How to change the default labels of a ForeignKey field whose form's widget is a RadioSelect-CodePudding

In my Django app, I have this form:

from django import forms
from main.models import Profile

class ProfileForm(forms.ModelForm):

    class Meta:
        model = Profile
        fields = ['picture']

        widgets = {
            'picture': forms.RadioSelect(),
        }

By default, the labels of the picture field within the template are generated according to the __str__() method of the Picture model. That's because the picture field on my Profile model is actually a ForeignKey field to a Picture model.

However, the value returned by that __str__() method doesn't make much sense in this particular template, and changing it is not possible because it's being used elsewhere.

Therefore, is there a way I might change the default labels for the picture field of my ProfileForm? For instance, changing from the __str__()'s default picture.src ' (' picture.description ')' to something like picture.description only?

I have checked what the docs had to say regarding the label_from_instance, but I didn't understand how to apply it. In fact, I couldn't even understand if that would be a solution for this case.

Some similiar questions here on Stack Overflow also mentioned that link of the docs, but the questions's forms were slightly different than mine, and I ended up not understanding them either.

CodePudding user response：

If you want to use label_from_instance, you should make a specific widget with:

class PictureField(forms.ModelChoiceField):
    def label_from_instance(self, obj):
        return obj.description


class ProfileForm(forms.ModelForm):
    picture = PictureField(Picture.objects)

    class Meta:
        model = Profile

On another hand, you can use your code and in init function of your form to change the choices attribute of the picture field

CodePudding user response：

It turns out I didn't need to use label_from_instance. I just needed to create a subclass of django.forms.RadioSelect and override its create_option() method, setting the label by giving option['label'] the value I'd like, just like this:

from django import forms
from main.models import Profile

class PictureRadioSelect(forms.RadioSelect):
    def create_option(self, name, value, label, selected, index, subindex=None, attrs=None):
        option = super().create_option(name, value, label, selected, index, subindex, attrs)
        option['label'] = value.instance.description
        return option

class ProfileForm(forms.ModelForm):

    class Meta:
        model = Profile
        fields = ['picture']

        widgets = {
            'picture': PictureRadioSelect(),
        }

This section of the docs explains this in greater detail. It actually describes how to override django.forms.Select options, but I tested it with django.forms.RadioSelect and it ran as expected.