GenericForeignKeys with fewer queries

This is a really great tip! Why don't you add the ready-made manager to djangosnippets.com? I Guess some people'd love it.

Cheers.

PS. Again: Great work.

Awesome. Great Manager! adding it to delicious

Martin,

I have been playing with your solution, but it seems that if you use ContentType object as a key in the .setdefault function, each object is unique and therefore resulting model_map has the same number of elements as there are items (=n). The following 'for' cycle is then repeated 'n' times, not 'm'. At least that is how it works for me in Django 0.96.1.

The solution could be to use item.content_type.name as a key for model_map, and translate it later back to ContentType object.

Also, it seems that this procedure will change the ordering of the list, so you basically can not control it. So far I do not know how to solve that.

The last two things, just a corrections, I think you can not use filter() and all() together in ct.model_class().objects.select_related() .filter(id__in=items_.keys()).all(), and in the snippet you posted you probably want to return item_map.items(), not the initial 'qs'.

I might be wrong, so please let me know what you think about that.

Anyway your code gave me some good ideas, if we will be able to fix/clarify above stuff, it will be perfect.

sorry, not Martin, but zerok, of course:)

Thanks for your reply, I realized I was wrong about 'ordering' and 'qs'. Regarding using 'all()' and 'filter()' together - that returns error in Django 0.96.1 (I do no use the latest SVN version, maybe I should:). The same for 'object as key' issue, again it is probably fixed in newer versions. To sum it up, I have put together a code that I believe works as meant in the release that I use, maybe it could help also some other guys. BTW, it makes me wandering, if it is necessary to use select_related('content_type'), as it could be done using content_type_id as a key and then get the correct ContentType object using that...

items = Item.objects.select_related('content_type').all()
model_map = {}
item_map = {}

for item in items:
  model_map.setdefault(item.content_type.name, {})[item.object_id] = item.id
  item_map[item.id] = item

    for ct_name, items_ in model_map.items():
      for o in ContentType.objects.get(name=ct_name).model_class().objects.select_related().filter(id__in=items_.keys()):
           node_map[nodes_[o.id]].content_object = o

## We do not have to do this, because if
## we modify item in item_map, the change is reflected in 'items' also (OOP:), so I take back my comment about 'ordering' and also about 'return qs'
#     for item in items:
#         item.content_object = item_map[item.id].content_object

return items

Thanks again, keep up good work.

fewer.