- Author:
- mpasternak
- Posted:
- July 19, 2017
- Language:
- Python
- Version:
- 1.10
- Score:
- 0 (after 0 ratings)
You can place this snippet in management/commands/ and have it upload files from disk into objects in your database.
Usage:
python manage.py upload_file_to_model myapp mymodel myfield 1 field_name /some/file/path
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 | # -*- 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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