Change representation of Superclass without field depending of Subclass object (str)

Question

I am running a django app and have a setup like this:

ModelSuper(models.Model):
   class Meta:
     abstract = False

ModelSub1(ModelA):
   name = models.CharField(...)
   
   def __str__:
      return self.name

ModelSub2(ModelA)
   name = models.CharField(...)

   def __str__:
      return self.name



ModelForeign(models.Model):
   element = models.ForeignKey(ModelA)

    def __str__:
      return self.name

So ModelForeign has a FK to ModelSuper. What happens now is that when I create an instance of ModelForeign I can choose if it belongs either to ModelSub1 or to ModelSub2. But the string representation is ModelSuper Onject (3) where (3) is the id.

Normally I can change this representation by overwriting the __str__ method on the model, but since I do not have any fields on the Supermodel I can't return anything.

What I tried:

I have already implemented the __str__ method in the Submodels but that does not help.
I wanted to make the Super model abstract. But this does not let me point FKs to the Supermodel, so I can't do this. My setup requires this FK
I used a generic FK with django's ContentType framework. This is also not an option because it messes completely with my app and is also not recommended from an SQL perspective.

Also when I do API-calls I get ModelSuper Onject (3) back instead of a human-readable name.

Is there a way to do what I intend to do? Thanks in advance for help and hints. Very much appreciated!

EDIT1: What I tried thanks to Abdul's help:

class ModelA(models.Model):
    class Meta:
        abstract = False

    TYPE_CHOICES = [('sub1', 'sub1'), ('sub2', 'sub2')]
    type_model = models.CharField(max_length=50, choices=TYPE_CHOICES, null=True, blank=True)

    def __str__(self):
        if self.type_model == "sub1":
            return "sub1"
        elif self.type_model == "sub2":
            return "sub2"
        else:
            return "unkown"

Abdul Aziz Barkat · Accepted Answer · 2021-01-13 15:54:18Z

1

I am not understanding how your foreign key works as model inheritance means the tables are separate. How about trying something like this:-

ModelA(models.Model):
   TYPE_CHOICES = [('Sub1', 'ModelSub1'), ('Sub2', 'ModelSub2')]
   model_type = models.CharField(max_length=4, choices=TYPE_CHOICES)
   def __str__:
      # Return string representation using if-else

   class Meta:
     abstract = False

ModelSub1(ModelA):
   name = models.CharField(...)
   model_a = models.ForeignKey(ModelA, on_delete=models.CASCADE)
   
   def __str__:
      return self.name

ModelSub2(ModelA)
   name = models.CharField(...)
   model_a = models.ForeignKey(ModelA, on_delete=models.CASCADE)

   def __str__:
      return self.name



ModelForeign(models.Model):
   element = models.ForeignKey(ModelA)

    def __str__:
      return self.name

answered Jan 13, 2021 at 15:54

Abdul Aziz Barkat

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5 Comments

Micromegas Over a year ago

thanks for your help. I tried what you suggested and added it to the question. Maybe to clarify: I never create instances of ModelA, but only of the Submodels. What I tried with your suggestion gives me unknown

Micromegas Over a year ago

Also, when I try to create a new instance of a submodel I then get __str__ returned non-string (type NoneType)... Weird... Any idea why that could be?

Abdul Aziz Barkat Over a year ago

With this solution you have to create instances of ModelA Along with the Submodels. You can simply add a few lines in the views to do this for you when the instances of the Submodels are made. Also to make the relation you need to point a foreign key / OnetoOnefield from the submodels to the ModelA

Abdul Aziz Barkat Over a year ago

do you have null=True on the CharField for name? If so try returning self.name or 'unknown'

Micromegas Over a year ago

Thank you again. hmmm I don't think it's an option for me to create an instance of ModelA every time I create a submodel... And I can't return self.name on the Superclass because I don't have a name field. And on the subclasses I already return self.name

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