>
> Hi,
>
> On Mon, 1 Jun 2020 at 23:50, Alban Hertroys <haram...@gmail.com> wrote:
> > On 1 Jun 2020, at 20:18, Shaheed Haque <shaheedha...@gmail.com> wrote:
> >
> > Hi,
> >
> > I'm using Django's ORM to access Postgres12. My "MyModel" table has a
> JSONB column called 'snapshot'. In Python terms, each row's 'snapshot'
> looks like this:
> >
> > ======================
> > snapshot = {
> > 'pay_definition' : {
> > '1234': {..., 'name': 'foo', ...},
> > '99': {..., 'name': 'bar', ...},
> > }
> > ======================
> >
> > I'd like to find all unique values of 'name' in all rows of MyModel. I
> have this working using native JSON functions from the ORM like this:
> >
> > =====================
> > class PayDef(Func):
> > function='to_jsonb'
> >
>
> template="%(function)s(row_to_json(jsonb_each(%(expressions)s->'pay_definition'))->'value'->'name')"
> >
> >
> MyModel.objects.annotate(paydef=PayDef(F('snapshot'))).order_by().distinct('paydef').values_list('paydef',
> flat=True)
> > =====================
> >
> > So, skipping the ordering/distinct/ORM parts, the core looks like this:
> >
> >
> to_jsonb(row_to_json(jsonb_each('snapshot'->'pay_definition'))->'value'->'name’)
>
>
> I do something like this to get a set of sub-paths in a JSONB field (no
> idea how to write that in Django):
>
> select snapshot->’pay_definition’->k.value->’name’
> from MyModel
> join lateral jsonb_object_keys(snapshot->’pay_definition’) k(value) on
> true
>
>
I was unaware of the LATERAL keyword, so thanks. After a bit of Googling
however, it seems that it is tricky/impossible to use from the ORM (barring
a full scale escape to a "raw" query). One question: as a novice here, I
think I understand the right hand side of your JOIN "... k(value)" is
shorthand for:
... AS table_name(column_name)
except that I don't see any clues in the docs that jsonb_object_keys() is a
"table function". Can you kindly clarify?
> I don’t know how that compares performance-wise to using jsonb_each, but
> perhaps worth a try. Obviously, the way it’s written above it doesn’t
> return distinct values of ’name’ yet, but that’s fairly easy to remedy.
>
> Indeed; this is what I managed to get to:
SELECT DISTINCT snapshot -> 'pay_definition' -> k.value -> 'name' AS name
FROM paiyroll_payrun
JOIN LATERAL jsonb_object_keys(snapshot -> 'pay_definition')
AS k(value) ON true
ORDER BY name;
At any rate, I'll have to ponder the "raw" route absent some way to "JOIN
LATERAL".
Thanks, Shaheed
> Alban Hertroys
> --
> If you can't see the forest for the trees,
> cut the trees and you'll find there is no forest.
>
>
>
