> > I'm curious. I've been on a sql course long time ago where the tutor
> > mentioned similiar problem as something special for sql smarties... And
> > yesterday when I faced this problem (which can be solved perfectly with
> > the outer join) I wasn't able to remember the solution... If there is
> > any... Just out of curiosity...
>
> well, you could do it like this I suppose:
>
> SELECT 1 AS day, COUNT(data) FROM table1 WHERE day=1
> UNION
> SELECT 2, COUNT(data) FROM table1 WHERE day=2
> UNION
> SELECT 3, COUNT(data) FROM table1 WHERE day=3
> UNION
> SELECT 4, COUNT(data) FROM table1 WHERE day=4
>
> (and so on ad infinitum)
>
> although I expect E.F. Codd will be turning in his grave ;-)
A little better, perhaps, but still a true hack:
select alldays.d, sum(c) from (select 1 as d union all select 2 union all select 3 union all select 4 union all
select5 ...) as alldays
left outer join d using (d) group by alldays.d;
If you could write a function in plpgsql that returned a query result, you
could use that as the from clause rather than the long union.
