Правильнее и проще было бы сделать UserInfoAdmin инлайном для UserAdmin. Но можно добиться и использования полей связанной модели:
from six import with_metaclass
from django.contrib import admin
from django.db import models
def getter_for_related_field(name, admin_order_field=None, short_description=None):
related_names = name.split('__')
def getter(self, obj):
for related_name in related_names:
obj = getattr(obj, related_name)
return obj
getter.admin_order_field = admin_order_field or name
getter.short_description = short_description or related_names[-1].title().replace('_',' ')
return getter
class RelatedFieldAdminMetaclass(type(admin.ModelAdmin)):
def __new__(cls, name, bases, attrs):
new_class = super(RelatedFieldAdminMetaclass, cls).__new__(cls, name, bases, attrs)
for field in new_class.list_display:
if '__' in field:
setattr(new_class, field, getter_for_related_field(field))
return new_class
class RelatedFieldAdmin(with_metaclass(RelatedFieldAdminMetaclass, admin.ModelAdmin)):
def get_queryset(self, request):
qs = super(RelatedFieldAdmin, self).get_queryset(request)
select_related = [field.rsplit('__',1)[0] for field in self.list_display if '__' in field]
model = qs.model
for field_name in self.list_display:
try:
field = model._meta.get_field(field_name)
except models.FieldDoesNotExist:
continue
if isinstance(field.rel, models.ManyToOneRel):
select_related.append(field_name)
return qs.select_related(*select_related)
@admin.register(UserInfo)
class UserInfoAdmin(RelatedFieldAdmin):
list_display = ['user', 'user__first_name', 'user__last_name', 'user__email', 'position']
list_display_links = list_display
list_filter = ['user__is_staff', 'position']
search_fields = ['user__username', 'user__first_name', 'user__last_name', 'user__email', 'position', 'phone', 'location']
