Re: Defining Parameters for PHP Calling UDF

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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