y need some help pls
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\xxxxxxxxx\index.php on line 7
No Template Selected


require("dbconfig.php");
$con =mysql_connect ($db_host,$db_user,$db_pass) or die("Error connecting");
mysql_select_db($db_name,$con);
$tem="select dir from template where active='1'";
$temr=mysql_query($tem,$con);
$temn=mysql_num_rows($temr); line 7
if($temn==0)

when im test in my local machine i dont have any error but when im trying to test in web i have this error, the same DB, the same tables, the same data in both structures any idea????

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site244/fst/var/www/html/categoria.php on line 107

PHP Syntax (Toggle Plain Text)

<?php
    require('conexion2.php');
    $SQLEmpresa= "SELECT empresa.cve_Empresa, Nombre, Direccion, pagina, telefono 
	             FROM empresa, empresa_Categoria  
		WHERE ".$_GET['cve_Categoria']." = empresa_Categoria.cve_Categoria 
		 AND empresa_Categoria.cve_Empresa = empresa.cve_Empresa";
    $rsEmpresa = mysql_query($SQLEmpresa,$conexion);
	$reg = array();
-->line 107	while ($reg = mysql_fetch_array($rsEmpresa))
	   {		
			$nombre =  $reg["Nombre"]; 
			$direccion =  $reg["Direccion"]; 
			$telefono =  $reg["telefono"]; 						
			$pagina = $reg["pagina"];
			$cve_Empresa = $reg["cve_Empresa"];	   					
?>

<?php require('conexion2.php'); $SQLEmpresa= "SELECT empresa.cve_Empresa, Nombre, Direccion, pagina, telefono FROM empresa, empresa_Categoria WHERE ".$_GET['cve_Categoria']." = empresa_Categoria.cve_Categoria AND empresa_Categoria.cve_Empresa = empresa.cve_Empresa"; $rsEmpresa = mysql_query($SQLEmpresa,$conexion); $reg = array(); -->line 107 while ($reg = mysql_fetch_array($rsEmpresa)) { $nombre = $reg["Nombre"]; $direccion = $reg["Direccion"]; $telefono = $reg["telefono"]; $pagina = $reg["pagina"]; $cve_Empresa = $reg["cve_Empresa"]; ?>

please help me...even i m getting this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\admin_loggedin_search_admin.php on line 35

PHP Syntax (Toggle Plain Text)

<?php
$con = mysql_connect("localhost","root","");
if(!$con)
  { 
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("polling",$con);
$result = mysql_query("SELECT * FROM administrator WHERE $_POST[searchby] like '%$_POST[query]%'");
$found=0;
[LINE 35]while($sri=mysql_fetch_array($result))
{
 if($sri['name']!="")
 {
 $found++;
 echo "&nbsp;&nbsp;&nbsp;&nbsp; ".$sri['name']."->".$sri['username']."<br />";
 }
}
if($found!=0)
{echo "Found ".$found." results!";}
mysql_close($con)
?>

<?php $con = mysql_connect("localhost","root",""); if(!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("polling",$con); $result = mysql_query("SELECT * FROM administrator WHERE $_POST[searchby] like '%$_POST[query]%'"); $found=0; [LINE 35]while($sri=mysql_fetch_array($result)) { if($sri['name']!="") { $found++; echo "&nbsp;&nbsp;&nbsp;&nbsp; ".$sri['name']."->".$sri['username']."<br />"; } } if($found!=0) {echo "Found ".$found." results!";} mysql_close($con) ?>

@venkat
doesnt help...now i am getting an parse error in the line select * .......

are you sure of the syntax in the sql query??
one more thing....$_post[query] ... returns the string entered in a search box named query.....