Django - model that has a count per CHOICES Field

Question

django 2.1, python 3.6

Let's say I have a set of things ['house_1','house_2',...] and I want to have a model that keeps a count of each item. I'd imagine that it'd look like the following.

class Card(models.Model):
    house_1_count = models.IntegerField(...
    house_2_count = models.IntegerField(...
    house_3_count = models.IntegerField(...

Is there a way to convert this model so that the items are not hard coded. Like by using a CHOICES field. The items in my list may change (slowly), but I don't want to have to create a field in the model every time it happens.

score 1 · Answer 1 · answered Jan 04 '19 at 03:36

1

In your situation, I don't think it would be wise to store the count of houses in a model. If you want to get count, then simply do it like this using count():

House.objects.filter(house_type=1).count()

answered Jan 04 '19 at 03:36

ruddra

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score 0 · Answer 2 · answered Jan 04 '19 at 07:18

You can create a choice field in your model rather than storing each count. Update you card model:

class Card(models.Model):
    house = models.IntegerField(choices=list(zip(range(1, 4), range(1, 4))), max_length=1)

To count Card with particular type house use count():

Card.objects.filter(house=1).count()

Django - model that has a count per CHOICES Field

2 Answers2