Re: Defining Parameters for PHP Calling UDF

I'm trying to use the return value from the UDF in the select statement. My
though was the "?" parameter marker was for the input variable since the
output variable is the return value.


On Wed, Feb 6, 2013 at 1:18 PM, Jon Paris <jon.paris@xxxxxxxxxxxxxx> wrote:

On Wed, 6 Feb 2013, at 12:42 PM, Michael Ryan <michaelrtr@xxxxxxxxx>
wrote:

When I try and use it like this:

$SQLStmt = "SELECT * from cusmstp where NEWSYS.MSP_ValidateStoreFN(cast

(?

as char (3) ) ) = '666'";
$prepare = db2_prepare($db2connection, $SQLStmt);
//db2_bind_param($prepare, 2, "$Wk_Error", DB2_PARAM_OUT);
db2_bind_param($prepare, 2, "$Wk_Store", DB2_PARAM_IN);
db2_bind_param($prepare, 1, "$Wk_Error", DB2_PARAM_OUT);
$stmt = db2_execute($prepare) or die("Failed SQL
Stmt:".$SQLStmt.":".db2_stmt_error().":".db2_stmt_errormsg());

Just noticed that you are binding two parms but I only see one "?" in the
SQL statement. Surely the return value will be in the result set - not a
parm? Haven't actually tried to do this myself but it just looks wrong.


Jon Paris

www.partner400.com
www.SystemiDeveloper.com




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