Question

I have a table with questions, where each row is a question and all questions has a asked_by and answered_by field. I want to select all unique combinations of “askers” and “answerers” and count how many times this combination of asking/answering have occurred.

question_tbl:

id      | asked_by | answered_by | session
--------+----------+-------------+---------
1       | AA       | JD          | 2011-2012
2       | JD       | AA          | 2011-2012
9       | AA       | JD          | 2011-2012
12      | AA       | JD          | 2009-2010

I want to return the unique combinations in in session, and the count for this combination.

E.g. for 2011-2012:

AA, JD, 2
JD, AA, 1

PS: Ideally I’d write this for the django orm, but as I can’t even seem to figure out how to formulate a sensible google search for this, to find the SQL-way is my goal.

Note after problem was solved:
As this turned out to be quite easy in SQL, it turned out to be quite easy in Djangos ORM too. I’ll leave it here for later record:

from django.db.models import Count
from myproject.models import Questions

Questions.objects.filter(sesjonid='2011-2012').values('sporsmal_til', 'sporsmal_fra').annotate(count=Count('besvart_av'))

By testing how that looks in SQL it seems to me to correct (in the ./manage.py shell):

print str(Questions.objects.filter(sesjonid='2011-2012').values('sporsmal_til', 'sporsmal_fra').annotate(count=Count('sporsmal_til')).query)

This prints SQL that does exactly the same as Bohemians’.

Answer 1

A straightforward COUNT(*) with a GROUP BY will do it:

select asked_by, answered_by, count(*)
from question_tbl
where session = '2011-2012'
group by 1, 2;

If you wanted a query to return data for all sessions, use this:

select session, asked_by, answered_by, count(*)
from question_tbl
group by 1, 2, 3;