CODEWITHSUNDEEP
djangomodelchoicefield
Garvit Jain
Garvit Jain

Reputation: 635

Django get choice display on using .values

For a Model with field type, is there any way to get the choice display value on using Model.objects.values() ? I tried Model.objects.values('get_type_display') but it doesn't work.

Upvotes: 4

Views: 6301

Answers (4)

Muhammadoufi
Muhammadoufi

Reputation: 95

This code example maybee help you:

 alertsfilter = list(Alerte.objects.filter(viewed__isnull=False).values(
    'id', 
    'type', 
    'created_at', 
    'viewed',
    'sent', 
    'agent_id', 
    'geofence_id', 
    'objet__id', 
    'objet__nom',
    'objet__marker',
    'geofence__id'))

for alert in alertsfilter:
    # alert_type = alert['type']
    alert['type'] = Alerte.objects.get(pk=alert['id']).get_type_display()

Upvotes: 0

msf
msf

Reputation: 96

I had a similar need and unfortunately you can't do so with only values; however, you can do something similar with some crafty annotations and django's custom enums (IntegerChoices/TextChoices) classes.

I created this from another SO question but I can't remember where I got the inspiration from. Basically, you can pass in the corresponding model or the choices you'd like to map back to labels via annotations on the queryset.

class ChoicesLabelCase(Case):
    def __init__(self, field: str, model: Model = None, choices: list = None, *args, **kwargs) -> None:
        if choices is None and model is None:
            raise ValueError("Either a model or a choices parameter must be provided.")
        elif choices is None:
            choices = model._meta.get_field(field).flatchoices

        cases_list = [When(**{field: val, "then": Value(label)}) for (val, label) in choices]
        super(ChoicesLabelCase, self).__init__(*cases_list, output_field=CharField(), *args, **kwargs)

As an example take this model:

class FruitType(models.IntegerChoices):
    APPLE = 1, 'Apple'
    BANANA = 2, 'Banana'
    ORANGE = 3, "Orange"

class Fruit(models.Model):
    type = models.IntegerField(choices=FruitType.choices)

You can annotate the labels like so:

>>> Fruit.objects.all().annotate(
    type_label=ChoicesLabelCase('type', FruitType)
).values("type", "type_label")

[
    {'type': 1, 'type_label': 'Apple'}, 
    {'type': 2, 'type_label': 'Banana'}, 
    ...
]

Upvotes: 1

GCru
GCru

Reputation: 516

Where I do call .values of choice fields into my queryset I deal with this in the following way:

Assume the following model

from enum import Enum

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

class MyModel(models.Model):
    type = models.CharField(max_length=1, choices=[(tag.name,tag.value) for tag in TypeChoice])

Using the query my_qset = MyModel.objects.values('type') the display values are available as:

for item in my_qset:
     print(TypeChoice[item].value)

To deal with this in my templates I write a custom template filter, say type_display:

from django import template
import TypeChoice
register = template.Library()

@register.filter
def type_display(var):
    return TypeChoice[var].value

Upvotes: 2

Nafees Anwar
Nafees Anwar

Reputation: 6608

You can't do that. values is a built in django queryset method which is used to get dictionaries of data instead of model instances you can read more about it here.

The conventional (and proper) way of attaching choices with model for a field is using static variable like this.

class MyModel(models.Model):
    TYPE_CHOICES = (
        # (<DB VALUE>, <DISPLAY_VALUE>)
        ('a', 'Choice A'),
        ('b', 'Choice B'),
    )

    type = models.CharField(max_length=1, choices=TYPE_CHOICES)

You can access choices for type field outside model like this.

MyModel.TYPE_CHOICES

Upvotes: 2

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