Defining A Model Class In Django Shell Fails
Solution 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 project\appname\models.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>>>classPoll(models.Model):... question = models.CharField(max_length=200)... pub_date = models.DateTimeField('date published')...classMeta:... app_label = 'test'
For explanation of why you can do that, look at that file mentioned in the traceback, D:\Python25\lib\site-packages\django\db\models\base.py
:
ifgetattr(meta, 'app_label', None) isNone:
# 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.)
Solution 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!
Solution 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.
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