How to fix this error:Notice: Undefined index: antari in C:\Users\Personal PC\Desktop\xamp\htdocs\phpushtrime\SMP PROJEKTI\functions.php on line 65

Question

I'm new at PHP and I'm having trouble with an error in a project, I am trying to get data from mysql database, I used the same method for another page and it worked, I don't know why it isn't working, this is the error:

Notice: Undefined index: antari in C:\Users\Personal PC\Desktop\xamp\htdocs\phpushtrime\SMP PROJEKTI\SMP_ver4-perfundimi\inc\functions.php on line 65

Notice: Trying to access array offset on value of type null in C:\Users\Personal PC\Desktop\xamp\htdocs\phpushtrime\SMP PROJEKTI\SMP_ver4-perfundimi\inc\functions.php on line 65

Notice: Undefined index: antari in C:\Users\Personal PC\Desktop\xamp\htdocs\phpushtrime\SMP PROJEKTI\SMP_ver4-perfundimi\inc\functions.php on line 68

Notice: Trying to access array offset on value of type null in C:\Users\Personal PC\Desktop\xamp\htdocs\phpushtrime\SMP PROJEKTI\SMP_ver4-perfundimi\inc\functions.php on line 68

I've tried everything I know to fix it, but the same results.

here is an image of the function

here is an image of the code

THE CODE FOR THE PAGE:

<?php
        $punet = merrPunet();
        if ($punet) {
            while ($puna = mysqli_fetch_assoc($punet)) {
                $pid = $puna['punaid'];
                echo "<tr>";
                echo "<td>" . $puna['emri'] . "</td>";
                echo "<td>" . $puna['data'] . "</td>";
                echo "<td>" . $puna['pershkrimi'] . "</td>";
                echo "<td><a href='modifikopune.php?pid=$pid'>Edit</a></td>";
                echo "<td><a href='fshijpune.php?pid=$pid'>Delete</a></td>";
                echo "</tr>";
            }
        }
        ?>

THE FUNCTION I USED TO GET DATA FROM DATABASE:

function merrPunet() {
global $dbconn;
$antariid = ($_SESSION['antari']['antariid']);
$sql = "SELECT p.punaid,p.data,p.pershkrimi,pr.emri";
$sql .= " FROM punet p INNER JOIN projektet pr ON p.projektiid=pr.projektiid";
if ($_SESSION['antari']['roli'] == 0) {
    $sql .= " WHERE antariid=$antariid";
}
return mysqli_query($dbconn, $sql);

}

Thank You in Advance!