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Home/ Questions/Q 496015
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T05:38:46+00:00 2026-05-13T05:38:46+00:00

For the following code: class Image(models.Model): alt_name = models.CharField(max_length=200) url = models.CharField(max_length=255, blank=True) class

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For the following code:

class Image(models.Model):
   alt_name = models.CharField(max_length=200)
   url = models.CharField(max_length=255, blank=True)

class Button(Image):
   source = models.ImageField(max_length=1024, upload_to='buttons')

class Snapshot(Image):
   source = models.ImageField(max_length=1024, upload_to='snapshots')

class Banner(Image):
   source = models.ImageField(max_length=1024, upload_to='banners')

In the above cases, I want to upload each of the different kind to its own upload folder. For example, banners will go under a folder called banners and snapshot will go under snapshots. The above works as expected but I’m repeating the ImageField for each sub-class. Is the above the only way to achieve my goal or is there a DRYer method?

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  1. Editorial Team
    Editorial Team
    2026-05-13T05:38:46+00:00Added an answer on May 13, 2026 at 5:38 am

    the upload_to argument is a callable, so you can pass in a function. you could do something like this:

    import os
    def upload_path(instance, filename):
         return os.path.join(instance.__class__.__name__ + 's', filename)
    
    class Image(models.Model):
        alt_name = models.CharField(max_length=200)
        url = models.CharField(max_length=255, blank=True)
        source = models.ImageField(max_length=1024, upload_to=upload_path)
    
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