Topic: new on Xajax
I have differents tutorial some referenced to 0.2 and the one here referenced to 0.5 but none had work on my page. I need some help, I have a form to insert new employees on a database, I need to validate if current emp code is alreade on database, so I did following function
<?
require('xajax_core/xajax.inc.php');
$objAjax = new xajax();
$objAjax->register(XAJAX_FUNCTION,'checkEmpCod');
function checkEmpCod($cod)
{
$cod=str_pad($cod,8,"0",STR_PAD_LEFT);
include('includes/conecta.php'); //connect to database
$query = "SELECT codemp from empleados WHERE codemp = '$cod'";
$result = mysql_query($query, $connection) or die ("Error in query: $query. " . mysql_error());
if (mysql_num_rows($result) == 1)
{
$strerror="Empleado Ya Existe";
}
else
{ $strerror="";
}
$objResponse = new xajaxResponse();
$objResponse->assign("chkcod", "innerHTML", $strerror);
return $objResponse;
}
$objAjax->processRequest();
printJavaScript();
?>
here is my html call
<td>codigo:</td>
<td><input type="text" name="codemp" id="codemp" style="empleado" onchange="$objAjax_checkEmpCod()" /></td>
<td><p id="chkcod"></p></td>
any idea what is my problem? please help me.