Thanks!
All of your workarounds involving filter didn't work, erroring with
"Cannot resolve keyword 'song_aggregation' into field", which makes
sense since I really wanted to go the other way around, I think.
Anyways, after some munging with 'extra', as per your suggestion, I
got it working!
Thanks again,
Eric
On Jul 24, 10:36 pm, "Russell Keith-Magee" <[EMAIL PROTECTED]>
wrote:
> On 7/25/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
>
>
> > An analogous operation would be:
> > playlist_aggregates = PlaylistAggregation.objects.order_by('-count')
> > playlists = [p.playlist for p in playlist_aggregates]
>
> The query itself looks like it's just a query over Playlist objects
> where all the query terms relate to Playlist_aggregate, e.g.,
>
> Playlist.objects.filter(playlist_aggregation__count=3)
>
> The ordering issue is a little more difficult. Ideally, it sounds like
> you want to be able to write:
>
> Playlist.objects.all().order_by('playlist_aggregation.count')
>
> but the query syntax doesn't currently support this; order_by isn't
> taken into account when the joins are determined. This is part of a
> task that Malcolm is currently looking at.
>
> In the interim, there are a few workarounds. In the end, the argument
> to order_by is just the SQL name of the attribute; it just happens
> that for attributes on a model, the SQL name matches the attribute
> name. However, you can use this to cheat a bit. If you substitute the
> SQL name for the joined column, you can order by joined attribute - as
> long as the related table is actually joined in the SQL query.
>
> A normal Playlist.objects.all() query won't join Playlist_aggregation
> - so you'll need to fake it. Either add a meaningless filter, like
> filter(playlist_aggregation__count__gte=0) (which will join the table,
> but not reject any rows), or use the 'extra' clause to manually add
> the join required.
>
> Yours,
> Russ Magee %-)
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