PHPFusion - tai laisvai platinama nemokama turinio valdymo sistema (TVS), kurios pagalba galite greitai, lengvai ir
patogiai susikurti norimą internetinį puslapį. Plačiau apie sistemą...
Sveiki, taigi meta errora, jau kelinta karta tikrinu koda bet nerandu problemos, tai prasausi pagalbos is patyrusiu:
Erroras kuri meta:
Warning:mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\IR\core\functions\users.php on line 23
mano php kodas
<?php function user_exists($username){ $username= sanitize($username); return(mysql_result(mysql_query("SELECT COUNT(`USER_id`) FROM `users` WHERE `username`='$username'"),0)==1) ? true:false;
}
function user_active($username){ $username= sanitize($username); return(mysql_result(mysql_query("SELECT COUNT(`USER_id`) FROM `users` WHERE `username`='$username' AND `active` = 1"),0)==1) ? true:false; } function user_id_from_username($username){ $username= sanitize($username); returnmysql_result(mysql_query("SELECT `USER_id` FROM `users` WHERE `username`='$username'"),0,'USER_id'); } function login($username,$password){ $USER_id= user_id_from_username($username);
return(mysql_result(mysql_query("SELECT COUNT (`USER_id`) FROM `users` WHERE `username`='$username' AND `password`='$password'"),0)==1) ? $USER_id:false; } ?>
kur cia problema ? Dar tureciau klausyma kaip cia i foruma parasyti koda kad rodytu eilutes nr kaireje, butu jum lengviau