php $this->view->$someVar 有可能吗？

Question

我正在创建一个 Model View Controller 框架，我对此很陌生。我想将变量从 Controller 传递到 View 。我的 View.php 构造函数如下所示:

    function __construct($file, $args) {
        $this->view = $file;
         foreach($args as $key => $arg) {
            $this->view->$key = 'awda';
         }
    }

它给了我错误原因

 $this->view->$key is not a valid statement. 

如果我从 Controller 做类似的事情

 $this->view->hello = 'hello world' 

我做 echo

 $this->hello 

鉴于它工作正常，但我希望能够传入多个变量。有没有人知道这样做的更好方法？谢谢

Answer 1

您正在尝试将属性分配给我怀疑是字符串 ( $file )。由于您在 View 的构造函数中，您可以简单地使用 $this引用观点:

function __construct($file, $args) {
    $this->view = $file;
     foreach($args as $key => $arg) {
        $this->view->$key = 'awda'; // HERE is the issue.. isn't $this->view a string?
     }
}


function __construct($file, $args) {
    $this->view = $file;
     foreach($args as $key => $arg) {
        $this->$key = 'awda'; // assign $key as property of $this instead...
     }
}