How to use model objects as choices for django-filter MultipleChoiceFilter

Question

I'm using Django-filter, and I would like one of the fields (supervisor) to be a ChoiceFilter where the choices are objects from the model. What's the most efficient way to do that? I tried following this post, but kept getting errors no matter what I changed (currently cannot unpack non-iterable int object).

# models.py
class people(models.Model):
    namelast = models.CharField(max_length=100, verbose_name='Last Name')
    namefirst = models.CharField(max_length=100, verbose_name='First Name')
    supervisor = models.ForeignKey('self', blank=True, null=True, on_delete=models.SET_NULL, verbose_name='Supervisor')
    
    def __str__(self):
        return "%s %s" % (self.namefirst, self.namelast)

# filters.py
class office_filter(django_filters.FilterSet):
    supervisor = django_filters.ChoiceFilter(choices=[], lookup_expr='icontains', label='Supervisor')
    # other fields

    class Meta:
        model = people
        fields = ['namelast', 'namefirst', 'supervisor']

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        try:
            self.filters['supervisor'].extra['choices'] = [x for x in
                people.objects.all().values_list('supervisor', flat=True).distinct()]
        except (KeyError, AttributeError):
            pass

The goal is to have the supervisor field be a nice menu of all of the people that have been added assigned as a supervisor in the people model.

Ananya · Accepted Answer · 2020-07-02 18:24:50Z

1

I am not a 100% sure, but could you try something?

[people.objects.get(pk=x[0]) for x in people.objects.all().values_list('supervisor', flat=True).distinct()]

or

[people.objects.get(id=x[0]) for x in people.objects.all().values_list('supervisor', flat=True).distinct()]

In the link that you mentioned I believe DatedResource.objects.all().values_list('date', flat=True).distinct()) returns an array of string

In your code people.objects.all().values_list('supervisor', flat=True).distinct() would return a string of int - the IDs of the records as it is a Foreign Key

answered Jul 2, 2020 at 18:24

Ananya

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2 Comments

Evan Over a year ago

That's good to know about what's returned from the link I posted. Unfortunately both versions of your suggestion give the error int object is not subscriptable, so I tried pk=x and got cannot unpack non-iterable people object. That whole line works okay in the shell, so I wonder if it's django-filter not liking whatever list is returned (not clear in error messages).

Evan Over a year ago

Oh, I think I figured it out...it has to return a tuple.

[(people.objects.get(pk=x).id, people.objects.get(pk=x)) for x in people.objects.all().values_list('supervisor', flat=True).distinct() if x is not None]

seems to be working. Thanks for pointing me in the right direction!

Evan · Accepted Answer · 2020-07-06 18:25:22Z

Ananya's answer helped me get on the correct track of what have that statement return, but after thinking about the errors and how choice lists are usually constructed, I realized I needed it to return a tuple (not just a value). Here is the relevant code that ended up working:

class office_filter(django_filters.FilterSet):
    supervisor = django_filters.ChoiceFilter(choices=[], label='Supervisor')
    #...
        def __init__(self, *args, **kwargs):
            super().__init__(*args, **kwargs)
            try:
                self.filters['supervisor'].extra['choices'] = [(people.objects.get(pk=x).id, people.objects.get(pk=x)) for x in people.objects.all().values_list('supervisor', flat=True).distinct() if x is not None]
            except (KeyError, AttributeError):
                pass

The important part being: (people.objects.get(pk=x).id, people.objects.get(pk=x)) rather than just people.objects.get(pk=x).

Also had to remove lookup_expr in the filter field.

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