Re: Group by range in hour of day

On Tuesday, March 17, 2015, Israel Brewster <israel(at)ravnalaska(dot)net> wrote:

>
>
> > On Mar 17, 2015, at 8:09 AM, Paul Jungwirth <pj(at)illuminatedcomputing(dot)com
> <javascript:;>> wrote:
> >
> >>> test=> select h, count(*) from start_end, generate_series(0, 23) as
> s(h) where h between extract(hour from start_time) and extract(hour from
> end_time) group by h order by h;
> >>>
> >>> h | count
> >>> ----+-------
> >>> 8 | 2
> >>> 9 | 3
> >>> 10 | 2
> >>> 11 | 2
> >
> > Note if you always want all 24 rows with a count of 0 when appropriate
> (which seems common in reports with tables or plots), you can just tweak
> the above query to use a left join: FROM generate_series(0, 23) AS s(h)
> LEFT OUTER JOIN start_end ON h BETWEEN ...
> >
> > Paul
>
> Right, thanks. That makes sense. So next question: how do I get the
> "active" time per hour from this? To use the same example that came up with
> this result set:
>

Which is why you do not (only?) want to convert your data to hour-of-day
but want to create timestamp end points. Then you simply do timestamp
subtraction to get durations which you can then sum together.

David J.