PHP将值传递给一个类

时间:2012-05-26 17:22:12

标签: php class constructor

我有一个名为DatabaseController的类,我将其设置为:

<?php

class DBController {

    public $dbName;

    function _controller($passedDBName) { 
        $dbName = $passedDBName;
    }

    function dbConnect() { 
        print $this->$dbName;
    }



} //end of class       
?>

我这样称呼它:

<?php

    //make new database connection
    $dbManager = new DBController("somevalue");
    $dbManager->dbConnect();


?>

但我一直收到这个错误:

<b>Fatal error</b>:  Cannot access empty property in 

我做错了什么?

谢谢

2 个答案:

答案 0 :(得分:3)

在构造函数中,$dbName = $passedDBName;应为$this->dbName = $passedDBName;

<强>更新

  1. $this->$dbName应为$this->dbName
  2. _controller()应为__construct()

答案 1 :(得分:3)

使用$this->dbName - 否则您尝试访问名称存储在$dbName中的字段。您还需要修复构造以分配给$this->dbName而不是$dbName

class DBController {
    public $dbName;
    function _controller($passedDBName) { 
        $this->dbName = $passedDBName;
    }
    function dbConnect() { 
        print $this->dbName;
    }
}