How to set up a callable for Django Imagefield's >> default << parameter

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So I have a Model that will contain a couple hundred instances and I now added an imagefield logo. As I don't want to upload each and every logo via the admin manually, I want to set up a callable that returns the correct file path so I can just push the logos to media/company_logos. Due to some technical hick-up I can push the logos to the folder only after creating their instance.

# models.py

def logo_directory_path(instance, filename):
    return f'company_logos/{instance.symbol}'

class Company(models.Model):

    symbol = models.CharField(max_length=50)
    logo = models.ImageField(upload_to='company_logos/', default=logo_directory_path(instance=symbol, filename=f'{symbol}.png'))

This returns an error at instance=symbol saying

Type 'str' doesn't have expected attribute 'symbol'

Another approach was

logo = models.ImageField(upload_to='company_logos/', default=f'company_logos/{symbol}')

which returns a URL as src="/media/company_logos/<django.db.models.fields.CharField>"

I am not quiet sure how to set this up as I cant fully understand the official docs.

My desired outcome is to have a URL for default like so:

company_logos/IBM.png with field symbol being the corresponding part (I tried it via f-string).

Update to Ankits request --> When trying to create a new instance with the code suggested it throws

logo_directory_path() missing 2 required positional arguments: 'instance' and 'filename'

django python
1 Answers

You use the logic in the uploaded_to=… parameter [Django-doc], so:

from pathlib import Path


class Company(models.Model):
    def logo_directory_path(self, filename):
        return f'company_logos/{self.symbol}{Path(filename).suffix}'

    symbol = models.CharField(max_length=50)
    logo = models.ImageField(upload_to=logo_directory_path)
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