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Jeff Muday
Jeff Muday
Treehouse Moderator 28,249 Points

Django: display ImageField() with admin.site.register(model.MyModel) trick?

If I declare an ImageField() to be part of a model, and then subsequently register that model in the admin site, is there a way to display that ImageField() as a thumbnail?

For example, if I have a UserProfile picture, I would like to see it displayed as a picture/thumbnail in the admin interface rather than just the filename (as it currently is).

Is there a trick I am missing about getting an ImageField() to display in the interface, or is this not built into the current version of Django (1.9.x)?

1 Answer

Jeff Muday
Jeff Muday
Treehouse Moderator 28,249 Points

I solved my issue after re-reading the Django project documentation on static files and admin site. This is probably not the most elegant solution, but it is working and can certainly use a few additional tweaks. I created a demo project and application as "proof of concept". In the sample "app" I have an image model. It works in the development environment, but haven't tested on a server environment, which may require a few additional tweaks.

from the sampleapp


import os
from django.db import models
from django.conf import settings
from django.utils.safestring import mark_safe

# the settings for MEDIA_ROOT and MEDIA_URL come from the project settings
# but could be overridden in the model
# MEDIA_ROOT = '/home/<user>/project/imgproject/media_cdn'
# MEDIA_URL = '/media'

# Create your models here.
class Image(models.Model):
   # allows for an image to be either stored in the MEDIA_ROOT path or
   # be a reference to an external URL to an image.
    created_at = models.DateTimeField(auto_now_add=True)
    title = models.CharField(max_length = 255)
    description = models.TextField(blank=True)
    image = models.ImageField(upload_to=settings.MEDIA_ROOT, blank=True)
    externalURL = models.URLField(blank=True)

    def url(self):
        # returns a URL for either internal stored or external image url
        if self.externalURL:
            return self.externalURL
            # is this the best way to do this??
            return os.path.join('/',settings.MEDIA_URL, os.path.basename(str(self.image)))

    def image_tag(self):
        # used in the admin site model as a "thumbnail"
        return mark_safe('<img src="{}" width="150" height="150" />'.format(self.url()) )
    image_tag.short_description = 'Image'    

    def __unicode__(self):
        # add __str__() if using Python 3.x
        return self.title

from sampleapp


from django.contrib import admin
from .models import Image

class ImageAdmin(admin.ModelAdmin):
    # explicitly reference fields to be shown, note image_tag is read-only
    fields = ( 'image_tag','title','description','image','externalURL', )
    readonly_fields = ('image_tag',)

admin.site.register(Image, ImageAdmin)