带有html5的自动图像滑块的javascript

Question

The following code is my html page for image slider but it will not continue after sliding last image. 以下代码是我的图像滑块的html页面，但在滑动最后一个图像后它不会继续。

<style type="text/css">
#slideshow {
position:relative;
height:300px;
}

#slideshow IMG {
position:absolute;
top:0;
left:0;
z-index:8;
}

#slideshow IMG.active {
z-index:10;
}

#slideshow IMG.last-active {
z-index:9;
}
</style>'


<script>
function slideSwitch() {
    var $active = $('div#slideshow IMG.active');
    var $next = $active.next();    

    $next.addClass('active');

}

$(function() {
    setInterval( "slideSwitch()", 5000 );
});
</script>'

<body>
<div id="slideshow" style="height:300px; width:100%">
<img src="${context:layout/images/images.jpeg}" title="Funky roots"     style="position:absolute; height:300px; width:100%;" class="active"/>
<img src="${context:layout/images/police.jpg}" title="The long and  winding road" style="position:absolute; height:300px; width:100%;"/>
<img src="${context:layout/images/viper_1.jpg}" title="Happy trees"   style="position:absolute; height:300px; width:100%;"/>
</div>
</body>

From this code, how can I get auto sliding and continue to next slide after a last image? 从这段代码中，如何在最后一张图像后自动滑动并继续下一张幻灯片？

Answer 1

You dont add the last-active class to the last element. 您不要将last-active类添加到last-active一个元素。 So every element has the active class and every element has z-index: 10 所以每个元素都有active类，每个元素都有z-index: 10

After some work on the code got this new JS for you: 在对代码进行一些工作之后，为您获得了这个新的JS：

function slideSwitch() {
  var $active = $('div#slideshow IMG.active');
  var $next = $active.next();    
  $active.addClass('last-active');
  $active.removeClass('active');
  if($next.length == 0) {
    $next = $('div#slideshow IMG').first(); //at the end we need to select the first element, because there is no next()
  }
  $next.removeClass('last-active');
  $next.addClass('active');
}

EDIT: 编辑：

TO answer your comment: 回答你的评论：

After the first round through your slider the classes look like this: 通过滑块的第一轮后，类看起来像这样：

<img ... class="last-active"/>
<img ... class="last-active"/>
<img ... class="active"/>

When we enter the code $active contains the last div and $next is empty, because there is no next element. 当我们输入代码时， $active包含最后一个div而$next是空的，因为没有下一个元素。

At this point we have to tell the code, that the next element will be the first one. 此时我们必须告诉代码，下一个元素将是第一个元素。 The $next = $('div#slideshow IMG').first(); $next = $('div#slideshow IMG').first(); just select the first div where we want to start over. 只需选择我们想重新开始的第一个div。

带有html5的自动图像滑块的javascript

问题描述

1 个解决方案

解决方案1
0 已采纳 2015-03-05 09:39:13

带有html5的自动图像滑块的javascript

问题描述

1 个解决方案

解决方案1 0 已采纳 2015-03-05 09:39:13

解决方案1
0 已采纳 2015-03-05 09:39:13