Setting url of Django Imagefield model attribute v

I have a Django web app where users upload images and others view them. I have a custom storage class in this app to upload image files to Azure Cloud Storage. Currently images are being uploaded successfully, but their urls are not being set. Thus, the following code in my template yields a broken image:

{% if entry.image_file %}
<img src="{{ entry.image_file.url }}"></img><br>
{% endif %}

Can you point out what my custom storage class is missing? Here's how it appears in my models.py currently:

from django.db import models
import os
from django.conf import settings
from django.core.files.storage import Storage
from azure.storage.blob import BlobService
accountName = 'accname'
accountKey = 'acckey'

class OverwriteStorage(Storage):
    container = 'containername'
    account_name = accountName
    account_key = accountKey

    def __init__(self, account_name=None, account_key=None, container=None):

        if account_name is not None:
            self.account_name = account_name

        if account_key is not None:
            self.account_key = account_key

        if container is not None:
            self.container = container
    def __getstate__(self):
        return dict(
            account_name=self.account_name,
            account_key=self.account_key,
            container=self.container
        )
    def _save(self,name,content):
        blob_service = BlobService(account_name=accountName, account_key=accountKey)
        import mimetypes
        content.open()
        content_type = None
        if hasattr(content.file, 'content_type'):
            content_type = content.file.content_type
        else:
            content_type = mimetypes.guess_type(name)[0]
        print content_type
        content_str = content.read()
        blob_service.put_blob(
            'containername',
            name,
            content_str,
            x_ms_blob_type='BlockBlob',
            x_ms_blob_content_type=content_type
        )
        content.close()
        return name
    def get_available_name(self,name):
        return name
    def _get_service(self):
        if not hasattr(self, '_blob_service'):
            self._blob_service = BlobService(
                account_name=self.account_name,
                account_key=self.account_key,
                protocol='http'
            )
        return self._blob_service
    def _open(self, name, mode='rb'):
        from django.core.files.base import ContentFile
        contents = self._get_service().get_blob(self.container, name)
        return ContentFile(contents)
    def _get_properties(self, name):
        return self._get_service().get_blob_properties(
            self.container,
            name
        )
    def _get_container_url(self):
        if not hasattr(self, '_container_url'):
            base_url = '{protocol}://{host}/{container}'
            if self.cdn_host:
                base_url = self.cdn_host
            self._container_url = base_url.format({
                'protocol': 'http',
                'host': self._get_service()._get_host(),
                'container': self.container,
            })
        return self._container_url
    def url(self, name):
        url = '%s/%s' % (self._get_container_url(), name)
        return url

class Entry(models.Model):
    description = models.TextField(validators=[MaxLengthValidator(500)])
    submitted_on = models.DateTimeField(auto_now_add=True)
    image_file = models.ImageField(upload_to=upload_to_location, storage=OverwriteStorage(), null=True, blank=True )

The example I'm following is here. I have looked at django documentation for custom file storage, and if you scroll through the code I've pasted above, I've defined a url(self, name): method. Yet this doesn't get called (I've tested it with a print statement). Please advise!

The Blobs in Azure Blob Storage have their own unique urls for accessing. The URL is in the format: http://<your_storage_name>.blob.core.windows.net/<container_name>/<blob_name>, you can directly access it in browser, if you set the access permission of blob to public.

To your issue, if it is not sensitive to your images in Blob Storage, you can simply set access permission to public blob to allow public read access to the blobs in the container, but not the container properties and metadata.

Login in Azure mange portal, click storage tab in left nav, click your storage name in the list to step in your storage manage page, click CONTAINERS tab, select the specific container name, click the EDIT button in bottom, change the access permission and click OK button to save the configuration:

Click the container name we can step in the list of blobs in this container. We can copy the URL of an item, visit in browser to have a check.

And per my understanding, if you want to show the images after uploading to Azure storage, we just need a few modification on the original code.

In the custom storage class, assume the function url() should return the correct the URL. In my test, I directly return the URL string for a quick test:

def geturl(self,name):
    return '%s/%s/%s' % ('http://garyteststorage.blob.core.windows.net','mycontainer', name)

And we can modify the return of function _save() to the URL of the image class instead of name:

url = self.geturl(name)
return url
#return name

In models.py:

def upload_path(instance, filename):
    return 'uploads-from-custom-storage-{}'.format(filename)

class Photo(models.Model):
    #description = models.TextField(validators=[MaxLengthValidator(500)])
    #submitted_on = models.DateTimeField(auto_now_add=True)
    image_file = models.ImageField(upload_to=upload_path, storage=OverwriteStorage(), null=True, blank=True )

As before, it will save the image name in database, and after modification, it will save the full url of blob in database.

code snippet In view.py:

if form.is_valid():
    newImage = Photo(image_file = request.FILES['image_file'])
    newImage.save()
    imageurl = newImage.image_file
    html = "<img src=%s></img><br>" %imageurl
    # Redirect to the document list after POST
    return HttpResponse(html)