[Django]-Download a remote image and save it to a Django model

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.

# this is my solution
from django.core import files
from django.core.files.base import ContentFile

import requests
from .models import MyModel

def download_img():
    r = requests.get("remote_file_url", allow_redirects=True)
    filename = "remote_file_url".split("/")[-1]

    my_model = MyModel(
        file=files.File(ContentFile(r.content), filename)
    )
    my_model.save()

    return

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.

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

Try doing it this way instead of assigning path to the image…

    import urllib2
    from django.core.files.temp import NamedTemporaryFile
    def handle_upload_url_file(url):
        img_temp = NamedTemporaryFile()
        opener = urllib2.build_opener()
        opener.addheaders = [('User-agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:15.0) Gecko/20120427 Firefox/15.0a1')]
        img_temp.write(opener.open(url).read())
        img_temp.flush()
        return img_temp

use the above function like this..

    new_image = images_data()
    #rest of the data in new_image and then do this.
    new_image.image.save(slug_filename,File(handle_upload_url_file(url)))
    #here slug_filename is just filename that you want to save the file with.

In case you are saving image by overriding models’ save method to modify the name of file and struggling with random invalid filename(like me) in django. You can follow up below code (Copied from Accepted answer):

lf = tempfile.NamedTemporaryFile()


for block in response.iter_content(1024*8):

        if not block:
            break

        lf.write(block)
    lf.name = name.  # Set your custom file name here
    dc = ImageFile(file=files.File(lf))

    dc.file.save()

I have configured my storage with django-storages, in order to directly upload media content to s3. For some reasons I wasn’t able to replace file name. After some R&D it worked.

Note: I have used FileField in the model, hence few line of code is not needed

def qrcodesave(request): 
    import urllib2;   
    url ="http://chart.apis.google.com/chart?cht=qr&chs=300x300&chl=s&chld=H|0"; 
    opener = urllib2.urlopen(url);  
    mimetype = "application/octet-stream"
    response = HttpResponse(opener.read(), mimetype=mimetype)
    response["Content-Disposition"]= "attachment; filename=aktel.png"
    return response