Mysql Makemigrations ValueError:要解压缩的值太多

当我尝试管理.py makemigrations时，出现一个错误：

我的模型是：

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)

回溯是：

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

BurrgerConfig

：

from django.apps import AppConfig
class BurrgerConfig(AppConfig):
    name = 'BURRGER'
    label = 'my.burger'

我删除了第行

bid=models.ForeignKey（'Burger'，on_delete=models.CASCADE）

，makemigrations结束时没有出现错误。但我的代码中完全需要这一行，因此问题仍然悬而未决。

应用程序配置中的不应包含点。你可以这样做：

class BurrgerConfig(AppConfig):
    name = 'BURRGER'
    label = 'burger'

您的代码更加混乱，因为您在代码和目录名中错误地拼写了“Burger”。Python/Django中通常的方法是

burger - directory name, name and lable in app config class
Burger - model name
BurgerConfig - App Config class name

---

发布实际文本，而不是发布文本图像。以及您尝试/研究的内容。您可以尝试manage.py makemigration djang_app_name吗？第一次迁移后，您必须键入应用程序名称。编辑问题时，请不要删除回溯，它会显示错误所在和发生位置。同样，您迄今为止尝试/研究了哪些内容。你可以在右边的任何相关问题中看到这个错误，你很可能在某个地方有一个错误的逗号。（感谢你的耐心，阿拉斯代尔）问题出在你的

AppConfig

课上，请看下面我的答案。

burger - directory name, name and lable in app config class
Burger - model name
BurgerConfig - App Config class name