Cannot assign "id": "Product.category" must be a "CategoryProduct" instance

user_7668302 0
Viewed 63

i'm working on a django project and i got this error (Cannot assign "'11'": "Product.category" must be a "CategoryProduct" instance.) anyone here can help me please.

Model:

class Product(models.Model):
    name = models.CharField("Nombre", max_length=150)
    category = models.ForeignKey(CategoryProduct, on_delete=models.SET_NULL, null=True, related_name='category')  

    def __str__(self):
        return self.name

View:

class ProductCreateView(CreateView):
model = Product
form_class = ProductForm
success_url = '/adminpanel/products/'

def post(self, request, *args, **kwargs):
    form = self.get_form()
    category = CategoryProduct.objects.get(id=request.POST['category'])
    
    if form.is_valid():
        product = form.save(commit=False)
        product.category = category
        product.save()

Form:

class ProductForm(forms.ModelForm):
    name = forms.CharField(max_length=150, label="Nombre")
    category = forms.ChoiceField(choices=[(obj.id, obj.name) for obj in CategoryProduct.objects.all()], label="Categoría")

class Meta:
    model = Product
    fields = ['name', 'category']
django-forms foreign-keys django-models django
1 Answers

You can let Django's ModelForm do its work, this will create a ModelChoiceField [Django-doc], which is where the system gets stuck: it tries to assign the primary key to category, but that should be a ProductCategory object, so you can let Django handle this with:

class ProductForm(forms.ModelForm):

    class Meta:
        model = Product
        fields = ['name', 'category']

If you want to specify a different label, you can use the verbose_name=… [Django-doc] from the model field, or specify this in the labels options [Django-doc] of the Meta of the ProductForm. So you can specify Categoria with:

class Product(models.Model):
    name = models.CharField('Nombre', max_length=150)
    category = models.ForeignKey(
        CategoryProduct,
        on_delete=models.SET_NULL,
        null=True,
        related_name='products',
        verbose_name='Categoria'
    )

    def __str__(self):
        return self.name

then the CreateView can just use its boilerplate logic:

class ProductCreateView(CreateView):
    model = Product
    form_class = ProductForm
    success_url = '/adminpanel/products/'

Note: The related_name=… parameter [Django-doc] is the name of the relation in reverse, so from the Category model to the Product model in this case. Therefore it (often) makes not much sense to name it the same as the forward relation. You thus might want to consider renaming the category relation to products.

user_675790
Related