如何通過在onchange事件兩個或多個參數

Question

我有顯示名稱的表，但它的價值是它的用戶id，電子郵件和水平

<?php if (count($users)): ?> 
    <table class="table table-hover"> 
    <thead> 
    <tr> 
     <th>Name</th> 
     <th>E-mail</th> 
     <th>Level</th> 
     <th></th> 
     </tr> 
    </thead> 
    <?php foreach ($users as $user): ?> 
    <tr> 
     <td><a href="<?php echo $this->url('user/edit', array('id' => $user->id)); ?>"><?php echo $this->escapeHtml($user->first_name . ' ' . $user->last_name); ?></a></td> 
     <td><?php echo $this->escapeHtml($user->email); ?></td> 
     <td> 
      <select name="level" id= "level" onchange="changeLevel(this);"> 

      <?php $level = $this->escapeHtml($user->level); ?> 

      <?php if ($level == 1): ?> 
       <option value="1" selected ="selected">Administrator</option> 
       <option value="2" <?php IsSelected($level,2); ?>> Manager</option> 
       <option value="3" <?php IsSelected($level,3); ?>>HR Staff</option> 
      <?php elseif ($level == 2): ?> 
       <option value="1" <?php IsSelected($level,1); ?>>Administrator</option> 
       <option value="2" selected ="selected">Manager</option> 
       <option value="3" <?php IsSelected($level,3); ?>>HR Staff</option> 
       <?php elseif ($level == 3): ?> 
       <option value="1" <?php IsSelected($level,1); ?>>Administrator</option> 
       <option value="2" <?php IsSelected($level,2); ?>>Manager</option> 
       <option value="3" selected ="selected">HR Staff</option> 
       <?php endif; ?> 
      </select> 
     </td> 
     <td style="text-align: right;"> 
      <a href="<?php echo $this->url('user/delete', array('id' => $user->id)); ?>" class="btn btn-mini btn-danger" rel="tooltip" title="Delete this user"><i class="icon-remove icon-white"></i></a> 
     </td> 
    </tr> 
    <?php endforeach; ?> 
    </table> 
<?php else: ?> 
<h3>There are no registered users available.</h3> 
<?php endif; ?> 

，現在來看，我的選擇下拉onchange事件僅傳遞值在下拉到AJAX函數調用一個Zend controlelr

這裏所選擇的值的是AJAX功能

<script> 

changeLevel = function(level){ 
     var level_value = level.value; 
     var user_id = 
     alert(level.value); 

    $.ajax({ 
     type: "POST", 
     url: 'http://localhost/ijm/public/user/level', 
     data: { function_params: level_value }, 
     Success: function(result){ 
      alert('Success'); 
     } 
    }); 
} 

</script> 

和這裏是用戶控制器功能

public function levelAction() 
    { 

     $storage = $this->getAuthService()->getStorage(); 
     $user = $storage->read(); 
     $request = $this->getRequest(); 
     $level = $request->getPost('function_params'); 
     echo $level; 
      // Check if form values are valid 
      $this->getUserTable()->editLevel($user->id, $level); 
      $this->redirect()->toRoute('user'); 
      echo "success";  

    } 

我要的是每次我觸發onchange事件我也希望用戶id要在onchange事件傳遞的功能，這樣我可以在我的Zend控制器上發送。我怎樣才能做到這一點？提前致謝。

Answer 1

  <select name="level" id= "level" onchange="changeLevel(this , '<?php echo $user->id?>');"> 

      ... 


      changeLevel = function(level, user_id){ 
... 

Answer 2

你選擇標籤可容納用戶ID太：

<select name="level" id="level" data-userid="<?=$user->id?>" onchange="changeLevel(this);"> 

比changeLevel功能，您可以訪問它：

changeLevel = function(level){ 
    var level_value = level.value; 
    var userid = $(level).data('userid'); 
    alert('Value: ' + level.value + ', userid: ' + userid); 

    $.ajax({ 
     type: "POST", 
     url: 'http://localhost/ijm/public/user/level', 
     data: { function_params: level_value, userid: userid }, 
     Success: function(result){ 
      alert('Success'); 
     } 
    }); 
} 

請檢查this Fiddle。