-2

我尝试获取具有显示块属性的 id slide-1,然后将一个类添加到活动的 id pagecount-1... 我试图为每张幻灯片获取显示为活动的指示器,这就是我所拥有的:

<div id="slider">
  <div class="sp active" id="slide-1" style="display: block"></div>
  <div class="sp" id="slide-2" style="display: none"></div>
  <div class="sp" id="slide-3" style="display: none"></div>
  <div class="sp" id="slide-4" style="display: none"></div>
</div>

<div id="page">
  <div id="page_count-1" class="indicate"></div>
  <div id="page_count-2" class="indicate"></div>
  <div id="page_count-3" class="indicate"></div>
  <div id="page_count-4" class="indicate"></div>
</div>
4

2 回答 2

1

此代码检查所有slide元素是否可见并将active类应用于相应的page_count元素:

for (var i = 1; i <= 4; i++) {
    if ($('#slide-' + i).is(':visible')) {
        $('#page_count-' + i).addClass('active');
    }
}
于 2013-09-24T14:42:27.090 回答
0
$('#slider div').each(function() {
    var currentIndex = parseInt($(this).attr('id').replace('slide-',''));
    if($(this).css('display') == 'block') {
        $("#page_count-" + currentIndex).addClass("active");  
    }
});

http://jsfiddle.net/jonigiuro/bZj2B/

于 2013-09-24T14:46:45.780 回答