copy file from one model to another

Question

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.

Adsorbent

Answer 1

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.

Calmas

I just realized, that this way can be very memory heavy - the whole file is loaded in memory. – Swamy

Copying the file could also be a one-liner: new_example.file = ContentFile(self.file.read(), name=self.file.name) – Orran

Answer 2

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

Gladdie

Note that using image.path won't work with any 'distant' file storages (like AWS, etc). – Clisthenes

Answer 3

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...

Walter

Answer 4

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()

Gunfight

Answer 5

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.

Llewellyn

Downvote for recommending any piece of code that uses eval – Tarrasa

ew, especially for file upload... oy vey! – Blackface

You can use getattr for the functionality you used with eval. – Commemorative