Tom Allison wrote:
> This should be a dumb question:
>
> --with-perl
>
> I don't see that I have to do this in order to load pl/perl as a
> function/trigger language option. So I should assume that this will
> compile pl/perl in rather than having it available as a loadable
> function. Nice for optimizations?
If you compile --without-perl you will not be able to use PL/Perl.
There will be no loadable module in that case.
It is not a question of optimization.
> I'm assuming this is the case because there is no option for something
> like PL/Ruby support, but Ruby is available as a loadable function.
Only if you download, compile and install PL/Ruby, which is a separate
module that is not in the PostgreSQL core distribution.
That will give you a loadable module for PL/Ruby.
Compiling and installing PL/Ruby is the equivalent of building
--with-perl (PL/Perl is included in PostgreSQL core).
> And should I also be able to assume that PL/PgSQL is compiled into
> postgresql?
PL/pgSQL is always built, there is no special flag to enable it.
> so I don't really need to use any particular flags, with the probable
> exception of --with-ssl?
Depends.
If you want only PL/pgSQL, you need no extra flags.
If you need PL/Perl, compile --with-perl.
If you want PL/Ruby, you must download, compile and install it.
Creating a procedural language in PostgreSQL involves two steps:
1) Build and install the loadable module (which is what my answers are about).
2) Run a CREATE LANGUAGE statement in the database where you want the
procedural language.
Yours,
Laurenz Albe
