format return of "age" to hh:mm

Started by David Gauthierabout 6 years ago5 messagesgeneral
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#1David Gauthier
davegauthierpg@gmail.com

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 !

#2Ray O'Donnell
ray@rodonnell.ie
In reply to: David Gauthier (#1)
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
ray@rodonnell.ie

#3Andrei Zhidenkov
andrei.zhidenkov@n26.com
In reply to: Ray O'Donnell (#2)
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 <interval>) and perform the conversion manually.

Show quoted text

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

#4David G. Johnston
david.g.johnston@gmail.com
In reply to: David Gauthier (#1)
Re: format return of "age" to hh:mm

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.

#5Adrian Klaver
adrian.klaver@aklaver.com
In reply to: David Gauthier (#1)
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.klaver@aklaver.com