copy file from one model to another

I have 2 simple models:

class UploadImage(models.Model):
   Image = models.ImageField(upload_to="temp/")

class RealImage(models.Model):
   Image = models.ImageField(upload_to="real/")

And one form

class RealImageForm(ModelForm):
    class Meta:
        model = RealImage 

I need to save file from UploadImage into RealImage. How could i do this. Below code doesn't work

realform.Image=UploadImage.objects.get(id=image_id).Image 
realform.save()

Tnx for help.

Answers


Inspired by Gerard's solution I came up with the following code:

from django.core.files.base import ContentFile

#...

class Example(models.Model):
    file = models.FileField()

    def duplicate(self):
        """
        Duplicating this object including copying the file
        """
        new_example = Example()
        new_file = ContentFile(self.file.read())
        new_file.name = self.file.name
        new_example.file = new_file
        new_example.save()

This will actually go as far as renaming the file by adding a "_1" to the filename so that both the original file and this new copy of the file can exist on disk at the same time.


Although this is late, but I would tackle this problem thus,

class UploadImage(models.Model):
    Image = models.ImageField(upload_to="temp/")

    # i need to delete the temp uploaded file from the file system when i delete this model      
    # from the database
    def delete(self, using=None):
        name = self.Image.name
        # i ensure that the database record is deleted first before deleting the uploaded 
        # file from the filesystem.
        super(UploadImage, self).delete(using)
        self.Image.storage.delete(name)


class RealImage(models.Model):
   Image = models.ImageField(upload_to="real/")


# in my view or where ever I want to do the copying i'll do this
import os
from django.core.files import File

uploaded_image = UploadImage.objects.get(id=image_id).Image
real_image = RealImage()
real_image.Image = File(uploaded_image, uploaded_image.name)
real_image.save()
uploaded_image.close()
uploaded_image.delete()
If I were using a model form to handle the process, i'll just do
# django model forms provides a reference to the associated model via the instance property
form.instance.Image = File(uploaded_image, os.path.basename(uploaded_image.path))
form.save()
uploaded_image.close()
uploaded_image.delete()

note that I ensure the uploaded_image file is closed because calling real_image.save() will open the file and read its content. That is handled by what ever storage system is used by the ImageField instance


Try doing that without using a form. Without knowing the exact error that you are getting, I can only speculate that the form's clean() method is raising an error because of a mismatch in the upload_to parameter.

Which brings me to my next point, if you are trying to copy the image from 'temp/' to 'real/', you will have to do a some file handling to move the file yourself (easier if you have PIL):

import Image
from django.conf import settings

u = UploadImage.objects.get(id=image_id)
im = Image.open(settings.MEDIA_ROOT + str(u.Image))
newpath = 'real/' + str(u.Image).split('/', 1)[1]
im.save(settings.MEDIA_ROOT + newpath)
r = RealImage.objects.create(Image=newpath)

Hope that helped...


I had the same problem and solved it like this, hope it helps anybody:

# models.py
class A(models.Model):
    # other fields...
    attachment = FileField(upload_to='a')

class B(models.Model):
    # other fields...
    attachment = FileField(upload_to='b')

# views.py or any file you need the code in
try:
    from cStringIO import StringIO
except ImportError:
    from StringIO import StringIO
from django.core.files.base import ContentFile
from main.models import A, B

obj1 = A.objects.get(pk=1)

# You and either copy the file to an existent object
obj2 = B.objects.get(pk=2)

# or create a new instance
obj2 = B(**some_params)

tmp_file = StringIO(obj1.attachment.read())
tmp_file = ContentFile(tmp_file.getvalue())
url = obj1.attachment.url.split('.')
ext = url.pop(-1)
name = url.pop(-1).split('/')[-1]  # I have my files in a remote Storage, you can omit the split if it doesn't help you
tmp_file.name = '.'.join([name, ext])
obj2.attachment = tmp_file

# Remember to save you instance
obj2.save()

Update Gerard's Solution to handle it in a generic way:

try:
    from cStringIO import StringIO
except ImportError:
    from StringIO import StringIO

from django.core.files.base import ContentFile

init_str = "src_obj." + src_field_name + ".read()"
file_name_str = "src_obj." + src_field_name + ".name"

try:
    tmp_file = StringIO(eval(str(init_str)))
    tmp_file = ContentFile(tmp_file.getvalue())
    tmp_file.name = os.path.basename(eval(file_name_str))
except AttributeError:
    tmp_file = None

if tmp_file:
    try:
        dest_obj.__dict__[dest_field_name] = tmp_file
        dest_obj.save()
    except KeyError:
        pass

Variable's Used:

  1. src_obj = source attachment object.
  2. src_field_name = source attachment object's FileField Name.
  3. dest_obj = destination attachment object.
  4. dest_field_name = destination attachment object's FileField Name.

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