[Django]-Django: Best way to implement "status" field in modules

by

49👍

✅

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.

👤Felix Kling

11👍

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']
👤fijter

8👍

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.

👤Serafeim

2👍

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

👤zaca

2👍

You don’t need your status_str method – Django automatically provides a get_status_display() which does exactly the same thing.

To reverse, you could use this:

def set_order_status(self, val):
    status_dict = dict(ORDER_STATUS)
    self.status = status_dict[val][0]

Now you can do:

order.set_order_status('Done')
👤Daniel Roseman

0👍

Maybe stick a method on the model, like:

def status_code(self, text):
    return [n for (n, t) in self.ORDER_STATUS if t == text][0]

Then you’d do:

order.status = order.status_code('Error')
👤Paul D. Waite

0👍

don’t do all those things.just make change in views.py as follows

  context['value'] = Model_name.objects.order_by('-choice')

where 

   choice = ('pending','solved','closed')
👤raghu

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