Question

我有一个显示名称的表，但其值是userId，email和level

<?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; ?>

现在，我的select dropdown onchange事件只将下拉列表中所选值的值传递给调用zend controlelr的ajax函数

这里是ajax函数

<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>

这是用户控制器功能

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";      

    }

我想要的是每当我触发onchange事件时我也希望userId传递onchange事件功能，以便我可以在我的zend控制器上发送它。我怎样才能做到这一点？提前谢谢。

Answer 1

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

            ...


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

Answer 2

您的select标签也可以包含用户ID：

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

与changeLevel功能相比，您可以通过以下方式访问它：

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');
        }
    });
}

请检查this Fiddle。

如何在onchange事件中传递两个或多个参数

2 个答案: