django-enum-choices 2.1.1

A custom Django field able to use subclasses of Python's internal `Enum` class as choices

django-enum-choices

A custom Django choice field to use with Python enums.

Table of Contents

• django-enum-choices
  • Table of Contents
  • Installation
  • Basic Usage
  • Choice builders
  • Usage inside the admin panel
  • Usage with forms
    • Usage with django.forms.ModelForm
    • Usage with django.forms.Form
  • Usage with django-filter
    • By using a Meta inner class and inheriting from EnumChoiceFilterMixin
    • By declaring the field explicitly on the FilterSet
  • Postgres ArrayField Usage
  • Usage with Django Rest Framework
    • Using serializers.ModelSerializer with EnumChoiceModelSerializerMixin
    • Using serializers.ModelSerializer without EnumChoiceModelSerializerMixin
    • Using a subclass of serializers.Serializer
    • Serializing PostgreSQL ArrayField
  • Implementation details
  • Using Python's enum.auto
  • Development

Installation

pip install django-enum-choices

Basic Usage

from enum import Enum

from django.db import models

from django_enum_choices.fields import EnumChoiceField


class MyEnum(Enum):
    A = 'a'
    B = 'b'


class MyModel(models.Model):
    enumerated_field = EnumChoiceField(MyEnum)

Model creation

instance = MyModel.objects.create(enumerated_field=MyEnum.A)

Changing enum values

instance.enumerated_field = MyEnum.B
instance.save()

Filtering

MyModel.objects.filter(enumerated_field=MyEnum.A)

Choice builders

EnumChoiceField extends CharField and generates choices internally. Each choice is generated using something, called a choice_builder.

A choice builder function looks like that:

def choice_builder(enum: Enum) -> Tuple[str, str]:
    # Some implementation

If a choice_builder argument is passed to a model's EnumChoiceField, django_enum_choices will use it to generate the choices. The choice_builder must be a callable that accepts an enumeration choice and returns a tuple, containing the value to be saved and the readable value.

By default django_enum_choices uses one of the four choice builders defined in django_enum_choices.choice_builders, named value_value.

It returns a tuple containing the enumeration's value twice:

from django_enum_choices.choice_builders import value_value

class MyEnum(Enum):
    A = 'a'
    B = 'b'

print(value_value(MyEnum.A))  # ('a', 'a')

You can use one of the four default ones that fits your needs:

• value_value
• attribute_value
• value_attribute
• attribute_attribute

For example:

from django_enum_choices.choice_builders import attribute_value

class MyEnum(Enum):
    A = 'a'
    B = 'b'

class CustomReadableValueEnumModel(models.Model):
    enumerated_field = EnumChoiceField(
        MyEnum,
        choice_builder=attribute_value
    )

The resulting choices for enumerated_field will be (('A', 'a'), ('B', 'b'))

You can also define your own choice builder:

class MyEnum(Enum):
    A = 'a'
    B = 'b'

def choice_builder(choice: Enum) -> Tuple[str, str]:
    return choice.value, choice.value.upper() + choice.value

class CustomReadableValueEnumModel(models.Model):
    enumerated_field = EnumChoiceField(
        MyEnum,
        choice_builder=choice_builder
    )

Which will result in the following choices (('a', 'Aa'), ('b', 'Bb'))

The values in the returned from choice_builder tuple will be cast to strings before being used.

Usage in the admin panel

Model fields, defined as EnumChoiceField can be used with almost all of the admin panel's standard functionallities.

One exception from this their usage in list_filter.

If you need an EnumChoiceField inside a ModelAdmin's list_filter, you can use the following options:

• Define the entry insite the list filter as a tuple, containing the field's name and django_enum_choices.admin.EnumChoiceListFilter

from django.contrib import admin

from django_enum_choices.admin import EnumChoiceListFilter

from .models import MyModel

@admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
    list_filter = [('enumerated_field', EnumChoiceListFilter)]

• Set DJANGO_ENUM_CHOICES_REGISTER_LIST_FILTER inside your settings to True, which will automatically set the EnumChoiceListFilter class to all list_filter fields that are instances of EnumChoiceField. This way, they can be declared directly in the list_filter iterable:

from django.contrib import admin

from .models import MyModel

@admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
    list_filter = ('enumerated_field', )

Usage with forms

There are 2 rules of thumb:

1. If you use a ModelForm, everything will be taken care of automatically.
2. If you use a Form, you need to take into account what Enum and choice_builder you are using.

