PHP解析错误：语法错误，意外'，'在

Question

嗯，基本上我正在制作 youtube 上的注册和登录教程。这是使用旧版本的PHP，我试图更新代码，但是我得到了这个错误：

解析错误：语法错误，第 23 行 C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\projects\Forum\forum\core\functions\users.php 中的意外 ','

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;
}
?>

Answer 1

线23是这个：return mysqli_affected_rows($con), 0, 'user_id';

必须是：return mysqli_affected_rows($con) ? 0 : 'user_id';，如果这是您的意思。

无法在 PHP 中返回多个值。