как получить $ id в функции addrequest как функцию getid

Question

function getid () {

global $ conn; $ username = $ _ SESSION ['username'];

$ result = $ conn-> prepare ('SELECT * FROM tbl_users WHERE username =?'); $ Result-> bindValue (1, $ имя пользователя); $ result-> execute ();

if ($ result-> rowCount ()> = 1) {$ row = $ result-> fetchAll (PDO :: FETCH_ASSO C); foreach ($ row как $ value) {return $ id = $ value ['id'];

    }


}

}

function addrequest () {

global $conn;
global $id;

if (isset ($ _ POST ['btn_send_request'])) {

$result=$conn->prepare('INSERT INTO tbl_request SET country=?,state=?,city=?,first_name=?,last_name=?,username=?,email=?,address=?,file=?,description=?,user_id=?');
$result->bindValue(1,$_POST['country']);
$result->bindValue(2,$_POST['state']);
$result->bindValue(3,$_POST['city']);
$result->bindValue(4,$_POST['first_name']);
$result->bindValue(5,$_POST['last_name']);
$result->bindValue(6,$_POST['username']);
$result->bindValue(7,$_POST['email']);
$result->bindValue(8,$_POST['address']);

$ file = $ _FILES ['file'];

    $_1 = explode('.', $file['name']);
    $_2 = end($_1);
    if (end($file) < 1000000) {
        if (in_array($_2, ['jpg', 'jpeg', 'pdf', 'dwg','png'])) {
            $name = 'img-' . rand(1, 6000) . '.' . $_2;
            move_uploaded_file($file['tmp_name'], 'panel/admin/cms/storage/' . $name);
            $result->bindValue(9,$name);

        }

    } else {

        header('location:send_request.php');

    }

$ result-> bindValue (10, $ _ POST ['description']);

$result->bindValue(11,$id);
$result->execute();
header('location:send_request.php?addrequest=ok');