PHP 实例化后类成员的闭包产生未定义的方法
我试图将 ACL 破解到模板中，而不使模板知道类定义中的 ACL 对象。下面的代码生成了一个未定义的方法 Template::isAllowed
这是为什么？蒂亚！
class ACL {
protected $allowed = array('anything');
public function isAllowed($what){
if(in_array($what, $this->allowed))
return true;
return false;
}
}
class Template extends stdClass { }
$Template = new Template;
$ACL = new ACL;
$Template->isAllowed = function($what) use($ACL) { return $ACL->isAllowed($what); };
if($Template->isAllowed('anything'))
echo 1;
else
echo 2;I am trying to hack an ACL into a Template without making the Template aware of the ACL object in the class definition. The following code generates an undefined method Template::isAllowed
Why is this? TIA!
class ACL {
protected $allowed = array('anything');
public function isAllowed($what){
if(in_array($what, $this->allowed))
return true;
return false;
}
}
class Template extends stdClass { }
$Template = new Template;
$ACL = new ACL;
$Template->isAllowed = function($what) use($ACL) { return $ACL->isAllowed($what); };
if($Template->isAllowed('anything'))
echo 1;
else
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这：
实际上告诉 PHP 调用一个方法
Template::isAllowed()，该方法显然不存在，如您的致命错误所示。您不能通过将闭包分配给属性来将
Template::isAllowed()视为真正的方法。但是，您仍然可以调用分配给$Template->isAllowed属性的闭包（这是闭包）。为此，您需要将该属性分配给变量，然后调用该变量：或者使用 call_user_func()：
This:
actually tells PHP to call a method
Template::isAllowed(), which obviously doesn't exist as given by your fatal error.You cannot treat
Template::isAllowed()as if it were a real method by assigning a closure to a property. However you can still call the closure that is assigned to the$Template->isAllowedproperty (which is an instance ofClosure). To do that, you need to either assign the property to a variable then call that:Or use
call_user_func():简单来说，这是行不通的——您不能在 PHP 中动态添加类方法，就这样。这在将闭包定义为类中的方法之类的地方进行了讨论。
Simply this does not work -- you can not add class methods dynamically in PHP, period. This was discussed at places like define a closure as method from class.