Re: time interval math
От | Edward W. Rouse |
---|---|
Тема | Re: time interval math |
Дата | |
Msg-id | 034901cce737$86614180$9323c480$@com обсуждение исходный текст |
Ответ на | Re: time interval math (Jasen Betts <jasen@xnet.co.nz>) |
Список | pgsql-sql |
> -----Original Message----- > From: pgsql-sql-owner@postgresql.org [mailto:pgsql-sql- > owner@postgresql.org] On Behalf Of Jasen Betts > Sent: Thursday, February 09, 2012 6:37 AM > To: pgsql-sql@postgresql.org > Subject: Re: [SQL] time interval math > > On 2012-02-08, Edward W. Rouse <erouse@comsquared.com> wrote: > > I'm still working on getting this to work, but the summary is this: > > > > I am getting several (many) intervals of hour, minutes and seconds. I > need a > > sum of the absolute value these intervals, similar to the SUM(ABS()) > > function for numbers; and I need to divide this sum by an integer > (bigint). > > Getting the intervals is no problem, but I can't find built in > functions for > > the rest. Currently on 8.3, want to upgrade to 9.x but I can't until > this is > > finished. > > the operation abs() is meaninless on the type interval > eg: what is abs( '1 month - 32 days + 24 hours'::interval ) If you need to add 30 intervals together, then +- is not meaningless. > > howevwer since all your intervals are in seconds (postgres pretends > that all > hours are 3600 seconds long) converting to seconds is probably the > best way to go. > > > Do these functions exist, or will I be forced to convert to seconds, > do the > > math and then convert back to hour-minute-second format (I am > assuming from > > current data that, after the divide, the result should be in the > minute: > > second range). > > Yeah, you need to do that, it's not hard, > > select ( sum(abs(extract('epoch' from YOUR_INTERVAL ))) / SOMETHING ) * > '1s'::interval > from ... > > -- > ⚂⚃ 100% natural > > > -- > Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org) > To make changes to your subscription: > http://www.postgresql.org/mailpref/pgsql-sql
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