salve ragazzi , ho fatto una funzione per cercare gli user , adesso mi restituisce constantemente un errore warning questa e la funzione
Codice PHP:
function ShearchStaff($staffsearch){
global $dbcore;
$sql = "select t.* , gs.id_groups ,gs.nome_group from tech as t left join grouptech as gs on t.groupid = gs.id_groups where t nome LIKE '%".addslashes($staffsearch)."%' or t cognome LIKE '%".addslashes($staffsearch)."%' or t username or t email LIKE '%".addslashes($staffsearch)."%' LIKE '%".addslashes($staffsearch)."%' order by t.staffid asc ";
$query = mysql_query($sql);
$result = mysql_num_rows($query);
if($result == 0){ return false;}
$result = array();
while ($row = mysql_fetch_assoc($query)){
$result[] = $row;
}
return $result;
}
questo e l errore
codice:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Programmi\Apache Software Foundation\Apache2.2\htdocs\supp\includes\function_staff.php on line 30