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    <p>I know question was about Django 1.3 however you mentioned on soon upgrading to 1.4. Also to people, like me who was looking for solution for 1.4, but found this entry, i decided to show full example of using <a href="https://docs.djangoproject.com/en/1.4/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_filter">SimpleListFilter</a> (available Django 1.4) to show only referenced (related, used) foreign key values</p> <pre class="lang-py prettyprint-override"><code>from django.contrib.admin import SimpleListFilter # admin.py class CountryFilter(SimpleListFilter): title = 'country' # or use _('country') for translated title parameter_name = 'country' def lookups(self, request, model_admin): countries = set([c.country for c in model_admin.model.objects.all()]) return [(c.id, c.name) for c in countries] # You can also use hardcoded model name like "Country" instead of # "model_admin.model" if this is not direct foreign key filter def queryset(self, request, queryset): if self.value(): return queryset.filter(country__id__exact=self.value()) else: return queryset # Example setup and usage # models.py from django.db import models class Country(models.Model): name = models.CharField(max_length=64) class City(models.Model): name = models.CharField(max_length=64) country = models.ForeignKey(Country) # admin.py from django.contrib.admin import ModelAdmin class CityAdmin(ModelAdmin): list_filter = (CountryFilter,) admin.site.register(City, CityAdmin) </code></pre> <p>In example you can see two models - City and Country. City has ForeignKey to Country. If you use regular list_filter = ('country',) you will have all the countries in the chooser. This snippet however filters only related countries - the ones that have at least one relation to city.</p> <p>Original idea from <a href="http://jmduke.net/post/39953950546/custom-admin-filters-in-django">here</a>. Big thanks to author. Improved class names for better clarity and use of model_admin.model instead of hardcoded model name.</p> <p>Example also available in Django Snippets: <a href="http://djangosnippets.org/snippets/2885/">http://djangosnippets.org/snippets/2885/</a></p>
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    1. POCan I make list_filter in django admin to only show referenced ForeignKeys?
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    1. COThis example is very close to what I'm looking for, but with one exception: I'm trying to add a list_filter for User objects by a field on the UserProfile model, named user_type. So I define a class, UserTypeFilter(SimpleListFilter): but I don't know what you put in the queryset function's return queryset.filter(<user_type?>=self.value()).
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      1. USAnthony Roberts
    2. COAnswer to my question is here: http://stackoverflow.com/questions/19187027/django-1-4-user-admin-list-filter-using-userprofile-field/19187261?noredirect=1#19187261
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A row detail

Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

Related data can be related in a to-one or a to-many fashion. Captions of data related in a to-many fashion link to a list view showing a filtered view of the table.

Try moving around until you find a non-empty to-many entry and click on the label to get to one. You can move back to the root by clicking on the database name in the header.

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