The Python Oracle

Programmatically saving image to Django ImageField

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Chapters
00:00 Programmatically Saving Image To Django Imagefield
01:46 Accepted Answer Score 201
02:47 Answer 2 Score 43
03:18 Answer 3 Score 18
03:54 Answer 4 Score 130
04:18 Thank you

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Full question
https://stackoverflow.com/questions/1308...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #django #djangomodels

#avk47



ACCEPTED ANSWER

Score 201


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
import urllib


# 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], 'rb'))
    )

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], 'rb'))
                    )
            self.save()



ANSWER 2

Score 130


Super easy if model hasn't been created yet:

First, copy your image file to the upload path (assumed = 'path/' in following snippet).

Second, use something like:

class Layout(models.Model):
    image = models.ImageField('img', upload_to='path/')

layout = Layout()
layout.image = "path/image.png"
layout.save()

tested and working in django 1.4, it might work also for an existing model.




ANSWER 3

Score 43


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.




ANSWER 4

Score 18


Here is a method that works well and allows you to convert the file to a certain format as well (to avoid "cannot write mode P as JPEG" error):

import urllib2
from django.core.files.base import ContentFile
from PIL import Image
from StringIO import StringIO

def download_image(name, image, url):
    input_file = StringIO(urllib2.urlopen(url).read())
    output_file = StringIO()
    img = Image.open(input_file)
    if img.mode != "RGB":
        img = img.convert("RGB")
    img.save(output_file, "JPEG")
    image.save(name+".jpg", ContentFile(output_file.getvalue()), save=False)

where image is the django ImageField or your_model_instance.image here is a usage example:

p = ProfilePhoto(user=user)
download_image(str(user.id), p.image, image_url)
p.save()

Hope this helps