Django Model - cannot serialize into migration files - django

Question

I have 2 models Car and Offer. The car table is an existing table filled with vehicle specifications. In the django admin I want to map an offer to an existing car in the specs table. This means that when a click on an offer in admin, I want to be able to see a list with all cars - find the correct one - and save it on the offer. Ive done this by populating the foreignkey field with a list of choices based on the the existing car objects.
models.py:
class Car(models.Model):
brand = models.TextField(max_length=300, default= "")
model = models.TextField(max_length=300, default= "")
edition = models.TextField(max_length=300, default= "")
engineVolume = models.FloatField(default=0.0)
def __unicode__(self):
return smart_unicode(self.brand)
carIds = []
idx = 1
for car in Car.objects.all():
carIds.append((idx, car))
idx = idx + 1
class Offer(models.Model):
stringUrl = models.TextField(max_length=300)
extractionDate = models.DateTimeField(default=datetime.datetime.now, blank=True)
cars = models.ForeignKey(Car, default= "", choices=carIds, null=True, to_field='id')
this works perfectly. I click on an offer in admin and I see a selectionbox populated with all existing cars. I find the correct car, save it, and the offers´s foreignkey car id in the database points to the correct vehicle.
But suddenly when I want to make later migrations django says it cannot serialze the car object?
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python2.7/site- packages/django/core/management/__init__.py", line 338, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 330, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python2.7/site-packages/django/core/management/base.py", line 390, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python2.7/site-packages/django/core/management/base.py", line 441, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python2.7/site-packages/django/core/management/commands/makemigrations.py", line 143, in handle
self.write_migration_files(changes)
File "/usr/local/lib/python2.7/site-packages/django/core/management/commands/makemigrations.py", line 171, in write_migration_files
migration_string = writer.as_string()
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 166, in as_string
operation_string, operation_imports = OperationWriter(operation).serialize()
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 124, in serialize
_write(arg_name, arg_value)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 87, in _write
arg_string, arg_imports = MigrationWriter.serialize(_arg_value)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 377, in serialize
return cls.serialize_deconstructed(path, args, kwargs)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 268, in serialize_deconstructed
arg_string, arg_imports = cls.serialize(arg)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 303, in serialize
item_string, item_imports = cls.serialize(item)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 303, in serialize
item_string, item_imports = cls.serialize(item)
File "/usr/local/lib/python2.7/site-packages/django/db/migrations/writer.py", line 465, in serialize
"topics/migrations/#migration-serializing" % (value, get_docs_version())
`enter code here`ValueError: Cannot serialize: <Car: Nissan>
There are some values Django cannot serialize into migration files.
For more, see https://docs.djangoproject.com/en/1.8/topics/migrations/#migration- serializing
I have no idea why this error occurs. I' m new to django and the just starting with the admin. I´ve been reading up on serialzation and deconstruction but cant see how to apply it here? Perhaps I should follow a different route to achieve what I want?

Answer 1

In my case I was tring to add a field like:
task = models.ForeignKey(Task, on_delete=models.CASCADE, default=Task.objects.first())
but corrected this by using:
task = models.ForeignKey(Task, on_delete=models.CASCADE, default=Task.objects.first().pk)

Answer 2

In a nutshell you don't need choices at all.
choices, although can be any iterable and can be modified, is more suited for static data.
Afterwards, this piece of code
carIds = []
idx = 1
for car in Car.objects.all():
carIds.append((idx, car))
idx = idx + 1
achieves nothing.
First, you are not filtering the choices in any way, just converting them to a list of tuples. Second, having a ForeignKey automatically provides choices from the referenced model's unfiltered queryset e.g. Car.objects.all().
So you could just drop choices, and if you need filtering in ModelForm and admin use ForeignKey.limit_choices_to instead.