Programmatically saving image to Django ImageField

Question

Ok, I've tried about near everything and I cannot get this to work.

• I have a Django model with an ImageField on it
• I have code that downloads an image via HTTP (tested and works)
• The image is saved directly into the 'upload_to' folder (the upload_to being the one that is set on the ImageField)
• All I need to do is associate the already existing image file path with the ImageField

I've written this code about 6 different ways.

The problem I'm running into is all of the code that I'm writing results in the following behavior: (1) Django will make a 2nd file, (2) rename the new file, adding an _ to the end of the file name, then (3) not transfer any of the data over leaving it basically an empty re-named file. What's left in the 'upload_to' path is 2 files, one that is the actual image, and one that is the name of the image,but is empty, and of course the ImageField path is set to the empty file that Django try to create.

In case that was unclear, I'll try to illustrate:

## Image generation code runs.... 
/Upload
     generated_image.jpg     4kb

## Attempt to set the ImageField path...
/Upload
     generated_image.jpg     4kb
     generated_image_.jpg    0kb

ImageField.Path = /Upload/generated_image_.jpg

How can I do this without having Django try to re-store the file? What I'd really like is something to this effect...

model.ImageField.path = generated_image_path

...but of course that doesn't work.

And yes I've gone through the other questions here like this one as well as the django doc on File

UPDATE After further testing, it only does this behavior when running under Apache on Windows Server. While running under the 'runserver' on XP it does not execute this behavior.

I am stumped.

Here is the code which runs successfully on XP...

f = open(thumb_path, 'r')
model.thumbnail = File(f)
model.save()

Answer 1

This is might not be the answer you are looking for. but you can use charfield to store the path of the file instead of ImageFile. In that way you can programmatically associate uploaded image to field without recreating the file.

Answer 2

I have some code that fetches an image off the web and stores it in a model. The important bits are:

from django.core.files import File  # you need this somewhere


# The following actually resides in a method of my model

result = urllib.urlretrieve(image_url) # image_url is a URL to an image

# self.photo is the ImageField
self.photo.save(
    os.path.basename(self.url),
    File(open(result[0]))
    )

self.save()

That's a bit confusing because it's pulled out of my model and a bit out of context, but the important parts are:

• The image pulled from the web is not stored in the upload_to folder, it is instead stored as a tempfile by urllib.urlretrieve() and later discarded.
• The ImageField.save() method takes a filename (the os.path.basename bit) and a django.core.files.File object.

Let me know if you have questions or need clarification.

Edit: for the sake of clarity, here is the model (minus any required import statements):

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0]))
                    )
            self.save()

Answer 3

Just a little remark. tvon answer works but, if you're working on windows, you probably want to open() the file with 'rb'. Like this:

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

or you'll get your file truncated at the first 0x1A byte.