Download a remote image and save it to a Django model

Similar to @boltsfrombluesky's answer above you can do this in Python 3 without any external dependencies like so:

from os.path import basename
import urllib.request
from urllib.parse import urlparse
import tempfile

from django.core.files.base import File

def handle_upload_url_file(url, obj):
    img_temp = tempfile.NamedTemporaryFile(delete=True)
    req = urllib.request.Request(
        url, data=None,
        headers={
            'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'
        }
    )
    with urllib.request.urlopen(req) as response:
        img_temp.write(response.read())
    img_temp.flush()
    filename = basename(urlparse(url).path)
    result = obj.image.save(filename, File(img_temp))
    img_temp.close()
    return result

As an example of what I think you're asking:

In forms.py:

imgfile = forms.ImageField(label = 'Choose your image', help_text = 'The image should be cool.')

In models.py:

imgfile =   models.ImageField(upload_to='images/%m/%d')

So there will be a POST request from the user (when the user completes the form). That request will contain basically a dictionary of data. The dictionary holds the submitted files. To focus the request on the file from the field (in our case, an ImageField), you would use:

request.FILES['imgfield']

You would use that when you construct the model object (instantiating your model class):

newPic = ImageModel(imgfile = request.FILES['imgfile'])

To save that the simple way, you'd just use the save() method bestowed upon your object (because Django is that awesome):

if form.is_valid():
    newPic = Pic(imgfile = request.FILES['imgfile'])
    newPic.save()

Your image will be stored, by default, to the directory you indicate for MEDIA_ROOT in settings.py.

Accessing the image in the template:

<img src="{{ MEDIA_URL }}{{ image.imgfile.name }}"></img>

The urls can be tricky, but here's a basic example of a simple url pattern to call the stored images:

urlpatterns += patterns('',
        url(r'^media/(?P<path>.*)$', 'django.views.static.serve', {
            'document_root': settings.MEDIA_ROOT,
        }),
   )

I hope it helps.

Django Documentation is always good place to start

class ModelWithImage(models.Model):
    image = models.ImageField(
        upload_to='images',
    )

UPDATED

So this script works.

• Loop over images to download
• Download image
• Save to temp file
• Apply to model
• Save model

.

import requests
import tempfile

from django.core import files

# List of images to download
image_urls = [
    'http://i.thegrindstone.com/wp-content/uploads/2013/01/how-to-get-awesome-back.jpg',
]

for image_url in image_urls:
    # Stream the image from the url
    response = requests.get(image_url, stream=True)

    # Was the request OK?
    if response.status_code != requests.codes.ok:
        # Nope, error handling, skip file etc etc etc
        continue
    
    # Get the filename from the url, used for saving later
    file_name = image_url.split('/')[-1]
    
    # Create a temporary file
    lf = tempfile.NamedTemporaryFile()

    # Read the streamed image in sections
    for block in response.iter_content(1024 * 8):
        
        # If no more file then stop
        if not block:
            break

        # Write image block to temporary file
        lf.write(block)

    # Create the model you want to save the image to
    image = Image()

    # Save the temporary image to the model#
    # This saves the model so be sure that it is valid
    image.image.save(file_name, files.File(lf))

Some reference links:

1. requests - "HTTP for Humans", I prefer this to urllib2
2. tempfile - Save temporay file and not to disk
3. Django filefield save

If you want to save downloaded images without saving them to disk first (without using NamedTemporaryFile etc) then there's an easy way to do that.

This will be slightly quicker than downloading the file and writing it to disk as it is all done in memory. Note that this example is written for Python 3 - the process is similar in Python 2 but slightly different.

from django.core import files
from io import BytesIO
import requests

url = "https://example.com/image.jpg"
resp = requests.get(url)
if resp.status_code != requests.codes.ok:
    #  Error handling here

fp = BytesIO()
fp.write(resp.content)
file_name = url.split("/")[-1]  # There's probably a better way of doing this but this is just a quick example
your_model.image_field.save(file_name, files.File(fp))

Where your_model is an instance of the model you'd like to save to and .image_field is the name of the ImageField.

See the documentation for io for more info.