The following function returns Resource id #11
. How can I fix this with my present code?
function GET_PAGE() {
global $page;
if(isset($_GET)) {
foreach($_GET as $key => $value){
$query = "SELECT * FROM pages WHERE title = '{$key}'";
$page = mysql_query($query);
confirm_query($page);
return $page;
}
}
}
You did not fetch any row.
mysql_query
returns an resource that has to be interpreted by one of the mysql_fetch_*
-functions. Depending on your preferences mysql_fetch_array
and mysql_fetch_object
would be most appropriate.
btw: The MySQL-extension is deprecated. Consider switching to PDO or MySQLi:
This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. (http://www.php.net/manual/en/function.mysql-query.php)
Both allow you to use Prepared Statements to properly secure your database statements. Your posted example is prone to SQL injections!!!