问题描述

我在一个地方渲染 Zend Navigation 对象的顶级元素，如下所示:

I am rendering the top-level elements of a Zend Navigation object in one place like this:

echo $this->navigation()->menu()->setMaxDepth(0);

如何为活动分支呈现从第二级向下的导航树?我已经尝试创建一个循环 $this->container 对象的部分，但我不知道如何确定我的当前项目是否是活动分支.一旦我确定它是活动分支，我该如何呈现菜单?我这样做是不是很困难并且遗漏了一些明显的东西?

How do I render the navigation tree from the second level on down for the active branch? I've tried creating a partial that loops the $this->container object, but I don't know how to determine if my current item is the active branch. Once I've determined that it's the active branch how do I render the menu? Am I doing this the hard way and missing something obvious?

谢谢！

更新:

我接受了一个解决方案，因为这是我使用的解决方案，但我也想提供我的实际问题的答案，以供参考.($this 是视图对象)

I accepted a solution because that's what I used, but I also would like to provide the answer to my actual question, for reference sake. ($this is the view object)

// Find the active branch, at a depth of one
$branch = $this->navigation()->findActive($this->nav, 1, 1);
if (0 == count($branch)) {
    // no active branch, find the default branch
    $pages = $this->nav->findById('default-branch')->getPages();
} else {
    $pages = $branch['page']->getPages();
}
$this->subNav = new Zend_Navigation($pages);

$this->subNav 然后可用于呈现子菜单.

$this->subNav can then be used to render the sub-menu.

推荐答案

我做了类似的事情.我的主要导航是用这样的方式处理的...

I do something similar. My main navigation is handled with something like this...

$this->navigation()->menu()->setPartial('tabs.phtml');
echo $this->navigation()->menu()->render();

然后在我的 tabs.phtml 中，我像这样遍历容器...

Then in my tabs.phtml I iterate over the container like so...

if (count($this->container)) {
  foreach($this->container as $page) {
    if ($page->isVisible()) {
      if ($page->isActive(true)) {
        $subcontainer = $page->getPages();
        foreach($subcontainer as $subpage) {
          // echo my link
        }
      }
    }
  }
}

希望对你有所帮助.

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