On Jun 29, 2007, at 13:17 , John D. Burger wrote:
I can't anything in the docs that explain how intervals print out.
They seem to show like this:
> select now() - '1990-01-01';
?column?
---
6388 days 13:06:26.3605600595
Without being anchored with a
"John D. Burger" <[EMAIL PROTECTED]> writes:
> Why do the first and third intervals print out differently?
The underlying storage is months, days, and seconds --- "1 year"
is the same as "12 months", but not the same as "365 days".
IIRC plain timestamp subtraction produces an interval with days an
I can't anything in the docs that explain how intervals print out.
They seem to show like this:
> select now() - '1990-01-01';
?column?
---
6388 days 13:06:26.3605600595
or like this:
> select now() - current_date;
?column?
-
14:06:4
am Thu, dem 28.06.2007, um 16:04:48 -0400 mailte Jasbinder Singh Bali
folgendes:
> Hi,
>
> I have a timestamp field in my talbe.
> I need to check its difference in days with the current date.
>
> field name is time_stamp and I did it as follows:
>
> select age(timestamp '2000-06-28 15:39:47.2
On Jun 28, 2007, at 15:13 , Raymond O'Donnell wrote:
Cast your result to type INTERVAL - something like this:
postgres=# select (current_timestamp - timestamp
'2007-05-01')::interval;
interval
--
58 days 21:10:36.748
(1 row)
The cast to interval is superfluous
On 28/06/2007 21:04, Jasbinder Singh Bali wrote:
How can i convert this result into absolute number of days.
Cast your result to type INTERVAL - something like this:
postgres=# select (current_timestamp - timestamp '2007-05-01')::interval;
interval
--
58 days 21:1
Hi,
I have a timestamp field in my talbe.
I need to check its difference in days with the current date.
field name is time_stamp and I did it as follows:
select age(timestamp '2000-06-28 15:39:47.272045')
it gives me something like
6 years 11 mons 29 days 08:20:12.727955
How can i convert th
rnd=# select extract(epoch from timestamp '2002-09-08 05:29:41');
date_part
1031477381
(1 row)
See the date/time functions documentation.
On Thursday 01 December 2005 04:07 am, Sterpu Victor saith:
> I need a function that will transform from the format '2002-09-08
> 05:29:41' to s
am 01.12.2005, um 11:07:59 +0200 mailte Sterpu Victor folgendes:
> I need a function that will transform from the format '2002-09-08 05:29:41'
> to seconds since 1970.
scholl=# select now();
now
---
2005-12-01 10:18:47.554236+01
(1 row)
scholl=# select
I need a function that will transform from the format '2002-09-08
05:29:41' to seconds since 1970.
Thank you.
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On Thu, Aug 28, 2003 at 23:09:44 +0900,
Alex <[EMAIL PROTECTED]> wrote:
> Hi,
> is there an easy way to get the Difference between 2 dates in Days... as
> an integer?
>
> age('date1','date2') results in x years y days or something similar.
> but I would like to
> have the result in number of
Alex <[EMAIL PROTECTED]> writes:
> is there an easy way to get the Difference between 2 dates in Days... as
> an integer?
Subtract 'em ...
regression=# select '2002-01-01'::date - '2001-01-01'::date;
?column?
--
365
(1 row)
regards, tom lane
Hi,
is there an easy way to get the Difference between 2 dates in Days... as
an integer?
age('date1','date2') results in x years y days or something similar.
but I would like to
have the result in number of days as an integer.
Thanks for your help
Alex
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