Django: modify model’s field before save

Question

I have a model, course, with an ImageField and FileField so I’d like to create a folder each time the user create a course. I think I can do this before the model is saved so here is my question.

How can I access a model’s fields in a method?

Models.py

Class Course(models.Model):
 Thumbnail = model.ImageField(upload_to=“...”,...)

    def save(self, *args, **kwargd):
        ... #How can I alter here the ImageField parameters?
        super().save(*args, **kwargs)

Access the model's fields or the instance of the model's fields? — Matthew Gaiser, Apr 11 '20 at 04:23

score 5 · Accepted Answer · edited May 09 '21 at 16:29

5

See here on getting the model's fields.

To get field of an object instance then it should just be as

def save(self, *args, **kwargs):
     #### How can I alter here the ImageField parameters?
     self.Thumbnail = #do whatever you want here
     super().save(*args, **kwargs)

It's not possible to alter field's params in a method. It must only be done in the field definitions because the model must be saved first

edited May 09 '21 at 16:29

alinajafi

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answered Apr 11 '20 at 04:34

AzyCrw4282

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self.Thumnail is the instance, I'm trying to modify the model's field parameters so this doesn't work – VicenteC Apr 11 '20 at 04:49
That's exactly how i mentioned it. To get the model's field simply follow the link above it. :) – AzyCrw4282 Apr 11 '20 at 04:50
I have the model's field! Thank you! By the way, how can I modify now the parameters of the fields? I know that in widgets its something like: `widgets.attrs.parameter(...')` – VicenteC Apr 11 '20 at 05:03
What do you exactly mean by params, you mean to alter this `(upload_to=“...”,...)`? – AzyCrw4282 Apr 11 '20 at 05:09
Yes, (upload_to=“...) and so on – VicenteC Apr 11 '20 at 05:11
I don't think that's possible to alter field's params in a method. It must only be done in the field definiitons – AzyCrw4282 Apr 11 '20 at 05:32

score 2 · Answer 2 · answered Apr 11 '20 at 04:39

2

Instance of the models field can be accessed using self like a normal instance variable.

For example in a model class below,

class DummyModel(models.Model):
    key = models.CharField()
    value = models.TextField()

    def save(self, *args, **kwargs):
        value = self.value # self.value is a model field.
        super().save(*args, **kwargs)

In your case, you can access it using self.Thumbnail

answered Apr 11 '20 at 04:39

Paras Mishra

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Is there any way to modify the parameters of the model's field? Not the instance – VicenteC Apr 11 '20 at 04:48

score 1 · Answer 3 · answered Apr 11 '20 at 06:17

Since it's not possible to modify the model field's parameters before create it, as @azyCrw4282 said in his answer, it's possible to create a directory with the name of the model's instance and upload there the files parsing a function to upload_to

models.py

def upload_to(instance, filename):
    return 'user_course/{0}/{1}'.format(instance.name, filename) #instance.name will be the name of the course


class Course(models.Model):
    name       = model.CharField(...)
    thumbnail  = models.ImageField(upload_to=upload_to, ...)

Django: modify model’s field before save

3 Answers3

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