我有一堆旋转木马,每个都包裹在.lazy_wrap
中。我想懒得加载它们以加快网站加载速度。我该如何处理?我看过一些懒惰的加载插件,但它们似乎只适用于图像。我找到了一个名为jquery-lazyloadanything的插件,但我无法弄清楚如何使用它。
http://jsfiddle.net/8bJUc/667/
<div class="lazy_wrap">
<div class="owl-carousel">
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
<div>
<img src="http://placehold.it/200x200/42bdc2/FFFFFF&text=1st Row" alt="" />
</div>
</div>
</div>
$(document).ready(function() {
$(".owl-carousel").owlCarousel({
navigation: true,
pagination: true
});
});
答案 0 :(得分:1)
http://jsfiddle.net/8bJUc/668/
您只需添加以下代码:
$('img').lazyloadanything({
'onLoad': function(e, LLobj) {
var $img = LLobj.$element;
var src = $img.attr('data-src');
$img.attr('src', src);
console.log("loading: ", src);
}
});
但是您不希望图片被自动加载#34;因此您需要在每个{{1}上将src="url"
更改为data-src="url"
标签,如:
img
您还需要通过html标签或css为图像添加高度,宽度,因为lazyloadanything只会在检测到它们进入屏幕时加载它们,但只有在它们有大小时才会发生。