Django FileField with upload_to determined at runtime

Question

I'm trying to set up my uploads so that if user joe uploads a file it goes to MEDIA_ROOT/joe as opposed to having everyone's files go to MEDIA_ROOT. The problem is I don't know how to define this in the model. Here is how it currently looks:

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to='.')

So what I want is instead of '.' as the upload_to, have it be the user's name.

I understand that as of Django 1.0 you can define your own function to handle the upload_to but that function has no idea of who the user will be either so I'm a bit lost.

Thanks for the help!

Answer 1

You've probably read the documentation, so here's an easy example to make it make sense:

def content_file_name(instance, filename):
    return '/'.join(['content', instance.user.username, filename])

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=content_file_name)

As you can see, you don't even need to use the filename given - you could override that in your upload_to callable too if you liked.

Answer 2

This really helped. For a bit more brevity's sake, decided to use lambda in my case:

file = models.FileField(
    upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)

Answer 3

A note on using the 'instance' object's pk value. According to the documentation:

In most cases, this object will not have been saved to the database yet, so if it uses the default AutoField, it might not yet have a value for its primary key field.

Therefore the validity of using pk depends on how your particular model is defined.

Answer 4

If you have problems with migrations you probably should be using @deconstructible decorator.

import datetime
import os
import unicodedata

from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str


@deconstructible
class UploadToPath(object):
    def __init__(self, upload_to):
        self.upload_to = upload_to

    def __call__(self, instance, filename):
        return self.generate_filename(filename)

    def get_directory_name(self):
        return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))

    def get_filename(self, filename):
        filename = default_storage.get_valid_name(os.path.basename(filename))
        filename = force_text(filename)
        filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
        return os.path.normpath(filename)

    def generate_filename(self, filename):
        return os.path.join(self.get_directory_name(), self.get_filename(filename))

Usage:

class MyModel(models.Model):
    file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)

Answer 5

I wanted to change the upload path in runtime, and none of the solutions were suitable for this need.

this is what I've done:

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=DynamicUploadPath.get_file_path)


class ContentSerializer(serializers.ModelSerializer):
    class Meta:
        model = Content
        fields = '__all__'


class UploadDir(models.TextChoices):
    PRODUCT = 'PRD', _('Product')
    USER_PROFILE = 'UP', _('User Profile')


class DynamicUploadPath:
    dir: UploadDir = None

    @classmethod
    def get_file_path(cls, instance, filename):
        return str(cls.dir.name.lower() + '/' + filename)


def set_DynamicUploadPath(dir: UploadDir):
    DynamicUploadPath.dir = dir


class UploadFile(APIView):
    parser_classes = (MultiPartParser, FormParser)

    def post(self, request):
        # file save path: MEDIA_ROOT/product/filename
        set_DynamicUploadPath(UploadDir.PRODUCT)

        # file save path: MEDIA_ROOT/user_profile/filename
        # set_DynamicUploadPath(UploadDir.USER_PROFILE)

        serializer = ContentSerializer(data=request.data)
        serializer.is_valid(raise_exception=True)
        serializer.save()

        return Response(serializer.data, status=status.HTTP_200_OK)

Answer 6

If you have a user instance, let there be a quick setup to generate

<model-slug>/<username>-<first_name>-<last_name>/filename-random.png

eg: /medias/content/ft0004-john-doe/filename-lkl9237.png

def upload_directory_name(instance, filename):

    user = getattr(instance, 'user', None)
    if user:
        name = f"{user.username}-{user.get_full_name().replace(' ', '-')}"
    else:
        name=str(instance)
    model_name = instance._meta.verbose_name.replace(' ', '-')
    return str(os.path.pathsep).join([model_name, name, filename])


class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=upload_directory_name)

[A Modified Version of @SmileyChris ]