Defining a model class in Django shell fails

Question

when I use the Django shell, it shows an error; this is the error:

>>> from django.db import models
>>> class Poll(models.Model):
...     question = models.CharField(max_length=200)
...     pub_date = models.DateTimeField('date published')
...
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "D:Python25libsite-packagesdjangodbmodelsbase.py", line 51, in __new__
    kwargs = {"app_label": model_module.__name__.split('.')[-2]}
IndexError: list index out of range

What can I do?

Asked By: zjm1126

||

Source

Answer 1

The model definition must come in an application – the error you’re seeing there is that it tries to take the __name__ model_module – which should be something like project.appname.models for projectappnamemodels.py – and get the app name, appname. In the interactive console, the module’s __name__ is '__main__' – so it fails.

To get around this, you’ll need to specify the app_label yourself in the Meta class;

>>> from django.db import models
>>> class Poll(models.Model):
...     question = models.CharField(max_length=200)
...     pub_date = models.DateTimeField('date published')
...     class Meta:
...         app_label = 'test'

For explanation of why you can do that, look at that file mentioned in the traceback, D:Python25libsite-packagesdjangodbmodelsbase.py:

    if getattr(meta, 'app_label', None) is None:
        # Figure out the app_label by looking one level up.
        # For 'django.contrib.sites.models', this would be 'sites'.
        model_module = sys.modules[new_class.__module__]
        kwargs = {"app_label": model_module.__name__.split('.')[-2]}
    else:
        kwargs = {}

(Where meta is the Meta class, see just above in that file.)

Answered By: Chris Morgan

Answer 2

That other answer definitely works for the interactive prompt, however, I don’t think that the intention of the first block of code was intended to actually be run. Immediately following that code in the models documentation, you are expected to put the next codes into your models.py file created during the previous tutorial… I guess that’s why they subtly labeled that section “Quick Example.” What a headache for me too!

Answered By: Bryan Wolfford

Answer 3

I ran into this problem using Eclipse, Django and PyDev. I needed to have the application (instead of some .py file for example) selected in the PyDev Package Explorer (left panel) before clicking Run for everything to work properly.

Answered By: Zaky German

Defining a model class in Django shell fails

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