我试图为zend导航创建一个自定义导航,但我有两个问题:
- 如何将变量传递到自定义的部分phtml,或者如果可能的话
- 如何通过整个活动菜单树设置类
这是我迄今为止的代码:
在控制器中:
$config = new Zend_Config($menu);
$nav = new Zend_Navigation();
$nav->addPages($config);
$this->view->nav = $nav;
在视图中:
<?php echo $this->navigation($this->nav)->menu()->setPartial('menu.phtml')->render(); ?>
和我的部分:
<?php
function genMenu($container)
{
foreach ($container as $page)
{
echo '<li>';
$href = $page->uri;
$target = '_self';
echo '<a href="' . $href . '" target="' . $target . '">' . $page->label . '</a>';
if (!empty($page->pages))
{
echo '<ul>';
genMenu($page->pages);
echo '</ul>';
}
echo '</li>';
}
}
echo '<ul>';
genMenu($this->container);
echo '</ul>';
提前感谢大家!
echo $this->navigation($this->nav)->menu()->setPartial('menu.phtml')->render(); ?>
不是很正确,你有正确的想法,但试试
//This will pass a valid container to your partial with the $this->nav
echo $this->navigation()->menu()->renderPartial($this->nav,'menu.phtml') ?>
这是api:
public function renderPartial(Zend_Navigation_Container $container = null,
$partial = null)
这一点看起来也有点不稳定:
$config = new Zend_Config($menu);
$nav = new Zend_Navigation();
$nav->addPages($config);
$this->view->nav = $nav;
我不认为->addPages()是你想要的,我认为你需要的是:
//where $menu is the container and is config(.ini) object not .xml
//for xml use Zend_Config_Xml or Zend_Config_Json for JSON
$config = new Zend_Config($menu);
$nav = new Zend_Navigation($config);
//assign the container to the view
$this->view->nav = $nav;
请参阅此处
如果使用ACL ,请将此行添加到有效ACL
if ($this->navigation()->accept($page))
其结果
...
foreach ( $iterator as $page ) {
//VALID ACL
if ($this->navigation()->accept($page)) {
...
...
}
}
..