Usage with django.forms.ModelForm

from .models import MyModel

class ModelEnumForm(forms.ModelForm):
    class Meta:
        model = MyModel
        fields = ['enumerated_field']

form = ModelEnumForm({
    'enumerated_field': 'a'
})

form.is_valid()

print(form.save(commit=True))  # <MyModel: MyModel object (12)>

Usage with django.forms.Form

If you are using the default value_value choice builder, you can just do that:

from django_enum_choices.forms import EnumChoiceField

from .enumerations import MyEnum

class StandardEnumForm(forms.Form):
    enumerated_field = EnumChoiceField(MyEnum)

form = StandardEnumForm({
    'enumerated_field': 'a'
})
form.is_valid()

print(form.cleaned_data)  # {'enumerated_field': <MyEnum.A: 'a'>}

If you are passing a different choice builder, you have to also pass it to the form field:

from .enumerations import MyEnum

def custom_choice_builder(choice):
    return 'Custom_' + choice.value, choice.value

class CustomChoiceBuilderEnumForm(forms.Form):
    enumerated_field = EnumChoiceField(
        MyEnum,
        choice_builder=custom_choice_builder
    )

form = CustomChoiceBuilderEnumForm({
    'enumerated_field': 'Custom_a'
})

form.is_valid()

print(form.cleaned_data)  # {'enumerated_field': <MyEnum.A: 'a'>}

Usage with django-filter

As with forms, there are 2 general rules of thumb:

1. If you have declared an EnumChoiceField in the Meta.fields for a given Meta.model, you need to inherit EnumChoiceFilterMixin in your filter class & everything will be taken care of automatically.
2. If you are declaring an explicit field, without a model, you need to specify the Enum class & the choice_builder, if a custom one is used.

By using a Meta inner class and inheriting from EnumChoiceFilterMixin

import django_filters as filters

from django_enum_choices.filters import EnumChoiceFilterMixin

class ImplicitFilterSet(EnumChoiceFilterSetMixin, filters.FilterSet):
    class Meta:
        model = MyModel
        fields = ['enumerated_field']

filters = {
    'enumerated_field': 'a'
}
filterset = ImplicitFilterSet(filters)

print(filterset.qs.values_list('enumerated_field', flat=True))
# <QuerySet [<MyEnum.A: 'a'>, <MyEnum.A: 'a'>, <MyEnum.A: 'a'>]>

The choice_builder argument can be passed to django_enum_choices.filters.EnumChoiceFilter as well when using the field explicitly. When using EnumChoiceFilterSetMixin, the choice_builder is determined from the model field, for the fields defined inside the Meta inner class.

import django_filters as filters

from django_enum_choices.filters import EnumChoiceFilter

def custom_choice_builder(choice):
    return 'Custom_' + choice.value, choice.value

class ExplicitCustomChoiceBuilderFilterSet(filters.FilterSet):
    enumerated_field = EnumChoiceFilter(
        MyEnum,
        choice_builder=custom_choice_builder
    )

filters = {
    'enumerated_field': 'Custom_a'
}
filterset = ExplicitCustomChoiceBuilderFilterSet(filters, MyModel.objects.all())

print(filterset.qs.values_list('enumerated_field', flat=True))  # <QuerySet [<MyEnum.A: 'a'>, <MyEnum.A: 'a'>, <MyEnum.A: 'a'>]>

