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A field for `IntegerChoices` which does not generate a new migration when the choices change
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| class IntegerChoicesField(models.IntegerField): | |
| """A field for `IntegerChoices` where changing the choices does not generate a new migration. To be used instead of an `IntegerField`. | |
| Based on a suggestion from https://mastodon.social/@bmispelon at https://mastodon.social/@bmispelon/114438157173090099. | |
| """ | |
| def deconstruct(self): | |
| deconstructed_field = super().deconstruct() | |
| if len(deconstructed_field) >= 4 and "choices" in deconstructed_field[3]: | |
| # Remove the `choices` from the dictionary so that new migrations are not created when the choices are modified. | |
| del deconstructed_field[3]["choices"] | |
| return deconstructed_field | |
| """ | |
| Example usage | |
| """ | |
| class Suit(models.IntegerChoices): | |
| CLUB = (1, "Club") | |
| DIAMOND = (2, "Diamond") | |
| HEART = (3, "Heart") | |
| SPADE = (4, "Spade") | |
| class Card(models.Model): | |
| suit = IntegerChoicesField(choices=Suit.choices) |
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Another (better?) approach is to use an intermediate function for
choicesas explained in detail here: https://adamj.eu/tech/2025/05/03/django-choices-change-without-migration/.