为什么包含的脚本中定义的变量没有值？

Question

I have this code in script.php:

<?php
session_start();

$user_session = $_SESSION['u_name'];

class check_sess {
    public function check_id_session($user_session, $db) { //email
        $stmt = $db->prepare('select id_user from users where email=?');
        $stmt->bind_param('s', $user_session);

        $stmt->execute();

        $stmt->bind_result($id);

        if ($stmt -> fetch()) {
            $id;
            echo $id; // here shows 20 for example
            return true;
        }
        else
        return false;

        $stmt->close();
        $db->close();
    }
}
?>

and in demo.php I have:

include 'script.php';
$val = new check_sess();
$val-> check_id_session($user_session, $db);
echo $id; //problem here. It is supposed echo 20

Why can't I do echo $id? There is no output in demo.php.

Answer 1

Read about variable scope. The variable $id does not exist outside the function. If you want the function to return $id, write return $id in the if ($stmt -> fetch()) { condition branch. You can then write:

include 'script.php';
$val = new check_sess();
$id = $val-> check_id_session($user_session, $db);
echo $id;