unpack all outer-most keys in json object as columns - json

Question

I have a postgres DB called sales with a json-object, data containing around 100 outer-keys, lets name them k1,k2,k3..,k100.
I want to write a query
select * from sales some_function(data)
which simply returns something like
k1 | k2 | .. | k100
--------------------
"foo" | "bar" | .. | 2
"fizz"| "buzz"| .. | 10
ie. just unpacks the keys as columsn and their values as row.
Note, k1,k2..k100 is not their real name thus I can't do a
data->> key loop

Answer 1

That's not possible. One restriction of the SQL language is, that all columns (and their data types) must be known to the database when parsing the statement - so before it is actually run.
You will have to write each one separately:
select data ->> 'k1' as k1, data ->> 'k2' as k2, ...
from sales
One way to make this easier, is to generate a view dynamically by extracting all JSON keys from the column, then using dynamic SQL to create the view. You will however need to re-create that view each time the number of keys change.
Something along the lines (not tested!)
do
$$
declare
l_columns text;
l_sql text;
begin
select string_agg(distinct format('data ->> %L as %I', t.key, t.key), ', ')
into l_columns
from sales s
cross join jsonb_each(s.data) as t(key, value);
-- l_columns now contains something like:
-- data ->> 'k1' as k1, data ->> 'k2' as k2
-- now create a view from that
l_sql := 'create view sales_keys as select '||l_columns||' from sales';
execute l_sql;
end;
$$
;
You probably want to add e.g. the primary key column(s) to the view, so that you can match the JSON values back to the original row(s).