[Django]-Get full label name of ChoiceField in QuerySet values list

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I think the easiest way would be to use only() instead of values(). Like values, only retrieves a subset of the database columns. The difference is that it creates model instance objects, which means that you can use get_X_display() as usual.

instances = MyModel.objects.filter(a='1').only('id', 'a', 'choicefield')
instances[0].get_choicefield_display()  # will work fine

Of course, you should check out the caveats in the documentation on only() and defer(), but for the purposes you’ve described I don’t see a problem.

If you want to use values() there are many ways to solve this problem. One approach is to define the mapping explicitly in your model:

from collections import OrderedDict

class MyModel(models.Model):
    CHOICES_DICT = OrderedDict((
        (1, 'Choice 1'),
        (2, 'Choice 2'),
    ))
    choicefield = models.IntegerField(choices=CHOICES_DICT.items())
    ...

Now you can explicitly look up up the string value in your view:

values = MyModel.objects.filter(a='1').values('id','a','choicefield')
name = MyModel.CHOICES_DICT[values[0]['choicefield']]

You could also create a custom template filter to do the same thing at the template level.

👤Kevin Christopher Henry

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