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@mpasternak
Created August 15, 2017 19:40
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Django - upload a file to a given model's FileField in a given application
# -*- encoding: utf-8 -*-
# save me as yourapp/management/commands/upload_file_to_model.py
from argparse import FileType
from pathlib import Path
from django.contrib.contenttypes.models import ContentType
from django.core.management.base import BaseCommand
from django.core.files.base import File
class Command(BaseCommand):
help = 'Uploads a file to a given field in a given model'
def add_arguments(self, parser):
parser.add_argument("app")
parser.add_argument("model")
parser.add_argument("pk")
parser.add_argument("field")
parser.add_argument("path", type=FileType('rb'))
def handle(self, *args, **options):
obj = ContentType.objects.get_by_natural_key(
options['app'].lower(),
options['model'].lower()
).get_object_for_this_type(pk=options['pk'])
try:
field = getattr(obj, options['field'])
except AttributeError as e:
fields = [field.name for field in obj._meta.get_fields()]
fields = ", ".join(fields)
e.args = (e.args[0] + f". Available names: {fields}", )
raise e
field.save(
name=Path(options['path'].name).name,
content=File(options['path']))
obj.save()
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