Django: Create model instance within model - django

Question

I have another probably quite basic Django question. I am writing a database app for my classes, and I would like to be able to set the current year from within the app. The idea behind this is that I can manually switch to the next academic year when all the marking etc is done. The function to switch the current year is then supposed to do other things as well - put the students in the new default courses etc.
So I created a model called MetaData:
class MetaData(models.Model):
data_id = models.IntegerField(default = 1, unique = True)
current_year = models.IntegerField(choices=ACADEMIC_YEARS)
It would be great if I could access the current_year in other models, for example to set it as default when I want to enter new courses/students. I tried this:
class Course(models.Model):
try:
meta = MetaData.objects.get(data_id=1)
current_year = meta.current_year
except MetaData.DoesNotExist:
current_year = 2010
year = models.IntegerField(choices=ACADEMIC_YEARS, default=current_year)
Unfortunately, I get the error "No such table: database_metadata" when I run syncdb for the first time - I would guess because syncdb has not created the tables when it tests the code.
Is there a better solution for this?

Answer 1

Either put the constant inside your code (if you could hard code the logic of computing academic years. Or always remember to change it manually, which sounds OK as you've already hard-coded ACADEMIC_YEARS), or in DB as danodonovan suggests. Or model academic years directly w/o storing a constant in MetaData
from django.core.exceptions import ValidationError
class AcademicYear(models.Model):
start = models.DateField() # or simply year
end = models.DateField()
is_current = models.BooleanField()
class Meta:
ordering = ('-is_current', 'start')
def clean(self): # only allow at most one current year
if self.is_current and \
self.__class__.objects.exclude(pk=self.pk).filter(is_current=True).exists():
raise ValidationError('Found another `current` academic year.'
def __unicode__(self):
return 'blahblah'
# then in Course
class Course(models.Model):
academic_year = models.ForeignKey(AcademicYear)
Thus in the Admin, you could easily step up AcademicYears and display them as choices of the academic_year field of the Course.
If you still would like to use MetaData, just change the current_year to be a function:
def get_current_year():
try:
return MetaData.objects.get(data_id=1).current_year
except MetaData.DoesNotExist:
return 2010
class Course(models.Model):
year = models.IntegerField(choices=ACADEMIC_YEARS, default=get_current_year)

Answer 2

This sounds like you need to use the ForeignKey field to link your models.
class MetaData(models.Model):
current_year = models.IntegerField(choices=ACADEMIC_YEARS)
class Course(models.Model):
year = models.ForeignKey(MetaData, default=this_year)
def this_year(self):
return MetaData.objects.get(current_year=datetime.now().year)
To set the default year I've added a this_year function to the MetaData model. The Course models year attribute will call this when setting the default year.
Having said that, it would be much simpler if you used a DateField for simple dates.

Answer 3

class Boat_Owner(models.Model):
username = models.CharField(max_length=200)
password = models.CharField(max_length=200)
name = models.CharField(max_length=200,unique=True,
verbose_name="Boat Owner Name")
address = models.CharField(max_length=200)
phone = models.CharField(max_length=200)
email = models.EmailField('email address',
unique=True, db_index=True)
avatar = models.ImageField('profile picture',
upload_to='static/media/images/avatars/',
null=True, blank=True)
def set_avatar(self):
self.has_picture = True
def __str__(self):
return u'%s %s %s %s %s %s' % (self.id,self.name,self.username,self.password,self.address,self.email)
class Meta:
verbose_name = "Boat Owner"
verbose_name_plural = "Boat Owner"