如何在CakePhp的当前页面中添加活动类

时间:2013-06-18 06:31:17

标签: php css cakephp cakephp-2.0 cakephp-2.1

我遇到类似这个问题的问题

How to identify active menu link in CakePHP

我的default.ctp文件中有一个页面,我想在链接上添加“活动”类。我如何识别页面的当前网址然后在链接上应用该类..我已经按照答案那里

      $url = $this->Html->url('INPUT_THE_URL') ;
     $active = $this->request->here == $url? true: false;

我不知道我怎么能在我的代码中这样做...很抱歉,因为我是cakephp的新手..这是我的代码

 **default.ctp file** 

 <li>
      <?php echo $this->Html->link('Dashboard', array('controller'=>'users','action' => 'controlpanel'), array('title' => 'Dashboard','class' => 'shortcut-dashboard'));?></li>



  <li> <?php echo $this->Html->link('Contacts', array('controller'=>'contacts','action' => 'index'), array('title' => 'Contacts','class' => 'shortcut-contacts'));?></li>

我想添加一个类似li的课程

   <li class = 'active''>

3 个答案:

答案 0 :(得分:12)

这是一个简单的逻辑如下

<li class="<?php echo (!empty($this->params['action']) && ($this->params['action']=='controlpanel') )?'active' :'inactive' ?>">
  <?php echo $this->Html->link('Dashboard', array('controller'=>'users','action' => 'controlpanel'), array('title' => 'Dashboard','class' => 'shortcut-dashboard'));?>
</li>

<li class="<?php echo (!empty($this->params['action']) && ($this->params['action']=='index') )?'active' :'inactive' ?>">
  <?php echo $this->Html->link('Contacts', array('controller'=>'contacts','action' => 'index'), array('title' => 'Contacts','class' => 'shortcut-contacts'));?></li>

答案 1 :(得分:7)

如果你有一个不同的控制器并且你声明了一个具有相同名称的方法,并且上面的代码不起作用,那么你可以执行以下操作:

<li class="<?php echo (($this->params['controller']==='hotels')&& ($this->params['action']=='view') )?'active' :'' ?>" >
   <?php echo $this->Html->link('Hotels', array('controller' => 'hotels', 'action' => 'view')); ?>
</li>

<li class="<?php echo (($this->params['controller']==='packages')&& ($this->params['action']=='view') )?'active' :'' ?>" >
   <?php echo $this->Html->link('Packages', array('controller' => 'packages', 'action' => 'view')); ?>
</li>

此处 view 方法在不同的控制器中声明。我希望它对你有所帮助。

答案 2 :(得分:1)

不要复活死去的帖子,但这就是我所做的(我相信它更清洁,更快,更易于管理)

我创建了一个包含页面数组的元素,然后检查数组中的每个项目以查看它是否是当前页面。如果是,我添加活动类。

然后我可以从任何地方调用此元素。

// Changed the line below to a multi-dimensional array to cater for different controllers and actions

//$mypages = array('Home','About','Pricing','FAQs','Contact');
$mypages = array(
 array('controller'=>'controller1','action'=>'action1','name'=>'name1'),
 array('controller'=>'controller2','action'=>'action2','name'=>'name2
')
);
foreach ($mypages as $page ){
// Changed to account for controller and action
//$currentPage = isset($this->params['pass'][0]) ?$this->params['pass'][0] : "";
$controller = isset($this->request->params['controller'])?$this->request->params['controller']: "";
$action= isset($this->request->params['action'])?$this->request->params['action']: "";

    if (strtolower($page['controller']) == $controller && strtolower($page['action']) == $action) {  
        echo "<li class='active'>" . $this->Html->link($page,array("controller"=>"pages", "action"=>strtolower($page))) . "</li>" ;  
    } 
    else  {
        echo "<li>" . $this->Html->link($page,array("controller"=>"pages", "action"=>strtolower($page)))  . "</li>"; 
    }
}