问题
I'm trying to copy a file using a hardlink, where the file is stored as a Django FileField. I'd like to use a hardlink to save space and copy time (no changes are expected to be made to the original file or copy). However, I'm getting some odd errors when I try to call new_file.save() from the snippet below.
AttributeError: 'file' object has no attribute '_committed'
My thinking is that after making the hardlink, I can just open the linked file and store it to the Django new File instance's FileFile. Am I missing a step here or something?
models.py
class File(models.Model):
stored_file = models.FileField()
elsewhere.py
import os
original_file = File.objects.get(id=1)
original_file_path = original_file.file.path
new_file = File()
new_file_path = '/path/to/new/file'
os.makedirs(os.path.realpath(os.path.dirname(new_file_path)))
os.link(original_file_path, new_file_path)
new_file.stored_file = file(new_file_path)
new_file.save()
回答1:
There is no need to create hardlink, just duplicate the file holder:
new_file = File(stored_file=original_file.stored_file)
new_file.save()
update
If you want to specify file to FileField or ImageField, you could simply
new_file = File(stored_file=new_file_path)
# or
new_file = File()
new_file.stored_file = new_file_path
# or
from django.core.files.base import File
# from django.core.files.images import ImageFile # for ImageField
new_file.stored_file = File(new_file_path)
the field accepts path in basestring or File() instance, the code in your question uses file() and hence is not accepted.
回答2:
I think I solved this issue, but not sure why it works. I wrapped the file object in a "DjangoFile" class (I imported as DjangoFile to avoid clashing with my previously defined File model).
from django.core.files.base import File as DjangoFile
...
new_file.stored_file = DjangoFile(file(new_file_path))
new_file.save()
This approached seemed to save the file OK.
来源:https://stackoverflow.com/questions/10092334/django-copy-filefields