Django: Adding objects to a related set without saving to DB - django

Question

I'm trying to write an internal API in my application without necessarily coupling it with the database.
class Product(models.Model):
name=models.CharField(max_length=4000)
price=models.IntegerField(default=-1)
currency=models.CharField(max_length=3, default='INR')
class Image(models.Model):
# NOTE -- Have changed the table name to products_images
width=models.IntegerField(default=-1)
height=models.IntegerField(default=-1)
url=models.URLField(max_length=1000, verify_exists=False)
product=models.ForeignKey(Product)
def create_product:
p=Product()
i=Image(height=100, widght=100, url='http://something/something')
p.image_set.add(i)
return p
Now, when I call create_product() Django throws up an error:
IntegrityError: products_images.product_id may not be NULL
However, if I call p.save() & i.save() before calling p.image_set.add(i) it works. Is there any way that I can add objects to a related object set without saving both to the DB first?

Answer 1

def create_product():
product_obj = Product.objects.create(name='Foobar')
image_obj = Image.objects.create(height=100, widght=100, url='http://something/something', product=product_obj)
return product_obj
Explanation:
Product object has to be created first and then assign it to the Image object because id and name here is required field.
I am wondering why wouldn't you not require to make a product entry in DB in first case? If there is any specific reason then i may suggest you some work around?
EDIT: Okay! i think i got you, you don't want to assign a product to an image object initially. How about creating a product field as null is equal to true.
product = models.ForeignKey(Product, null=True)
Now, your function becomes something like this:
def create_product():
image_obj = Image.objects.create(height=100, widght=100, url='http://something/something')
return image_obj
Hope it helps you?

Answer 2

I got same issue with #Saurabh Nanda
I am using Django 1.4.2. When I read in django, i see that
# file django/db/models/fields/related.py
def get_query_set(self):
try:
return self.instance._prefetched_objects_cache[rel_field.related_query_name()]
except (AttributeError, KeyError):
db = self._db or router.db_for_read(self.model, instance=self.instance)
return super(RelatedManager,self).get_query_set().using(db).filter(**self.core_filters)
# file django/db/models/query.py
qs = getattr(obj, attname).all()
qs._result_cache = vals
# We don't want the individual qs doing prefetch_related now, since we
# have merged this into the current work.
qs._prefetch_done = True
obj._prefetched_objects_cache[cache_name] = qs
That 's make sese, we only need to set property _prefetched_objects_cache for the object.
p = Product()
image_cached = []
for i in xrange(100):
image=Image(height=100, widght=100, url='http://something/something')
image_cached.append(image)
qs = p.images.all()
qs._result_cache = image_cached
qs._prefetch_done = True
p._prefetched_objects_cache = {'images': qs}

Answer 3

Your problem is that the id isn't set by django, but by the database (it's represented in the database by an auto-incremented field), so until it's saved there's no id. More about this in the documentation.
I can think of three possible solutions:
Set a different field of your Image model as the primary key (documented here).
Set a different field of your Production model as the foreign key (documented here).
Use django's database transactions API (documented here).