id


foo





You're right, I mispoke. I sometimes get confused switching over from C++ and other languages (where the naming convention seems to be different) to Python. method is implemented in the parent class.
– Homunculus Reticulli
Sep 5 '18 at 17:48




Thanks for contributing an answer to Stack Overflow!



But avoid



To learn more, see our tips on writing great answers.



Some of your past answers have not been well-received, and you're in danger of being blocked from answering.



Please pay close attention to the following guidance:



But avoid



To learn more, see our tips on writing great answers.



Required, but never shown



Required, but never shown




By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

Edmonton

Crossroads (UK TV series)