In django, how to limit choices of a foreignfield based on another field in the same model?

I have these models (I have limited the number of fields to just those needed)

class unit(models.Model):
    name = models.CharField(max_length=200)

class project(models.Model):
    name = models.CharField(max_length=200)

class location(address):
    project = models.ForeignKey(project)

class project_unit(models.Model):
    project = models.ForeignKey(project)         
    unit = models.ForeignKey(unit)

class location_unit(models.Model):
    project = models.ForeignKey(project)    
      #Limit the selection of locations based on which project has been selected
    location = models.ForeignKey(location)
      #The same here for unit. But I have no idea how.
    unit = models.ForeignKey(project_unit)       

My newbie head just cannot grasp how to limit the two fields, location and unit, in the location_unit model to only show the choices which refers to the selected project in location_unit. Should I override the modelform and make a query there or can I use the limit_choices_to. Either way I have failed trying both

Edit: Just to clarify, I want this to happen in the Django Admin. I have also tried formfield_for_foreignkey, but still a no go for me.

EDIT 2:

def formfield_for_foreignkey(self, db_field, request, **kwargs):
    if db_field.name == "开发者_运维知识库unit":
        kwargs["queryset"] = project_unit.objects.filter(project=1)
        return db_field.formfield(**kwargs)
    return super(location_unit_admin, self).formfield_for_foreignkey(db_field, request, **kwargs)

The above code snippet works. But of course I don't want the project to point to 1. How do I reference to the models project_id? I tried this:

kwargs["queryset"] = project_unit.objects.filter(project=self.model.project.project_id)

But that doesn't work (actually I have tried a lot of variations, yes I am a django newbie)

This is the answer, it is brilliant: https://github.com/digi604/django-smart-selects

Your formfield_for_foreignkey looks like it might be a good direction, but you have to realize that the ModelAdmin (self) won't give you a specific instance. You'll have to derive that from the request (possibly a combination of django.core.urlresolvers.resolve and request.path)

If you only want this functionality in the admin (and not model validation in general), you can use a custom form with the model admin class:

forms.py:

from django import forms

from models import location_unit, location, project_unit

class LocationUnitForm(forms.ModelForm):
    class Meta:
        model = location_unit

    def __init__(self, *args, **kwargs):
        inst = kwargs.get('instance')
        super(LocationUnitForm, self).__init__(*args, **kwargs)
        if inst:
            self.fields['location'].queryset = location.objects.filter(project=inst.project)
            self.fields['unit'].queryset = project_unit.objects.filter(project=inst.project)

admin.py:

from django.contrib import admin

from models import location_unit
from forms import LocationUnitForm

class LocationUnitAdmin(admin.ModelAdmin):
    form = LocationUnitForm

admin.site.register(location_unit, LocationUnitAdmin)

(Just wrote these on the fly with no testing, so no guarantee they'll work, but it should be close.)