Implementing Cascaded Filters in Django Admin

For your hierarchical model structure (Org → Branch → Store → Product), you can implement cascaded filtering in the Django Admin using the formfield_for_foreignkey method. This approach will help filter the available choices based on previously selected values.

Here's how you can implement it for your models:

Solution for Product Admin

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class ProductAdmin(admin.ModelAdmin):
    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "store":
            if request.GET.get('store'):
                kwargs["queryset"] = Store.objects.filter(id=request.GET.get('store'))
            elif request.GET.get('branch'):
                kwargs["queryset"] = Store.objects.filter(branch_id=request.GET.get('branch'))
            elif request.GET.get('org'):
                kwargs["queryset"] = Store.objects.filter(branch__org_id=request.GET.get('org'))
        return super().formfield_for_foreignkey(db_field, request, **kwargs)

admin.site.register(Product, ProductAdmin)

Solution for Store Admin

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class StoreAdmin(admin.ModelAdmin):
    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "branch":
            if request.GET.get('branch'):
                kwargs["queryset"] = Branch.objects.filter(id=request.GET.get('branch'))
            elif request.GET.get('org'):
                kwargs["queryset"] = Branch.objects.filter(org_id=request.GET.get('org'))
        return super().formfield_for_foreignkey(db_field, request, **kwargs)

admin.site.register(Store, StoreAdmin)

Solution for Branch Admin

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class BranchAdmin(admin.ModelAdmin):
    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "org" and request.GET.get('org'):
            kwargs["queryset"] = Org.objects.filter(id=request.GET.get('org'))
        return super().formfield_for_foreignkey(db_field, request, **kwargs)

admin.site.register(Branch, BranchAdmin)

For More Complex Filtering

For more complex scenarios, you can use a custom ModelForm with JavaScript to create dynamic dropdowns. Here's a simplified approach:

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class ProductAdminForm(forms.ModelForm):
    class Meta:
        model = Product
        fields = '__all__'
        
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        # If we have an instance, filter the store field
        if self.instance.pk:
            if self.instance.store:
                self.fields['store'].queryset = Store.objects.filter(
                    branch=self.instance.store.branch
                )

class ProductAdmin(admin.ModelAdmin):
    form = ProductAdminForm

This approach works well for editing existing records, but for creating new records, you'll need to add JavaScript to your admin to dynamically update the dropdowns as selections are made.

Remember that for a complete solution, you'll need to:

1. Register all your model admins with their respective admin classes
2. Consider adding list filters to make the list views more manageable
3. Potentially add JavaScript for a more dynamic user experience when creating new records

This implementation should significantly improve the usability of your Django Admin interface by reducing the number of choices in each dropdown based on the hierarchical relationship between your models.