> On 28 Feb 2023, at 3:54, Thorsten Glaser <tg@xxxxxxxxxxx> wrote:
(…)
>> Well, that may be what you want, but it's not what you wrote in
>> the query. Follow David's advice and do
> […]
>> I'm pretty sure that this will only incur one evaluation of the
>> common subexpression, so even though it's tedious to type it's not
>> inefficient.
>
> Thanks. But I fear it’s not as simple as you wrote. More like:
>
> jsonb_build_object('opening_times',
> jsonb_agg(DISTINCT jsonb_build_object(
> 'weekday', cot.weekday,
> 'from_hour', cot.from_hour,
> 'to_hour', cot.to_hour)
> ORDER BY
> jsonb_build_object(
> 'weekday', cot.weekday,
> 'from_hour', cot.from_hour,
> 'to_hour', cot.to_hour)->>'weekday',
> jsonb_build_object(
> 'weekday', cot.weekday,
> 'from_hour', cot.from_hour,
> 'to_hour', cot.to_hour)->>'from_hour',
> jsonb_build_object(
> 'weekday', cot.weekday,
> 'from_hour', cot.from_hour,
> 'to_hour', cot.to_hour)->>'to_hour')
> )
>
> Isn’t that more like it?
Perhaps you can use a lateral cross join to get the result of jsonb_build_object as a jsonb value to pass around?
Alban Hertroys
--
If you can't see the forest for the trees,
cut the trees and you'll find there is no forest.
