在ddSlick下拉列表中获取所选<option>的值

Question

这当前返回undefined.注释行应该注意什么来警告当前<option>标记的值(1,2,3或4)？

<select id="dropdown" name="dropdown">
        <option value="0" data-imagesrc="images/icons/all.png">All Questions</option>
        <option value="1" id="friends" data-imagesrc="images/icons/friends.png">Friends</option>
        <option value="2" data-imagesrc="images/icons/friends_of_friends.png">Friends of Friends</option>
        <option value="3" data-imagesrc="images/icons/network.png"><?php echo $network; ?></option>
        <option value="4" data-imagesrc="images/icons/location.png"><?php echo $location ?></option>
</select>

<script type="text/javascript">
$('#dropdown').ddslick({
showSelectedHTML: false,
    onSelected: function(selectedData){
    var str = $(this).attr('id'); // WHAT SHOULD GO HERE?
    alert(str);
    }   
});
</script>

编辑

如果它是相关的,我正在使用这个插件.

也许这个问题可能有所帮助 我想弄清楚它.

管理好解决这个问题.最终的工作代码是:

<select id="dropdown" name="dropdown" value="hello">
        <option value="0" data-imagesrc="images/icons/all.png">All Questions</option>
        <option value="1" id="friends" data-imagesrc="images/icons/friends.png">Friends</option>
        <option value="2" data-imagesrc="images/icons/friends_of_friends.png">Friends of Friends</option>
        <option value="3" data-imagesrc="images/icons/network.png"><?php echo $network; ?></option>
        <option value="4" data-imagesrc="images/icons/location.png"><?php echo $location ?></option>
</select>

<script type="text/javascript">
$('#dropdown').ddslick({
    showSelectedHTML: false,
    onSelected: function(data){
        alert(data.selectedData.value);
    }   
});
</script>

Answer 1

用来$(this).val()代替$(this).attr('id')

当前选择的值<option>,当你调用返回.val()的上<select>元素.