Set Django IntegerField by choices=... name
Here's a field type I wrote a few minutes ago that I think does what you want. Its constructor requires an argument 'choices', which may be either a tuple of 2-tuples in the same format as the choices option to IntegerField, or instead a simple list of names (ie ChoiceField(('Low', 'Normal', 'High'), default='Low') ). The class takes care of the mapping from string to int for you, you never see the int.
class ChoiceField(models.IntegerField):
def __init__(self, choices, **kwargs):
if not hasattr(choices[0],'__iter__'):
choices = zip(range(len(choices)), choices)
self.val2choice = dict(choices)
self.choice2val = dict((v,k) for k,v in choices)
kwargs['choices'] = choices
super(models.IntegerField, self).__init__(**kwargs)
def to_python(self, value):
return self.val2choice[value]
def get_db_prep_value(self, choice):
return self.choice2val[choice]
As of Django 3.0, you can use:
class ThingPriority(models.IntegerChoices):
LOW = 0, 'Low'
NORMAL = 1, 'Normal'
HIGH = 2, 'High'
class Thing(models.Model):
priority = models.IntegerField(default=ThingPriority.LOW, choices=ThingPriority.choices)
# then in your code
thing = get_my_thing()
thing.priority = ThingPriority.HIGH
Do as seen here. Then you can use a word that represents the proper integer.
Like so:
LOW = 0
NORMAL = 1
HIGH = 2
STATUS_CHOICES = (
(LOW, 'Low'),
(NORMAL, 'Normal'),
(HIGH, 'High'),
)
Then they are still integers in the DB.
Usage would be thing.priority = Thing.NORMAL
I'd probably set up the reverse-lookup dict once and for all, but if I hadn't I'd just use:
thing.priority = next(value for value, name in Thing.PRIORITIES
if name=='Normal')
which seems simpler than building the dict on the fly just to toss it away again;-).