Hello
2013/2/25 Stefan Keller <sfkeller@gmail.com>:
> Hi,
>
> I have a simple void function:
>
> CREATE OR REPLACE FUNCTION myfn(myparam OUT int)
> AS $$
> BEGIN
> pnr := 1;
> END;
> $$ LANGUAGE plpgsql;
>
> How do I access myparam?
> I thought this should work with 9.1/9.2: SELECT (myfn()).myparam;
> Or inside another function?
>
you cannot access to out parameters outside function - because they
doesn't exist - postgresql cannot pass parameters by ref.
your example is exactly same as int returning function - you can use
it in plpgsql
variable := myfn(); -- variable is scalar int type
if function has more out parameters, then return type is record type.
CREATE OR REPLACE FUNCTION public.f1(a integer, b integer, OUT c
integer, OUT d integer)
RETURNS record
LANGUAGE plpgsql
AS $function$
begin
c := a + b;
d := c * 2;
end;
$function$
postgres=# select f1(10,20);
f1
---------
(30,60)
(1 row)
postgres=# select * from f1(10,20);
c | d
----+----
30 | 60
(1 row)
create or replace function foo()
returns void as $$
declare r record;
begin
r := f1(10,20);
raise warning 'c=%, d=%', r.c, r.d;
end;
$$ language plpgsql;
CREATE FUNCTION
postgres=# select foo();
WARNING: 01000: c=30, d=60
foo
-----
(1 row)
Regards
Pavel Stehule
> Yours, Stefan
>
>
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