我想在使用ZF 1.X
的网站构建中实现下拉菜单在我的应用程序/ layouts / layout.phtml中,我接到了电话
$this->navigation()->menu()->renderMenu(
$this->navigation()->findByLabel('My Account'),
array('maxDepth' => 0)
);
在myapp / Controller / Plugin / Navigation.php中,我看到标有“我的帐户”的菜单:
$account_container = new Zend_Navigation_Page_Mvc(
array(
'route' => 'account_index',
'label' => 'My Account',
'pages' => array(
new Zend_Navigation_Page_Mvc(
array(
'route' => 'productions_list',
'label' => 'My Productions',
)
),
new Zend_Navigation_Page_Mvc(
array(
'route' => 'productions_create',
'label' => 'Create a Production',
)
),
new Zend_Navigation_Page_Mvc(
array(
'route' => 'account_inbox',
'label' => sprintf('My Inbox (%s)', $ident->getAllUnreadMessagesCount()),
'id' => 'inbox-count'
)
),
new Zend_Navigation_Page_Mvc(
array(
'route' => 'search_search_productions',
'label' => 'Search Productions'
)
),
new Zend_Navigation_Page_Mvc(
array(
'route' => 'search_search_users',
'label' => 'Search Users'
)
),
)
)
);
并生成
<ul class="navigation">
<li class="active"><a href="/news">Blog</a></li>
...
...
</ul>
我需要使用UL添加嵌套级别,如下所示:
<ul class="navigation">
<li class="active">
<a href="/news">Blog</a>
<ul>
<li><a href="/somethingelse">link</a></li>
...
...
</ul>
</li>
...
...
</ul>
这样做的方法是什么?我需要这个来生成javascript中的下拉菜单
答案 0 :(得分:0)
我找到了解决方案, 'maxDepth'=&gt; 1还不够,
需要它在子数组中实现新页面:
$account_container = new Zend_Navigation_Page_Mvc(
array(
'route' => 'account_index',
'label' => 'My Account',
'pages' => array(
new Zend_Navigation_Page_Mvc(
array(
'route' => 'productions_list',
'label' => 'My Productions',
'pages' => array(
new Zend_Navigation_Page_Mvc(
array(
'route' => 'mychildurl',
'label' => 'My Child Label',
)
),
)
),
答案 1 :(得分:-1)
我想你只需要设置:
'maxDepth' => 1
而不是0。