使用下面的HTML,我需要只针对有4个或更少的项目的旋转木马,即本例中的第二个:
<div class="book_carousel">
<h2 class="type_one">Carousel 1</h2>
<a class="prev disabled">Previous Item</a>
<div class="scrollable">
<div class="items">
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
</div>
</div>
<a class="next">Next Item</a>
</div>
<div class="book_carousel">
<h2 class="type_one">Carousel 1</h2>
<a class="prev disabled">Previous Item</a>
<div class="scrollable">
<div class="items">
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
<div class="item">
<a href="http://www.place.com" class="thumb"><img src="image.jpg"></a>
<p>Info</p>
<a class="button" href="#">LOOK</a>
</div>
</div>
</div>
<a class="next">Next Item</a>
</div>
一旦我得到了包含4个或更少元素的div,我的想法是关闭carousel,因为这是不必要的
试试这个:-
过滤掉div.item
小于等于4的.book_carousel
并做一些事情。过滤器将返回小于或等于4个.items
的carouseldiv。
$('.book_carousel').filter(function(){
return $(this).find('div.item').length <= 4;
}).hide(); // Do hide or remove or something with them.