Re: Interval output format
От | Stefan Waidele jun. |
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Тема | Re: Interval output format |
Дата | |
Msg-id | 5.0.2.1.0.20010118221111.00aad0b0@imap.Krone-Neuenburg.de обсуждение исходный текст |
Ответ на | Re: Interval output format (Tom Lane <tgl@sss.pgh.pa.us>) |
Ответы |
Re: Re: Interval output format
Re: Re: Interval output format Re: Re: Interval output format |
Список | pgsql-novice |
At 13:37 18.01.2001 -0500, Tom Lane wrote: >"Stefan Waidele jun." <St.Waidele.jun@Krone-Neuenburg.de> writes: > > How can I tell Postgres to return an interval value in an format like > hhh:mm? > >See to_char(), >http://www.postgresql.org/devel-corner/docs/postgres/functions-formatting.htm > > regards, tom lane Thanks Tom, I have been at this page before and did not get a clue. Have I overlooked something? (Note: I am using an interval, not a date) It would be great to have an example, since to_char('1 14:30', 'hh:mm') (and others) gave me errors. On the next page I have found dat_part(), and solved my problem using: select date_part('day', sum(gearbeitet))*24+date_part('hour', sum(gearbeitet))||':'||date_part('min', sum(gearbeitet)) from stunden; But here comes the next problem: When I tried to do a SQL-Function create function sum_gearbeitet(text) returns text as "select date_part('day',sum(gearbeitet))*24+date_part('hour',sum(gearbeitet))||':'||date_part('min',sum(gearbeitet)) from stunden;" language 'sql'; psql said: NOTICE: identifier "select date_part('day',sum(gearbeitet))*24+date_part('hour',sum(gearbeitet))||':'||date_part('min',sum(gearbeitet)) from stunden;" will be truncated to "select date_part('day',sum(gear" ERROR: parser: parse error at or near """ What am I doing wrong? (shorter functions are also truncated) Stefan
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