By declaring the field explicitly on the FilterSet

import django_filters as filters

from django_enum_choices.filters import EnumChoiceFilter

class ExplicitFilterSet(filters.FilterSet):
    enumerated_field = EnumChoiceFilter(MyEnum)


filters = {
    'enumerated_field': 'a'
}
filterset = ExplicitFilterSet(filters, MyModel.objects.all())

print(filterset.qs.values_list('enumerated_field', flat=True))  # <QuerySet [<MyEnum.A: 'a'>, <MyEnum.A: 'a'>, <MyEnum.A: 'a'>]>

Postgres ArrayField Usage

You can use EnumChoiceField as a child field of an Postgres ArrayField.

from django.db import models
from django.contrib.postgres.fields import ArrayField

from django_enum_choices.fields import EnumChoiceField

from enum import Enum

class MyEnum(Enum):
    A = 'a'
    B = 'b'

class MyModelMultiple(models.Model):
    enumerated_field = ArrayField(
        base_field=EnumChoiceField(MyEnum)
    )

Model Creation

instance = MyModelMultiple.objects.create(enumerated_field=[MyEnum.A, MyEnum.B])

Changing enum values

instance.enumerated_field = [MyEnum.B]
instance.save()

Usage with Django Rest Framework

As with forms & filters, there are 2 general rules of thumb:

1. If you are using a ModelSerializer and you inherit EnumChoiceModelSerializerMixin, everything will be taken care of automatically.
2. If you are using a Serializer, you need to take the Enum class & choice_builder into acount.

Using serializers.ModelSerializer with EnumChoiceModelSerializerMixin

from rest_framework import serializers

from django_enum_choices.serializers import EnumChoiceModelSerializerMixin

class ImplicitMyModelSerializer(
    EnumChoiceModelSerializerMixin,
    serializers.ModelSerializer
):
    class Meta:
        model = MyModel
        fields = ('enumerated_field', )

By default ModelSerializer.build_standard_field coerces any field that has a model field with choices to ChoiceField which returns the value directly.

Since enum values resemble EnumClass.ENUM_INSTANCE they won't be able to be encoded by the JSONEncoder when being passed to a Response.

That's why we need the mixin.

When using the EnumChoiceModelSerializerMixin with DRF's serializers.ModelSerializer, the choice_builder is automatically passed from the model field to the serializer field.

Using serializers.ModelSerializer without EnumChoiceModelSerializerMixin

from rest_framework import serializers

from django_enum_choices.serializers import EnumChoiceField

class MyModelSerializer(serializers.ModelSerializer):
    enumerated_field = EnumChoiceField(MyEnum)

    class Meta:
        model = MyModel
        fields = ('enumerated_field', )

# Serialization:
instance = MyModel.objects.create(enumerated_field=MyEnum.A)
serializer = MyModelSerializer(instance)
data = serializer.data  # {'enumerated_field': 'a'}

# Saving:
serializer = MyModelSerializer(data={
    'enumerated_field': 'a'
})
serializer.is_valid()
serializer.save()

If you are using a custom choice_builder, you need to pass that too.

def custom_choice_builder(choice):
    return 'Custom_' + choice.value, choice.value

class CustomChoiceBuilderSerializer(serializers.Serializer):
    enumerted_field = EnumChoiceField(
        MyEnum,
        choice_builder=custom_choice_builder
    )

serializer = CustomChoiceBuilderSerializer({
    'enumerated_field': MyEnum.A
})

data = serializer.data # {'enumerated_field': 'Custom_a'}

Using a subclass of serializers.Serializer

from rest_framework import serializers

from django_enum_choices.serializers import EnumChoiceField

class MySerializer(serializers.Serializer):
    enumerated_field = EnumChoiceField(MyEnum)

# Serialization:
serializer = MySerializer({
    'enumerated_field': MyEnum.A
})
data = serializer.data  # {'enumerated_field': 'a'}

# Deserialization:
serializer = MySerializer(data={
    'enumerated_field': 'a'
})
serializer.is_valid()
data = serializer.validated_data  # OrderedDict([('enumerated_field', <MyEnum.A: 'a'>)])

If you are using a custom choice_builder, you need to pass that too.

Serializing PostgreSQL ArrayField

django-enum-choices exposes a MultipleEnumChoiceField that can be used for serializing arrays of enumerations.

Using a subclass of serializers.Serializer

from rest_framework import serializers

from django_enum_choices.serializers import MultipleEnumChoiceField

class MultipleMySerializer(serializers.Serializer):
    enumerated_field = MultipleEnumChoiceField(MyEnum)

# Serialization:
serializer = MultipleMySerializer({
    'enumerated_field': [MyEnum.A, MyEnum.B]
})
data = serializer.data  # {'enumerated_field': ['a', 'b']}

# Deserialization:
serializer = MultipleMySerializer(data={
    'enumerated_field': ['a', 'b']
})
serializer.is_valid()
data = serializer.validated_data  # OrderedDict([('enumerated_field', [<MyEnum.A: 'a'>, <MyEnum.B: 'b'>])])

Using a subclass of serializers.ModelSerializer

class ImplicitMultipleMyModelSerializer(
    EnumChoiceModelSerializerMixin,
    serializers.ModelSerializer
):
    class Meta:
        model = MyModelMultiple
        fields = ('enumerated_field', )

# Serialization:
instance = MyModelMultiple.objects.create(enumerated_field=[MyEnum.A, MyEnum.B])
serializer = ImplicitMultipleMyModelSerializer(instance)
data = serializer.data  # {'enumerated_field': ['a', 'b']}

# Saving:
serializer = ImplicitMultipleMyModelSerializer(data={
    'enumerated_field': ['a', 'b']
})
serializer.is_valid()
serializer.save()

The EnumChoiceModelSerializerMixin does not need to be used if enumerated_field is defined on the serializer class explicitly.

Implementation details

• EnumChoiceField is a subclass of CharField.
• Only subclasses of Enum are valid arguments for EnumChoiceField.
• max_length, if passed, is ignored. max_length is automatically calculated from the longest choice.
• choices are generated using a special choice_builder function, which accepts an enumeration and returns a tuple of 2 items.
  • Four choice builder functions are defined inside django_enum_choices.choice_builders
  • By default the value_value choice builder is used. It produces the choices from the values in the enumeration class, like (enumeration.value, enumeration.value)
  • choice_builder can be overriden by passing a callable to the choice_builder keyword argument of EnumChoiceField.
  • All values returned from the choice builder will be cast to strings when generating choices.

For example, lets have the following case:

class Value:
    def __init__(self, value):
        self.value = value

    def __str__(self):
        return self.value


class CustomObjectEnum(Enum):
    A = Value(1)
    B = Value('B')

	# The default choice builder `value_value` is being used

class SomeModel(models.Model):
    enumerated_field = EnumChoiceField(CustomObjectEnum)

We'll have the following:

• SomeModel.enumerated_field.choices == (('1', '1'), ('B', 'B'))
• SomeModel.enumerated_field.max_length == 3

Using Python's enum.auto

enum.auto can be used for shorthand enumeration definitions:

from enum import Enum, auto

class AutoEnum(Enum):
    A = auto()  # 1
    B = auto()  # 2

class SomeModel(models.Model):
    enumerated_field = EnumChoiceField(Enum)

This will result in the following:

• SomeModel.enumerated_field.choices == (('1', '1'), ('2', '2'))

Overridinng auto behaviour Custom values for enumerations, created by auto, can be defined by subclassing an Enum that defines _generate_next_value_:

class CustomAutoEnumValueGenerator(Enum):
    def _generate_next_value_(name, start, count, last_values):
        return {
            'A': 'foo',
            'B': 'bar'
        }[name]


class CustomAutoEnum(CustomAutoEnumValueGenerator):
    A = auto()
    B = auto()

The above will assign the values mapped in the dictionary as values to attributes in CustomAutoEnum.

Development

Prerequisites

• SQLite3
• PostgreSQL server
• Python >= 3.5 virtual environment

git clone https://github.com/HackSoftware/django-enum-choices.git
cd django_enum_choices
pip install -e .[dev]

Linting and running the tests:

tox