Обсуждение: format return of "age" to hh:mm
Hi:
How does one reformat the output of the "age" function to always be in terms of hours:mins.
E.g.
dvdb=> select age('2020-03-05 01:40:32-05','2020-03-01 21:56:05-05');
age
-----------------
3 days 03:44:27
(1 row)
age
-----------------
3 days 03:44:27
(1 row)
I want...
"75:44"
I'm not married to "age" If there's a better way to do this that's fine too.
Thanks in advance !
On 05/03/2020 15:50, David Gauthier wrote: > Hi: > > How does one reformat the output of the "age" function to always be in > terms of hours:mins. Hi there, age() returns an interval, so without having tried it I'm guessing you could use to_char() to format it whatever way you want. Ray. -- Raymond O'Donnell // Galway // Ireland ray@rodonnell.ie
However, you cannot use to_char() to display the count of days for a given interval. In this case, if your interval is largerthan 24 hours, you might use extract(epoch from <interval>) and perform the conversion manually. > On 5. Mar 2020, at 17:07, Ray O'Donnell <ray@rodonnell.ie> wrote: > > On 05/03/2020 15:50, David Gauthier wrote: >> Hi: >> >> How does one reformat the output of the "age" function to always be in >> terms of hours:mins. > > Hi there, > > age() returns an interval, so without having tried it I'm guessing you > could use to_char() to format it whatever way you want. > > Ray. > > -- > Raymond O'Donnell // Galway // Ireland > ray@rodonnell.ie > >
On Thu, Mar 5, 2020 at 8:50 AM David Gauthier <davegauthierpg@gmail.com> wrote:
Hi:How does one reformat the output of the "age" function to always be in terms of hours:mins.
Custom function.
Use justify_hours(interval) to normalize the input in terms of days
Use extract(field from interval) to get the components, including days
Multiply the days result by 24, add it to the hours result
Deal with fractional hours
Combine and return
There is no justify_minutes function unfortunately which, if implemented to the behavior of justify_hours, would do what you are looking for. You basically want to write one, though I suspect in SQL instead of C.
David J.
On 3/5/20 7:50 AM, David Gauthier wrote:
> Hi:
>
> How does one reformat the output of the "age" function to always be in
> terms of hours:mins.
>
> E.g.
>
> dvdb=> select age('2020-03-05 01:40:32-05','2020-03-01 21:56:05-05');
> age
> -----------------
> 3 days 03:44:27
> (1 row)
>
> I want...
>
> "75:44"
>
> I'm not married to "age" If there's a better way to do this that's fine
> too.
Not sure it's better, but it will give you idea of what needs to be done:
SELECT
floor(
extract(
epoch FROM ('2020-03-05 01:40:32-05'::timestamptz -
'2020-03-01 21:56:05-05'::timestamptz))
/ 3600)::varchar || ':' ||
((mod(
extract(
epoch FROM ('2020-03-05 01:40:32-05'::timestamptz -
'2020-03-01 21:56:05-05'::timestamptz))::numeric,
3600::numeric) / 60)::int)::varchar;
?column?
----------
75:44
(1 row)
>
> Thanks in advance !
--
Adrian Klaver
adrian.klaver@aklaver.com