Django how to check if user uploaded an image file when saving model with ImageField?

Question

I am looking for a way to resize, compress, and optimize the uploaded image when saving an ImageField.

class Image(models.Model):
    name = models.CharField(max_length=254, blank=True)
    caption = models.TextField(max_length=1000, blank=True)
    height = models.IntegerField()
    width = models.IntegerField()
    image = models.ImageField(upload_to='', height_field='height', width_field='width', storage=S3MediaStorage())

My first thought was to override the model's save() and implement this logic there, but I don't want the resize/compression/optimization to run again if the user doesn't update the image file (i.e. if he only updates name or caption on an existing object and saves it).

1. What is a proper way to check when a new image file is uploaded to the ImageField, but not when the user only changes another field in the Model, eg. the user updates caption but leaves everything else as-is?

2. How can the uploaded image file be accessed in code? I.e. what is the variable that contains the actual image file that can be passed to Pillow?

edit: This is unique from the suspected duplicate. I am not asking if the field has changed, because that would always cause false positives. I am asking if the user has uploaded an image file, which I will immediately change (resize/optimize/compress), so if the user immediately downloads his uploaded image he'll find that has a different binary with a randomly generated filename, and therefore comparing the filename or binary are not valid methods to determine if the user is uploading a different image.

Answer 1

Your model could use a different name.

Nevertheless, you can try manipulating the image through a post_save signal (https://docs.djangoproject.com/en/1.9/ref/signals/#post-save)

from PIL import Image
from django.db.models.signals import post_save

@receiver(post_save, sender=Image)
def crop_image(sender, instance, **kwargs):
    img = instance.image
    original = Image.open(img.src.path)
    # ... your code here...

EDIT: Apologies. Jumped the gun a bit. One of your actual problems was to not manipulate the image if it's the same. You can do it on save() like this (UNTESTED):

def save(self, **kwargs):
    try:
        related_img = Image.objects.get(id=self.id)
        if related_img.image != self.image:
            crop_me(self.image)
    except Image.DoesNotExist:
        # object doesn't exist. Passing...
        pass

    return super(Image, self).save(**kwargs)

def crop_me(img):
    original_img = Image.open(img.src.path)
    # ... your code here...

EDIT 2: If the name changes you could save the original filename in an helper field

class Image(models.Model):
    image = models.ImageField(upload_to='', height_field='height', width_field='width', storage=S3MediaStorage())

    __original_image_filename = None

    def __init__(self, *args, **kwargs):
        super(Image, self).__init__(*args, **kwargs)
        self.__original_image_filename = self.image.name

    def save(self, force_insert=False, force_update=False, *args, **kwargs):
        if self.image.name != self.__original_image_filename:
        # name changed - do something here

        super(Image, self).save(force_insert, force_update, *args, **kwargs)
        self.__original_image_filename = self.image.name

I am modifying another answer on the fly so there could be an error or two. Please check the original answer. There are other methods on that question that could help you.