尝试回显值时未定义的索引错误，我正在使用类外的函数

Question

Hey guys I have a question and I still consider myself pretty new at coding, so forgive me if I come off foolish.

I am studying in school as of now and we have a project to build a full stack recreation of craigslist. Any who the problem I am having deals with PHP. I have created an account page with text areas. I would like to echo out the user's information on their so the user can see what he put on and update as he likes. Since my navbar is included on every page, I added the code:

if(isset($_SESSION['logged_in_user'])){
      var_dump($_SESSION['logged_in_user']);
      $user = $_SESSION['logged_in_user'];
      var_dump($user);
 }

on my account page I figured I can echo it out as <?= $attributes['first_name']?> within the placeholders. But I keep getting:

Undefined index: first_name

Also when I var_dump($user) I get an protected $attributes array.

In My Auth class is where I first defined $user as such:

public static function attempt($attemptedUsername, $attemptedPassword) {

$user = User::findByUserName($attemptedUsername);
    if ($user == null) {
        return false;
    }

    $validPassword = password_verify($attemptedPassword,$user->password);
    if ($validPassword == true) {
        $_SESSION['logged_in_user'] = $user;            
    }    
    return false;       
}   

and my findByUserName function is in the user class. the code is:

public static function findByUserName($user_name){

        // Get connection to the database
        self::dbConnect();

        $stmt = self::$dbc->prepare('SELECT * FROM users WHERE user_name = :user_name');

        $stmt->bindValue(':user_name', $user_name , PDO::PARAM_STR);

        //execute gets its own line, t or false
        $stmt->execute();

        $result=$stmt->fetch(PDO::FETCH_ASSOC);

        // @TODO: Create select statement using prepared statements

        // @TODO: Store the result in a variable named $result

        // The following code will set the attributes on the calling object based on the result variable's contents
        $instance = null;
        if ($result) {
            $instance = new static($result);
        }
        return $instance;
    }

Answer 1

Your problem seems to be with not being able to access the variable $user outside of the static method attempt() this can be fixed by declaring the variable globally at the beginning of the method attempt() like this:

public static function attempt($attemptedUsername, $attemptedPassword) {

global $user;
$user = User::findByUserName($attemptedUsername);
    if ($user == null) {
        return false;
    }

    $validPassword = password_verify($attemptedPassword,$user->password);
    if ($validPassword == true) {
        $_SESSION['logged_in_user'] = $user;

    }

    return false;


}

More information can be found on this in the PHP documentation here.