Bootstrapping Zend布局和包含导航资源

Question

当我使用以下引导程序时，我在呈现导航菜单时遇到了问题：

public function _initViewHelpers()
{
    $this->bootstrap('layout');
    $layout = $this->getResource('layout');
    $view = $layout->getView(); // Never inits navigation resource?

    $view->headTitle()->setSeparator(' - ')
         ->headTitle('Test');

    $role = ($this->_auth->getStorage()->read() === null) ? 'guest' : $this->_auth->getStorage()->read()->role;
    $view->navigation()->setAcl($this->_acl)->setRole($role); 
}

在我的layout.phtml中，我有：

echo $this->navigation()->menu();

在我的application.ini中，我有：

resources.navigation.pages.index.label = "Home"
resources.navigation.pages.index.title = "Go Home"
resources.navigation.pages.index.controller = "index"
resources.navigation.pages.index.action = "index"
resources.navigation.pages.index.order = -100
resources.navigation.pages.index.route = "default"

发出$view = $layout->getView();会导致我的导航菜单不会被呈现。如果我注释掉这一部分，它呈现得很好。

如何在引导程序中设置标题和acl角色，并仍然正确呈现菜单？

Answer 1

您是否尝试过将视图资源添加到application.ini并直接检索资源？

application.ini

resources.view[] =

引导：

public function _initViewHelpers()
{
    $this->bootstrap('layout');
    $this->bootstrap('view');
    $this->bootstrap('navigation');
    $layout = $this->getResource('layout');
    $view = $this->getResource('view');
    ....