Question

我有以下代码：

    return array(
        ULogt::UPDATE => '

 <div>     
    <a href="#link">Navigate</a>
 </div>
'

此代码位于名为links.php的类中。当用户按下导航按钮时，我需要导航到('viewform', array('model'=>$this->loadJson($id))。我不知道如何插入此代码而不是#link。我该怎么办？

Answer 1

CHtml::link( 
   "Navigate", 
   "javascript:void(0);", // link for destination or you can handle same with jQuery
   array(
    'id' => 'navigation-id', // id for handeling in jQuery
    'key' => $data, // Required data will be appeared as attributes for this link
   )
);

您可以创建

之类的链接

Yii::app()->createUrl(
         '/profile/membership/view', // your link
         array(
             'id'=> 1  // data to be sent
         )
 )

Check for URL formating

如何在yii 1中诋毁链接？

1 个答案: