Return display_name in ChoiceField

Question

I'm implementing some REST API in DRF with ModelViewSet and ModelSerializer. All my APIs use the JSON format and some of my models use ChoiceField field, like that:

MyModel(models.Model):
     KEY1 = 'Key1'
     KEY2 = 'Key2'
     ATTRIBUTE_CHOICES = (
         (KEY1, 'Label 1'),
         (KEY2, 'Label 2'))
     attribute = models.CharField(max_length=4, 
                                  choices=ATTRIBUTE_CHOICES, default=KEY1)

My problem is that by default DRF always returns (and accept) the key of these choices for JSON messages (see here), but I would like to use the label instead, because I think it's more consistent and clear to unterstand for who will use those APIs. Any suggestion?

score 5 · Accepted Answer · answered Dec 30 '15 at 21:21

5

I found a possible solution, namely defining my own field as follow:

class MyChoiceField(serializers.ChoiceField):

    def to_representation(self, data):
        if data not in self.choices.keys():
            self.fail('invalid_choice', input=data)
        else:
            return self.choices[data]

    def to_internal_value(self, data):
        for key, value in self.choices.items():
            if value == data:
                 return key
        self.fail('invalid_choice', input=data)

It works the same way as to ChoiceField, but returns and accepts labels instead of keys.

answered Dec 30 '15 at 21:21

SimoV8

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Any reason why you're saving label instead of key? – mariodev Dec 31 '15 at 08:10
@mariodev I'm not saving the label, in the db will be stored the key. I'm using the label only for communication with the client, `MyChoiceField` just map values with keys and viceversa. I think in this way it's more comprehensible for clients that don't have to know the internal representation of these data. – SimoV8 Dec 31 '15 at 10:45
I guess this is correct for your particular case. I'm just not convinced that passing a label (instead of a key) into `to_internal_value` is the best practice to follow. In general passing key representation from the client would make more sense. Unless this is somehow impossible in your case, therefore your solution is acceptable. +1 ;) – mariodev Dec 31 '15 at 11:46
Hi @SimoV8 The answer works, I am just wondering if there is an easier way to achieve this since the answer was posted? – Vikram Bankar Aug 12 '21 at 13:57

score 1 · Answer 2 · edited May 23 '17 at 12:33

1

There's no way other than overriding your serializer. Please take a look here, to see how it can be done.

edited May 23 '17 at 12:33

Community

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answered Dec 30 '15 at 18:49

mariodev

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Hi, thank you, I read it. It works in getting resources but not to create/update a new entity, any idea? – SimoV8 Dec 30 '15 at 19:01
Oh you haven't mentioned that.. so maybe you can define the [ChoiceField serializer field](http://www.django-rest-framework.org/api-guide/fields/#choicefield) in your serializer instead.. – mariodev Dec 30 '15 at 19:23

score 1 · Answer 3 · answered Jun 29 '20 at 16:04

previous answers helped me a lot but not worked for me as I use Django version 3 and DRF version 3.11 , so I came up with this:

# models.py
class Ball(models.Model):
    
    class Types(models.TextChoice):
        VOLLYBALL = 'VB', gettext_lazy('VollyBall')
        FOOTBALL = 'FB', gettext_lazy('FootBall')

    type = models.CharField(max_length=2, choices=Types.choices)

# serializers.py

class CustomChoiceField(serializers.ChoiceField):

    def to_representation(self, value):
        if value in ('', None):
            return value

        choice_dict = {str(key): key.label for key in self.choices}
        return choice_dict.get(str(value), value)

    def to_internal_value(self, data):
        if data == '' and self.allow_blank:
            return ''

        try:
            choice_dict = {key.label: str(key) for key in self.choices}
            return choice_dict[str(data)]
        except KeyError:
            self.fail('invalid_choice', input=data)

class BallSerializer(serializers.ModelSerializer):
    type = CustomChoiceField(choices=Book.Types)
    
    class Meta:
        model = Book
        fields = ['type']

Return display_name in ChoiceField

3 Answers3

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