Обсуждение: Difficulty with quotes

Hi,

I'm trying to write a very simple SQL function, but I'm having problems
with it. My function is as follows:

CREATE OR REPLACE FUNCTION fnCPUpdateVolunteerType(int, text) RETURNS
INTEGER AS '
    UPDATE vblVolunteerType
       SET vchDisplayName = ''$2''
     WHERE intVolunteerTypeID= $1;

    SELECT 1;
' LANGUAGE SQL;

A very simple update function which gets accepted with no problems.

Then when I run something like this:

SELECT fnCPUpdateVolunteerType(17, 'Test');

The value of vchDisplayName in the applicable row gets set to '$2'.

I know I'm getting something very simple wrong but I can't figure it out
and I can't find anything in the documentation. Any help would be
greatly appreciated.

Russell Brown.