Re: format return of "age" to hh:mm
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.kla...@aklaver.com
Re: format return of "age" to hh:mm
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 r...@rodonnell.ie
Re: format return of "age" to hh:mm
On Thu, Mar 5, 2020 at 8:50 AM David Gauthier 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.
Re: format return of "age" to hh:mm
However, you cannot use to_char() to display the count of days for a given interval. In this case, if your interval is larger than 24 hours, you might use extract(epoch from ) and perform the conversion manually. > On 5. Mar 2020, at 17:07, Ray O'Donnell 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 > r...@rodonnell.ie > >
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)
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 !
