Django get display name choices

Question

I\'m trying to find a solution to my problem.

models.py

class Article(models.Model):
    title = models.CharField(max_length=100)
    slug = models.S         


        

Answer 1

Option #1:

models.py

CHOICES_QUALITY = (
    ('1', 'HD YB'),
    ('2', 'HD BJ'),
    ('3', 'HD POQD'),
    ('4', 'HD ANBC'),
)

class Article(models.Model):
    title = models.CharField(max_length=100)
    slug = models.SlugField()
    description = models.TextField()

    def archive_quality(self):
        quality = self.archive_set.order_by('-quality').distinct().values_list(
            'quality', flat=True)
        lists = []
        for q in quality:
            for choice in CHOICES_QUALITY:
                if choice[0] == q:
                    lists.append({'quality': choice[1]})
        return lists

class Archive(models.Model):
    article = models.ForeignKey(Article)
    quality = models.CharField(max_length=100, choices=CHOICES_QUALITY)

template

{% for article in articles %}
    {% for item in article.archive_quality %}
        {{ item.quality }},
    {% endfor %}
{% endfor %}

Option #2:

archive_tag.py

from django import template
from app_name.models import CHOICES_QUALITY

register = template.Library()

@register.filter
def quality(q):
    for choice in CHOICES_QUALITY:
        if choice[0] == q:
            return choice[1]
    return ''

template

{% load archive_tag %}

{% for article in articles %}
    {% for item in article.archive_quality %}
        {{ item|quality }},
    {% endfor %}
{% endfor %}