在 View 中使用 Zend_Acl 来显示/隐藏部分视图的方法是什么

发布于 2024-10-10 15:08:35 字数 418 浏览 2 评论 0原文

我想知道使用 Zend_Acl 来显示/隐藏部分视图的方法是什么？我想我会

创建一个控制器插件来传递登录用户+ acl 来查看

 $this->view->loggedInUser = Zend_Auth::getIdentity();
 $this->view->acl = Zend_Registry::get('acl');

然后在视图脚本中执行类似的操作

$this->acl->isAllowed($this->view->loggedInUser, '资源', '权限');

或者有更好的方法吗？或者我应该使用视图助手？返回一个布尔值是否允许登录用户？

原文

I am wondering whats the way to use Zend_Acl to show/hide parts of view? I am thinking I will

Create a Controller Plugin that passes the logged in user + acl to view

 $this->view->loggedInUser = Zend_Auth::getIdentity();
 $this->view->acl = Zend_Registry::get('acl');

Then in view scripts do something like

$this->acl->isAllowed($this->view->loggedInUser, 'resource', 'privilege');

Or is there a better way? Or should I use a View Helper? That returns a boolean whether the logged in user is allowed?

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哭了丶谁疼 2024-10-17 15:08:35

您在视图中使用它，所以对我来说 ViewHelper 是正确的位置 - 我已经这样做过一次：

class Zend_View_Helper_HasAccess extends Zend_View_Helper_Abstract
{
    private $_acl;
    public function hasAccess($role, $controller, $action)
    {
        if (!$this->_acl) {
            $this->_acl = Zend_Controller_Front::getInstance()->getPlugin('Acl'); 
            //In yout case registry, but front controller plugin is better way to implement ACL
        }
        return $this->_acl->isAllowed($role, $controller, $action);
    }
}

You are using it in the view, so for me ViewHelper is correct place for that - I've done it once that way:

class Zend_View_Helper_HasAccess extends Zend_View_Helper_Abstract
{
    private $_acl;
    public function hasAccess($role, $controller, $action)
    {
        if (!$this->_acl) {
            $this->_acl = Zend_Controller_Front::getInstance()->getPlugin('Acl'); 
            //In yout case registry, but front controller plugin is better way to implement ACL
        }
        return $this->_acl->isAllowed($role, $controller, $action);
    }
}

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