upload image to a folder with model's title name django [duplicate] - django

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 ]