PHP分析错误：语法错误，意外''；在里面

基本上，我正在youtube上做注册和登录教程。它使用的是旧版本的PHP，我试图更新代码，但出现以下错误：

分析错误：语法错误，在第23行的C:\Program Files（x86）\EasyPHP-DevServer-14.1VC11\data\localweb\projects\Forum\Forum\core\functions\users.php中出现意外“，”

users.php

<?php
function user_exists($username, $con) {
    $data = $username;
    $username = sanitize($data, $con); 
    $username = $data;
    mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");
    return(mysqli_affected_rows($con) == 1) ? true : false;
}

function user_active($username, $con) {
    $data = $username;
    $username = sanitize($data, $con); 
    $username = $data;
    mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username' AND `active` = 1");
    return(mysqli_affected_rows($con) == 1) ? true : false;
}

function user_id_from_username ($username, $con) {
    $data = $username;
    $username = sanitize($data, $con); 
    $username = $data;
    mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username'");
    return mysqli_affected_rows($con), 0, 'user_id';
}

function login($username, $password, $con) {
    $user_id = user_id_from_username($username, $con);
    $data = $username;
    $username = sanitize($data, $con); 
    $username = $data;
    $password = md5($password); 
    return (mysqli_affected_rows(mysqli_query($con, "SELECT `user_id` FROM `users` WHERE `username` = '$username' AND `password` = '$password'"), 0) == 1) ? $user_id : false;
}
?>

---

行
23
是这样的：
返回mysqli\u受影响的行（$con），0，'user\u id'

必须是：
返回mysqli\u受影响的行（$con）？0:'user_id'如果这就是你的意思

无法在PHP中返回多个值。
返回mysqli\u受影响的行（$con），0，'user\u id'Huuu？那应该是什么？