How to use Django ImageField, and why use it at all? - django

Question

Up until now, I've been storing my image filenames in a CharField and saving the actual file directly to S3. This was a fine solution for my own usage. I'd like to reconsider using an ImageField, since now there will be other users and file input validation would be appropriate.
I have a couple of questions that weren't exactly answered after reading the docs and the source code for FileField (which appears to be essentially ImageField minus the Pillow check and dimension field updating functionality).
1) Why use an ImageField at all? Or rather, why use a FileField? Sure, it's convenient for quick-and-easy forms and convenient for inserting to Django templates. But are there any substantial reasons, eg. Is it evidently secured against exploits and malicious uploads?
2) How to write to the field file? If it is correct that the file can be read by instance.imagefield (or is it instance.imagefield.file?), if I want to write to it can I simply do the following?
#receiver(pre_save, sender=Image)
def pre_save_image(sender, instance, *args, **kwargs):
instance.imagefield = process_image(instance.imagefield)
3) How to try saving with a specific filename, then try again with a new filename if that randomly generated filename already exists? For example with my code right now I do this, how can it be done with ImageField? I want to do it at the model layer, because if I do repeated tries at the view layer then the pre_save processing would run again which is ghetto (even though it's unlikely that it'll have a second try ever in the lifetime of the service).
for i in range(tries):
try:
name = generate_random_name()
media_storage.save(name + '.jpg', ContentFile(final_bytes))
break
except:
pass
4) In the models.py pre_save and post_save signals and in the actual model's save(), how can I tell if a file came in with the request? i.e. I want to know if a new image is incoming to be saved, or if there is no image (some other field in the object is being updated and the image itself remains unchanged).

Answer 1

I don't see any advantage of FileField or ImageField over what you are doing today. In fact, as I see it, the proper/modern/scalable way to deal with uploads is to have the client (browser) upload files directly to S3.
If done correctly (from a security stand point), this scheme allows you to scale in an incredible way without the need to add more computer power on your side. As an example, consider 100 people uploading a picture at the same time. Your server will need to receive all these data, only to upload it again to S3. On the other side, you can have a 1000 people upload at the same time, and I can assure you AWS can handle it. Your server only needs to handle the signing of the URL, which is a lot less work.
Take a look at fine-uploader, as a good technology to use to handle the efficient upload to s3 (loading in chunks, error checking, etc): http://docs.fineuploader.com/endpoint_handlers/amazon-s3.html. Google "django fineuploader" to find a sample application for Django.
In my case, I use a Model with a couple CharFields (bucket, key) plus a few other things specific to my application. My data flow is as follows:
Django services a page with the fine-uploader widget, configured based on my settings.
Fineuploader requests a signed URL from the django server (endpoint), and uses that to upload to S3 directly.
When the upload is complete, fineUploader makes another request to my server to register the completion of the upload, at which time, I create my object on the database. In this case, if the upload fails, I never create an object on the database.
On the AWS side, S3 triggers a Lambda function, which I use to create a thumbnail, and store it back to S3. So, I don't even use my own CPU (e.g. Celery) for resizing. So you see, not only can I have thousands of users uploading at the same time, but I can resize those thousand pictures in parallel, and for less than what an EC2 worker will cost me.
My Django Model is also used as a wrapper to manage the business logic (e.g. functions like get_original_url() and get_thumbnail_url()), so after the uploads, it is easy for my templates to get the signed read-onlly URLs.
In short, you can implement your own version of Fineuploader if you want, or use many of the alternative, but assuming you follow the recommended security best practices on the AWS side (e.g. create a special IAM with only write permission for the client, even if you are using signed URLs), this, IMO, is the best practice for dealing with uploads, especially if you are using S3 or similar to store these files.
Sorry if I am only really answering question 1, but questions 2 and 3 don't apply if you accept my answer for 1.

Answer 2

1) Why use an ImageField at all? Or rather, why use a FileField?
It's convenient for quick-and-easy forms and convenient for inserting
to Django templates.
But are there any substantial reasons, eg. Is it evidently secured against exploits and malicious uploads?
Yes. I daresay your own code probably does it too, but for a newby using the FileField will probably ensure that your important system files are not getting overwritten by a malicious upload.
2) How to write to the field file?
In your situation you would need to use a special storage backend that makes it possible to write directly to the Amazon S3. As you know, the storage backend for FileFile and ImageField are plugable. Here is one example plugin: `http://django-storages.readthedocs.io/en/latest/backends/amazon-S3.html
There is sample code which demonstrates how it can be written to. So I wll not go into that.`
3) How to try saving with a specific filename, then try again with a new filename if that randomly generated filename already exists?
ImageField and FileField takes care of this for you automatically. It will create a new filename if the old one exists. The code in my answer here did that automatically when I called it over and over again. here are some sample filenames produces (input being bada.png)
"4", "media/bada.png"
"5", "media/bada_aH0gV7t.png"
"7", "media/bada_XkzthgK.png"
"8", "media/bada_YzZuwDi.png"
"9", "media/bada_wpkasI3.png"
4) In the models.py pre_save and post_save signals and in the actual model's save(), how can I tell if a file came in with the request?
Your instance.pk will be None
If this is a modification to an existing file the PK will be set.
If this is a new image upload in the pre_save

Answer 3

Took me forever to learn how to save an image using ImageField. Turns out it's crazy easy -- once you know how to do it, it is, at least. I mean, it all comes together sensibly after you see it.
So basically, you're working with a FileField. I already looked into the differences between ImageField and FileField:
ImageField takes everything FileField takes in terms of attributes,
but ImageField also takes a width and height attribute if indicated.
ImageField, unlike FileField, validates an upload, making sure it's
an image.
Using ImageField comes down to most of the same constructs as FileField does. The biggest things to remember:
request.FILES['name_of_model']
So a form is generated from something in forms.py (or wherever your forms are) like this:
imgfile = forms.ImageField(label = 'Choose your image',
help_text = 'The image should be cool.')
In the model, you might have this in correspondence:
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.
The tough part, which isn't really so tough when you catch on, is accessing the image.
In your template, you could have something like this:
<img src="{{ MEDIA_URL }}{{ image.imgfile.name }}"></img>
Where {{ MEDIA_URL }} is something like /media/, as indicated in settings.py and {{ image.imgfile.name }} is the name of the file and the subdirectory you indicated in the model. "image" in this case is just the current image in a loop of images you might create to access each image in the database:
{% for image in images %}
{% endfor %}
Make SURE you configure your urls properly to handle the image or the image won't work. Add this to your urls:
urlpatterns += patterns('',
url(r'^media/(?P<path>.*)$', 'django.views.static.serve', {
'document_root': settings.MEDIA_ROOT,
}),
)