PHP类未定义？明确规定

Question

我正在使用包含php类并初始化它。这真的是非常直接和容易......这就是为什么这是如此令人沮丧。

无论如何我得到一个Acmedb是未定义的错误，这是我的代码

<?php
    include("acmedb.php");

    $database = new Acmedb();

    $database->connect();
?>

这里是acmedb.php

<?php

class Acmedb {
    private $connectstr_dbhost = '';
    private $connectstr_dbname = '';
    private $connectstr_dbusername = '';
    private $connectstr_dbpassword = '';

    public function __construct(){

    }

    public function connect(){
        foreach ($_SERVER as $key => $value)
        {
            if (strpos($key, "MYSQLCONNSTR_") !== 0)
            {
                continue;
            }

            $connectstr_dbhost = preg_replace("/^.*Data Source=(.+?);.*$/", "\\1", $value);
            $connectstr_dbname = preg_replace("/^.*Database=(.+?);.*$/", "\\1", $value);
            $connectstr_dbusername = preg_replace("/^.*User Id=(.+?);.*$/", "\\1", $value);
            $connectstr_dbpassword = preg_replace("/^.*Password=(.+?)$/", "\\1", $value);

        }
    }
    public function create_user($username, $pass){
        global $connectstr_dbhost, $connectstr_dbname, $connectstr_dbpassword, $connectstr_dbusername;


        if($link=mysqli_connect($connectstr_dbhost, $connectstr_dbusername, $connectstr_dbpassword,$connectstr_dbname)) {
            $sql = "INSERT INTO `users` (`id`,`username`,`password`) VALUES (1, $username, $pass)";
            if (mysqli_query($link, $sql)) {
                if (($aff_rows = mysqli_affected_rows($link)) > 0) {
                    echo "New record created successfully ($aff_rows)";
                } else {
                    echo "Query Logic Error: $sql";
                }
            } else {
                echo "Syntax Error:: ", mysqli_error($link);
            }
            mysqli_close($link);
        }
        else{
            echo "Connection Error: ", mysqli_connect_error();
        }
    }

}
?>

任何帮助都会很大。我确信这是一件非常简单的事情，我不知道。