Php 突出显示来自阵列的活动链接

我想应用

class=“active”

，具体取决于您所处的页面。最好的方法是什么

例如：

在构造函数中，它创建一个包含链接名称和地址的数组

class OrderController extends Controller {

   public $data = array();

   private __construct() {
        $this->data['sidebar'] = array(
            'New Orders' => '/admin/order',
            'Processing Orders' => '/admin/order/processing',
            'Completed Orders' => '/admin/order/completed',
    );


   public function actionIndex($status) {
        $data = $this->data;
        if ($status == "processing") {
         //how to apply class="active" when you on the order/processing page?
         //Apply active in the data['sidebar'] array 
        }

        $view = new view('orders.php', $data);
        $view->render();
    }
  } 

数组传递到

actionIndex（）

和render orders.php中

在orders.page中，我有如下内容：

<?php foreach ($sidebar as $name => $link): ?>
    <li>
      <strong><a href="<?php echo $link; ?>"><?php echo $name; ?></a></strong>
   </li>
<?php endforeach; ?>

我想在

li

标签中添加

class=“active”

。

我的基本想法是：

if ($tatus == "processing") {
    data['active'] = 'processing';
}

<?php foreach ($sidebar as $name => $link): ?>
    <li <?php if ($active == $name) echo class="active"; ?>>
      <strong><a href="<?php echo $link; ?>"><?php echo $name; ?></a></strong>
   </li>
<?php endforeach; ?>

条件基于必须处理的名称。。。您可以根据需要随意修改。

我的基本想法是：

if ($tatus == "processing") {
    data['active'] = 'processing';
}

<?php foreach ($sidebar as $name => $link): ?>
    <li <?php if ($active == $name) echo class="active"; ?>>
      <strong><a href="<?php echo $link; ?>"><?php echo $name; ?></a></strong>
   </li>
<?php endforeach; ?>

---

条件基于必须处理的名称。。。您可以根据需要进行修改。

您可以在阵列上构建：

class OrderController extends Controller {

   public $data = array();

   private __construct() {
        $this->data['sidebar'] = array(
            'New Orders' => array( 'url'=>'/admin/order', 'class'=>'' ),
            'Processing Orders' => array( 'url'=>'/admin/order/processing', 'class'=>'' )
            'Completed Orders' => array( 'url'=>'/admin/order/completed', 'class'=>'' ),
    );


   public function actionIndex($status) {
        $data = $this->data;
        if ($status == "processing") {
          $data['sidebar']['Processing Orders']['class'] = 'active';
        }

        $view = new view('orders.php', $data);
        $view->render();
    }
  }

现在您必须更改foreach：

<?php foreach ($sidebar as $name => $link): ?>
   <li class="<?php echo $link['class']; ?>">
      <strong><a href="<?php echo $link['url']; ?>"><?php echo $name; ?></a></strong>
   </li>
<?php endforeach; ?>

您可以在阵列上构建：

class OrderController extends Controller {

   public $data = array();

   private __construct() {
        $this->data['sidebar'] = array(
            'New Orders' => array( 'url'=>'/admin/order', 'class'=>'' ),
            'Processing Orders' => array( 'url'=>'/admin/order/processing', 'class'=>'' )
            'Completed Orders' => array( 'url'=>'/admin/order/completed', 'class'=>'' ),
    );


   public function actionIndex($status) {
        $data = $this->data;
        if ($status == "processing") {
          $data['sidebar']['Processing Orders']['class'] = 'active';
        }

        $view = new view('orders.php', $data);
        $view->render();
    }
  }

现在您必须更改foreach：

<?php foreach ($sidebar as $name => $link): ?>
   <li class="<?php echo $link['class']; ?>">
      <strong><a href="<?php echo $link['url']; ?>"><?php echo $name; ?></a></strong>
   </li>
<?php endforeach; ?>


您是否有机会使用URL重写？首先将
if（$tatus==
…更改为
if（$status
）。您忘记了$tatus的s，您是否有机会使用URL重写？首先将
if（$tatus=
…更改为
if（$status
）。您忘记了$tatus的s