Django: Distinct foreign keys

Queries don't work like that - either in Django's ORM or in the underlying SQL. If you want to get unique IDs, you can only query for the ID. So you'll need to do two queries to get the actual Log entries. Something like:

id_list = Log.objects.order_by('-date').values_list('project_id').distinct()[:4]entries = Log.objects.filter(id__in=id_list)

Actually, you can get the project_ids in SQL. Assuming that you want the unique project ids for the four projects with the latest log entries, the SQL would look like this:

SELECT project_id, max(log.date) as max_dateFROM logsGROUP BY project_idORDER BY max_date DESC LIMIT 4;

Now, you actually want all of the log information. In PostgreSQL 8.4 and later you can use windowing functions, but that doesn't work on other versions/databases, so I'll do it the more complex way:

SELECT logs.*FROM logs JOIN (    SELECT project_id, max(log.date) as max_date    FROM logs    GROUP BY project_id    ORDER BY max_date DESC LIMIT 4 ) as latestON logs.project_id = latest.project_id   AND logs.date = latest.max_date;

Now, if you have access to windowing functions, it's a bit neater (I think anyway), and certainly faster to execute:

SELECT * FROM (   SELECT logs.field1, logs.field2, logs.field3, logs.date       rank() over ( partition by project_id                      order by "date" DESC ) as dateorder   FROM logs ) as logsortWHERE dateorder = 1ORDER BY logs.date DESC LIMIT 1;

OK, maybe it's not easier to understand, but take my word for it, it runs worlds faster on a large database.

I'm not entirely sure how that translates to object syntax, though, or even if it does. Also, if you wanted to get other project data, you'd need to join against the projects table.

I know this is an old post, but in Django 2.0, I think you could just use:

Log.objects.values('project').distinct().order_by('project')[:4]