Makemigrations ValueError: too many values to unpack - mysql

Question

When I'm trying to manage.py makemigrations there is a error:
My models are:
class Burger(models.Model):
title = models.CharField(max_length=100)
mt = models.CharField(max_length=50)
description = models.TextField(max_length=255)
price = models.IntegerField()
img = models.ImageField()
class Comment(models.Model):
author = models.CharField(max_length=50)
email = models.CharField(max_length=100)
text = models.TextField(max_length=255)
date = models.DateTimeField()
bid = models.ForeignKey('Burger', on_delete=models.CASCADE)
The traceback is:
Traceback (most recent call last):
File "C:\Burger123\manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line 353, in execute_from_command_line
utility.execute()
File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line 327, in execute
django.setup()
File "C:\Python27\lib\site-packages\django\__init__.py", line 18, in setup
apps.populate(settings.INSTALLED_APPS)
File "C:\Python27\lib\site-packages\django\apps\registry.py", line 108, in populate
app_config.import_models(all_models)
File "C:\Python27\lib\site-packages\django\apps\config.py", line 202, in import_models
self.models_module = import_module(models_module_name)
File "C:\Python27\lib\importlib\__init__.py", line 37, in import_module
__import__(name)
File "C:\Burger123\BURRGER\models.py", line 15, in <module>
class Comment(models.Model):
File "C:\Python27\lib\site-packages\django\db\models\base.py", line 158, in __new__
new_class.add_to_class(obj_name, obj)
File "C:\Python27\lib\site-packages\django\db\models\base.py", line 299, in add_to_class
value.contribute_to_class(cls, name)
File "C:\Python27\lib\site-packages\django\db\models\fields\related.py", line 703, in contribute_to_class
super(ForeignObject, self).contribute_to_class(cls, name, virtual_only=virtual_only)
File "C:\Python27\lib\site-packages\django\db\models\fields\related.py", line 308, in contribute_to_class
lazy_related_operation(resolve_related_class, cls, self.remote_field.model, field=self)
File "C:\Python27\lib\site-packages\django\db\models\fields\related.py", line 85, in lazy_related_operation
return apps.lazy_model_operation(partial(function, **kwargs), *model_keys)
File "C:\Python27\lib\site-packages\django\db\models\fields\related.py", line 83, in <genexpr>
model_keys = (make_model_tuple(m) for m in models)
File "C:\Python27\lib\site-packages\django\db\models\utils.py", line 13, in make_model_tuple
app_label, model_name = model.split(".")
ValueError: too many values to unpack
My INSTALLED_APPS setting is:
INSTALLED_APPS = [
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'BURRGER.apps.BurrgerConfig'
]
BurrgerConfig:
from django.apps import AppConfig
class BurrgerConfig(AppConfig):
name = 'BURRGER'
label = 'my.burger'
I deleted line bid = models.ForeignKey('Burger', on_delete=models.CASCADE) and makemigrations ended without errors. But I totally need this line in my code so the question is still open.

Answer 1

The label in your app config should not contain a dot. You could do something like:
class BurrgerConfig(AppConfig):
name = 'BURRGER'
label = 'burger'
Your code is more confusing because you have mispelled 'BURRGER' in your code and directory name. The usual approach in Python/Django would be
burger - directory name, name and lable in app config class
Burger - model name
BurgerConfig - App Config class name