Django - How to get self.id when saving a new object? - django

Question

I have a problem in one of my models. I'm uploading an image, and I want to store the id (pk in the database table) but I need to know at which point Django will have access to self.id.
models.py
class BicycleAdItemKind(MPTTModel):
def url(self, filename):
pdb.set_trace()
url = "MultimediaData/HelpAdImages/ItemKind/%s/%s" % (self.id, filename)
return url
def item_kind_image(self):
return '<img align="middle" src="/media/%s" height="60px" />' % self.image
item_kind_image.allow_tags = True
# Bicicleta completa, Componentes para bicicleta, Acessorios para ciclista
n_item_kind = models.CharField(max_length=50)
parent = TreeForeignKey('self', null=True,
blank=True, related_name='children')
description = models.TextField(null=True, blank=True)
image = models.ImageField(upload_to=url, null=True, blank=True)
date_inserted = models.DateTimeField(auto_now_add=True)
date_last_update = models.DateTimeField(auto_now=True)
def __unicode__(self):
return self.n_item_kind
class MPTTMeta:
order_insertion_by = ['n_item_kind']
The problem is in the url() method; I can only get self.id when updating an object, I don't get the self.id when creating a new object. How can I modify this model so that I get self.id when creating a new object?
With the current code, when I'm creating a new object I will end up with a url like:
MultimediaData/HelpAdImages/ItemKind/None/somefile.jpg
And I need to have something like:
MultimediaData/HelpAdImages/ItemKind/35/somefile.jpg
Any clues?

Answer 1

If it's a new object, you need to save it first and then access self.id, because
"There's no way to tell what the value of an ID will be before you call save(),
because that value is calculated by your database, not by Django."
Check django's document https://docs.djangoproject.com/en/dev/ref/models/instances/

Answer 2

You might need to save this file/instance twice:
You can use a post_save signal on the model that looks for the created flag, and re-saves the instance updating the url (and moving/renaming the file as necessary), since the instance will now have an ID. Make sure you only do this conditioned on created, though, otherwise you will continuously loop in saving: saving kicks off a post-save signal, which does a save, which kicks off a post-save signal...
See https://docs.djangoproject.com/en/dev/ref/signals/#post-save

Answer 3

There is actually a way to trick this out.
class Test(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=150)
def __str__(self):
return self.name
def update_model(self):
# You now have both access to self.id and self.name
test_id = Test.objects.get(name=self.name).id
print(test_id)
# Do some stuff, update your model...
Test.objects.filter(id=test_id).update(name='New Name')
def save(self, *args, **kwargs):
super(Test, self).save(*args, **kwargs)
self.update_model() # Call the function

Answer 4

Note: You need to have the models.AutoField(primary_key=True) attribute set, otherwise the database will be updated with a new id but Django will not recognize it.
models.AutoField(primary_key=True)

Answer 5

I understand this is old but for anyone that stumbles across this in the future, here's actually how you do it now in Django.
def url(instance, filename):
pdb.set_trace()
url = "MultimediaData/HelpAdImages/ItemKind/%s/%s" % (instance.id, filename)
return url

Answer 6

q = Order.objects.values_list('id', flat=True).order_by('-id')[:1]
if len(q):
self.number = str(self.id) if self.id else str(int(q.get()) + 1)
else:
self.number = 1