Bona:

You're using OCIParse incorrectly. It's used to parse SQL commands. The var
statement that you quoted is a SQL*Plus command. That is, you're defining a
variable in SQL*Plus. That's why you're getting an Invalid SQL Command
error.

Use the OCIBindByName to bind a PHP variable (you're using test) to an
Oracle placeholder, which is denoted by a preceding colon.

e.g., 

$sql = "INSERT INTO emp (ename) VALUES (:ename_placeholder);";
$stmt = OCIParse($conn, $sql);

OCIBindByName($stmt,":ename_placeholder",$empno,32);

$empno = "Bona";

OCIExecute($stmt, OCI_DEFAULT);


Please let me know if this makes sense,


Anthony Carlos


> From: [EMAIL PROTECTED]
> Date: Mon, 10 Sep 2001 11:10:04 -0800
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] creating an Oracle variable...
> 
> Hi,
> 
> I creat the variable "test" using the sql  plus:
> 
> var test varchar2(10);
> 
> Ive been trying to do it wiht PHP:
> 
> ================================================
> $comando1 = "exec var test varchar2(10);";
> 
> $controle = OCIParse ($connection, $comando1);
> if ($controle == false){
> echo OCIError($controle)."<BR>";
> }
> 
> $result1 = OCIExecute ($controle);
> ================================================
> 
> I tryed $comando1 using begin and without the exec... with and without the
> ";" on the end...
> 
> but I allways have the same message: Invalid SQL Command...
> 
> can anybody help me???
> 
> thanks
> 
> Bona
> 
> __________________________
> Visite http://www.trama.com.br
> 
> 
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