| From: | Andrei Zhidenkov <andrei(dot)zhidenkov(at)n26(dot)com> |
|---|---|
| To: | Ray O'Donnell <ray(at)rodonnell(dot)ie> |
| Cc: | David Gauthier <davegauthierpg(at)gmail(dot)com>, Postgres General <pgsql-general(at)postgresql(dot)org> |
| Subject: | Re: format return of "age" to hh:mm |
| Date: | 2020-03-05 16:12:16 |
| Message-ID: | B8FA4081-EB2C-4017-B36B-3B2B6F326CDC@n26.com |
| Views: | Raw Message | Whole Thread | Download mbox | Resend email |
| Thread: | |
| Lists: | pgsql-general |
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.
> On 5. Mar 2020, at 17:07, Ray O'Donnell <ray(at)rodonnell(dot)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(at)rodonnell(dot)ie
>
>
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