Re: Constant value for a partitioned table query inside a plpgsql function

Em 26 de fevereiro de 2012 12:45, Clodoaldo Neto <
clodoaldo(dot)pinto(dot)neto(at)gmail(dot)com> escreveu:

> When I explain a query using a partitioned table the result is the
> expected. That is, only the corrected partition is scanned. But when the
> query is inside a plpgsql function it takes forever to complete suggesting
> it is scanning all partitions.
>
> create table p (c integer);
> create table p1 (like p);
> alter table p1 add constraint p1c check (c = 1);
> create table p2 (like p);
> alter table p2 add constraint p2c check (c = 2);
> insert into p1 values (1);
> insert into p2 values (2);
> alter table p1 inherit p;
> alter table p2 inherit p;
>
> The explain shows the expected plan and the select is also very fast:
> (obviously the real query and table are more complex)
>
> explain select c from p where c = 1;
>
> A function like this takes very long to complete:
>
> create or replace function pf() returns integer as
> $body$
> declare
> v constant integer := 1;
> begin
> return (select c from p where c = v);
> end
> $body$
> language plpgsql stable
> cost 100;
>
> Isn't the "constant" option to a variable declaration enough to the
> planner? Or else what is the limitation here? Is there some way to see the
> plan for a plpgsql function?
>
>
It seems that the only solution is to make the query dynamic:

create or replace function pf() returns integer as
$body$
declare
v constant integer := 1;
r integer;
begin
execute 'select c from p where c = $1' into r using v;
return r;
end
$body$
language plpgsql stable
cost 100;

Using the dynamic solution the actual function executes very fast.

Clodoaldo