Django override save for model only in some cases?

Question:

Before saving model I’m re-size a picture. But how can I check if new picture added or just description updated, so I can skip rescaling every time the model is saved?

class Model(model.Model):
    image=models.ImageField(upload_to='folder')
    thumb=models.ImageField(upload_to='folder')
    description=models.CharField()


    def save(self, *args, **kwargs):
        if self.image:
            small=rescale_image(self.image,width=100,height=100)
            self.image_small=SimpleUploadedFile(name,small_pic)
        super(Model, self).save(*args, **kwargs)

I want to rescale only if new image loaded or image updated, but not when description updated.

Asked By: Pol

||

Answers:

Query the database for an existing record with the same PK. Compare the file sizes and checksums of the new and existing images to see if they’re the same.

You may supply extra argument for confirming a new image is posted.
Something like:

def save(self, new_image=False, *args, **kwargs):
    if new_image:
        small=rescale_image(self.image,width=100,height=100)
        self.image_small=SimpleUploadedFile(name,small_pic)
    super(Model, self).save(*args, **kwargs)

or pass request variable

def save(self, request=False, *args, **kwargs):
    if request and request.FILES.get('image',False):
        small=rescale_image(self.image,width=100,height=100)
        self.image_small=SimpleUploadedFile(name,small_pic)
    super(Model, self).save(*args, **kwargs)

I think these wont break your save when called simply.

You may put this in your admin.py so that this work with admin site too (for second of above solutions):

class ModelAdmin(admin.ModelAdmin):

    ....
    def save_model(self, request, obj, form, change): 
        instance = form.save(commit=False)
        instance.save(request=request)
        return instance
Answered By: crodjer

Some thoughts:

class Model(model.Model):
    _image=models.ImageField(upload_to='folder')
    thumb=models.ImageField(upload_to='folder')
    description=models.CharField()

    def set_image(self, val):
        self._image = val
        self._image_changed = True

        # Or put whole logic in here
        small = rescale_image(self.image,width=100,height=100)
        self.image_small=SimpleUploadedFile(name,small_pic)

    def get_image(self):
        return self._image

    image = property(get_image, set_image)

    # this is not needed if small_image is created at set_image
    def save(self, *args, **kwargs):
        if getattr(self, '_image_changed', True):
            small=rescale_image(self.image,width=100,height=100)
            self.image_small=SimpleUploadedFile(name,small_pic)
        super(Model, self).save(*args, **kwargs)

Not sure if it would play nice with all pseudo-auto django tools (Example: ModelForm, contrib.admin etc).

Answered By: petraszd

Check the model’s pk field. If it is None, then it is a new object.

class Model(model.Model):
    image=models.ImageField(upload_to='folder')
    thumb=models.ImageField(upload_to='folder')
    description=models.CharField()


    def save(self, *args, **kwargs):
        if 'form' in kwargs:
            form=kwargs['form']
        else:
            form=None

        if self.pk is None and form is not None and 'image' in form.changed_data:
            small=rescale_image(self.image,width=100,height=100)
            self.image_small=SimpleUploadedFile(name,small_pic)
        super(Model, self).save(*args, **kwargs)

Edit: I’ve added a check for ‘image’ in form.changed_data. This assumes that you’re using the admin site to update your images. You’ll also have to override the default save_model method as indicated below.

class ModelAdmin(admin.ModelAdmin):
    def save_model(self, request, obj, form, change):
        obj.save(form=form)
Answered By: DM Graves

What I did to achieve the goal was to make this..

# I added an extra_command argument that defaults to blank
def save(self, extra_command="", *args, **kwargs):

and below the save() method is this..

# override the save method to create an image thumbnail
if self.image and extra_command != "skip creating photo thumbnail":
    # your logic here

so when i edit some fields but not editing the image, I put this..

Model.save("skip creating photo thumbnail")

you can replace the "skip creating photo thumbnail" with "im just editing the description" or a more formal text.

Hope this one helps!

Answered By: bonbon.langes

In new version it is like this:

def validate(self, attrs):
    has_unknown_fields = set(self.initial_data) - set(self.fields.keys())
    if has_unknown_fields:
        raise serializers.ValidationError("Do not send extra fields")
    return attrs
Answered By: Dan Goriaynov

I have found one another simple way to store the data into the database

models.py

class LinkModel(models.Model):
    link = models.CharField(max_length=500)
    shortLink = models.CharField(max_length=30,unique=True)

In database I have only 2 variables

views.py

class HomeView(TemplateView):
    def post(self,request, *args, **kwargs):
        form = LinkForm(request.POST)

        if form.is_valid():
            text = form.cleaned_data['link'] # text for link

        dbobj = LinkModel()
        dbobj.link = text
        self.no = self.gen.generateShortLink() # no for shortLink
        dbobj.shortLink = str(self.no)
        dbobj.save()         # Saving from views.py

In this I have created the instance of model in views.py only and putting/saving data into 2 variables from views only.

Answered By: Devendra Bhat
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