如何在另一个类中创建一个类的实例

我不确定这里可能漏了什么。谷歌搜索也没有真正的帮助。

我想做的是调用databaseServer类并在userControl类中使用它的方法。这是我的图书馆_类.php文件:

<?php

include('definitions.php');

class databaseServer {

    var $con;
    var $db;
    var $close;
    var $qry;
    var $sql;

    function connect($host,$user,$pw,$db) {
        $this->con = mysql_connect($host,$user,$pw);
        if (!$this->con) {
            die('Could not connect: ' . mysql_error());
            }
        else {
            echo "Database Connected";
            }
        $this->selectDb($db);
        }

    function selectDb($database) {
        $this->db = mysql_select_db($database,$this->con);
        if (!$this->db) {
            echo "Could not Select database";
            }
        else {
            echo "Database Selected";
            }
        }

    function disconnect() {
        $this->close = mysql_close($this->con);
        if ($this->close) {
            echo "Disconnected";
            }
        }

    function query($test) {
        if (!mysql_query($test)) {
            die("Error: " . mysql_error());
            }
        }

} // databaseServer

class cookie {

    var $expireTime;

    function set($name,$value,$expiry) {
        $this->expireTime = time()+60*60*24*$expiry;
        setcookie($name,$value,$expireTime);
        }

    function delete($name) {
        setcookie($name,"",time()-3600);
        }

    function check($name) {
        if (isset($_COOKIE["$name"]))
            echo "Cookie Set";
        else
            echo "Cookie failed";
        }

} //cookie

class userControl {

    public function __construct(databaseServer $server) {
        $this->server = new databaseServer();
    }

    function createUser($uname,$pword) {

        $this->server->connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
        $result = $this->server->query("SELECT * FROM user_list WHERE uname='" . $this->server->real_escape_string($uname) . "'");
        if ($this->result->num_rows() === 0) {

            if ($this->server->query("INSERT INTO user_list (uname, pword) 
            VALUES ('" . $this->server->real_escape_string($uname) . "','" . $this->server->real_escape_string($pword) . "')") {
                echo "User Added Successfully!";
            }
            else {
                echo "Error Adding User!";
            }
        }

        else {
            echo "User Already Exists!";
        }

    } // createUser

} // userControl

?>

然而,这不起作用,我不明白为什么。当我从文件中省略userControl类时,我的databaseServer和cookie类工作正常,所以我知道错误一定在该类的某个地方。哎呀,我一直在努力学习,但我一直在磕磕绊绊。

<?php

include('definitions.php');
include('class_lib.php');

$bmazed = new databaseServer();

$bmazed->connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);

$sql = "INSERT INTO blah
VALUES ('testing 92')";

$bmazed->query($sql);

$bmazed->disconnect();

// $control = new userControl();

// $uname = "Test1";
// $pword = "test1";

// $control->createUser($uname,$pword);

echo "<br />";
echo "<br />";

?>

为了测试的目的,行被注释掉了,所以我不必继续重新编写代码。

我真的不知道问题出在哪里,我检查了语法,一切似乎都很好。