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.

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.

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