PostgreSQL: fill records using fields from JSONB with jsonb_to_record()

I have a function returning jsonb

SELECT get_my_records( 'some', 'args' );

get_my_records
{ "a": 1, "b": 3.14 }
{ "a": 2, "b": 2.71 }
...

and trying to use jsonb_to_record() to convert the records to

a	b
1	3.14
2	2.71
...	...

Tried

SELECT * 
FROM jsonb_to_record( val ) as x( a int, b double precision ) 
FROM get_my_records( 'some', 'arg' ) AS val;`

and a few variations but without success. How does one go about it?

(probably lacking proper terminology to get meaningful results from the web search)

Solution

Your attempt was mostly correct, but you added multiple FROM lists for some reason, which isn't allowed. You can get away with a single FROM item if you use ->> accessors instead: demo

select (j->>'a')::int as a, 
       (j->>'b')::double precision as b 
from get_my_records( 'some', 'args' ) as jdata(j);

a	b
1	3.14
2	2.71

You can't have multiple from lists in a single query - you'd have to split that into CTEs/subqueries

SELECT * 
FROM jsonb_to_record( val ) as x( a int, b double precision ) 
FROM get_my_records( 'some', 'arg' ) AS val;

ERROR:  syntax error at or near "FROM"
LINE 4: FROM get_my_records( 'some', 'arg' ) AS val;
        ^

Or, you can just extend your from list by joining the sources:

SELECT x.* 
FROM get_my_records( 'some', 'arg' ) AS val 
LEFT JOIN LATERAL jsonb_to_record(val) as x(a int,b double precision) ON TRUE;

a	b
1	3.14
2	2.71

You can use a comma , to join implicitly, since lateral is implied by dependence between these set-returning functions posing as sources, and on true just lets them join freely.

SELECT x.* 
FROM get_my_records( 'some', 'arg' ) AS val,
     jsonb_to_record( val ) as x( a int, b double precision );