ช่วยดู Error Warning: mysql_result(): supplied argument is not a valid MySQL result resource

<?php
function logged_in(){
  return isset($_SESSION['user_id']);  
}

function login_check($email, $password){
    $email = mysql_real_escape_string($email);
    $login_query = mysql_query("SELECT COUNT(user_id) as count, user_id FROM users WHERE email = '$email' AND password = '".md5($password)."'");
    return (mysql_result($login_query, 0) == 1) ? mysql_result($login_query, 0, 'user_id') : false;
}

function user_data(){
    $args = func_num_args();
    $fields = '`'.implode('`, `', $args).'`';
    
    $query = mysql_query("SELECT $fields FROM `users` WHERE `user_id` = ".$_SESSION['user_id']);
    $query_result = mysql_fetch_assoc($query);
    foreach ($args as $fields){
        $args[$fields] = $query_result[$fields];
    }
    return $args;
}

function user_signup($email, $name, $password){
    $email = mysql_real_escape_string($email);
    $name = mysql_real_escape_string($name);
    mysql_query("INSERT INTO users VALUES ('', '$email', '$name', '".md5($password)."')");
    return mysql_insert_id();
    
}

function user_exists($email){
    $email = mysql_real_escape_string($email);
    $query = mysql_query("SELECT COUNT(user_id) FROM users WHERE email = '$email'");
    return (mysql_result($query, 0) ==1) ? true : false;
    
}

?>