Django: add image in an ImageField from image url

Question

I just created http://www.djangosnippets.org/snippets/1890/ for this same problem. The code is similar to pithyless' answer above except it uses urllib2.urlopen because urllib.urlretrieve doesn't perform any error handling by default so it's easy to get the contents of a 404/500 page instead of what you needed. You can create callback function & custom URLOpener subclass but I found it easier just to create my own temp file like this:

from django.core.files import File
from django.core.files.temp import NamedTemporaryFile

img_temp = NamedTemporaryFile(delete=True)
img_temp.write(urllib2.urlopen(url).read())
img_temp.flush()

im.file.save(img_filename, File(img_temp))

from myapp.models import Photo
import urllib
from urlparse import urlparse
from django.core.files import File

img_url = 'http://www.site.com/image.jpg'

photo = Photo()    # set any other fields, but don't commit to DB (ie. don't save())
name = urlparse(img_url).path.split('/')[-1]
content = urllib.urlretrieve(img_url)

# See also: http://docs.djangoproject.com/en/dev/ref/files/file/
photo.image.save(name, File(open(content[0])), save=True)

Combining what Chris Adams and Stan said and updating things to work on Python 3, if you install Requests you can do something like this:

from urllib.parse import urlparse
import requests
from django.core.files.base import ContentFile
from myapp.models import Photo

img_url = 'http://www.example.com/image.jpg'
name = urlparse(img_url).path.split('/')[-1]

photo = Photo() # set any other fields, but don't commit to DB (ie. don't save())

response = requests.get(img_url)
if response.status_code == 200:
    photo.image.save(name, ContentFile(response.content), save=True)

More relevant docs in Django's ContentFile documentation and Requests' file download example.

ImageField is just a string, a path relative to your MEDIA_ROOT setting. Just save the file (you might want to use PIL to check it is an image) and populate the field with its filename.

So it differs from your code in that you need to save the output of your urllib.urlopen to file (inside your media location), work out the path, save that to your model.

I do it this way on Python 3, which should work with simple adaptations on Python 2. This is based on my knowledge that the files I’m retrieving are small. If yours aren’t, I’d probably recommend writing the response out to a file instead of buffering in memory.

BytesIO is needed because Django calls seek() on the file object, and urlopen responses don’t support seeking. You could pass the bytes object returned by read() to Django's ContentFile instead.

from io import BytesIO
from urllib.request import urlopen

from django.core.files import File


# url, filename, model_instance assumed to be provided
response = urlopen(url)
io = BytesIO(response.read())
model_instance.image_field.save(filename, File(io))

Recently I use the following approach within python 3 and Django 3, maybe this might be interesting for others aswell. It is similar to Chris Adams solution but for me it did not work anymore.

import urllib.request
from django.core.files.uploadedfile import SimpleUploadedFile
from urllib.parse import urlparse

from demoapp import models


img_url = 'https://upload.wikimedia.org/wikipedia/commons/f/f7/Stack_Overflow_logo.png'
basename = urlparse(img_url).path.split('/')[-1]
tmpfile, _ = urllib.request.urlretrieve(img_url)

new_image = models.ModelWithImageOrFileField()
new_image.title = 'Foo bar'
new_image.file = SimpleUploadedFile(basename, open(tmpfile, "rb").read())
new_image.save()