Django: Best way to implement "status" field in modules

Question

I have a field in my module that is used to hold the status of the object. So far I have used:

ORDER_STATUS = ((0, 'Started'), (1, 'Done'), (2, 'Error'))
status = models.SmallIntegerField(choices=ORDER_STATUS)

Its easy to convert one way:

def status_str(self): return ORDER_STATUS[self.status][1]

The problem is when updating. I find myself having code like this:

order.status = 2 # Error Status

Which is quite awful and gets really hard to synchronize. I guess a solution would be something similar to C's enum{}. Or perhaps there is a whole different way to tackle this problem ?

Thanks

Answer 1

Maybe this question helps you: Set Django IntegerField by choices=… name.
I quote from the accepted answer (with adjustments ;)):
Put this into your class (STATUS_CHOICES will be the list that is handed to the choices option of the field):

PENDING = 0
DONE = 1
STATUS_CHOICES = (
    (PENDING, 'Pending'),
    (DONE, 'Done'),
)

Then you can do order.status = Order.DONE.

---

Note that you don't have to implement an own method to retrieve the (readable) value, Django provides the method get_status_display itself.

Answer 2

what I usually do for this situation is:

models.py

from static import ORDER_STATUS    
status = models.PositiveSmallIntegerField(choices=ORDER_STATUS)

static.py

ORDER_STATUS = ((0, 'Started'), (1, 'Done'), (2, 'Error'))
ORDER_STATUS_DICT = dict((v, k) for k, v in ORDER_STATUS)

Now you can do:

from static import ORDER_STATUS_DICT
order.status = ORDER_STATUS_DICT['Error']

Answer 3

This is a very late answer, however for completeness I should mention that django-model-utils already contains a StatusField and even better a StatusModel. I am using it everywhere I need to have a status.

Answer 4

you can try enum package: http://pypi.python.org/pypi/enum/