我的问题是我有多个图标触发不同的旋转木马。默认情况下,第一个旋转木马是可见的,其余的是隐藏的。下面是图标(在我的例子中是织物样本):
<div id="home" class="fabrics">
<h2>Fabrics Available</h2>
<a class="charcoal" href="javascript:void(0)">
<div class="magnify-3">
<div class="large-zoom-3"></div>
<img class="small-zoom-3" src="<?php echo $this->getSkinUrl('images/recliners/fabrics/charcoal.png');?>" alt="Charcoal" />
</div>
</a>
<a class="mushroom" href="javascript:void(0)">
<div class="magnify-2">
<div class="large-zoom-2"></div>
<img class="small-zoom-2" src="<?php echo $this->getSkinUrl('images/recliners/fabrics/mushroom.png');?>" alt="Mushroom" />
</div>
</a>
</div>这是我的传令:
<div class="imgs slider charcoal">
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/charcoal-1.png');?>" alt="Charcoal" />
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/charcoal-2.png');?>" alt="Charcoal" />
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/charcoal-3.png');?>" alt="Charcoal" />
</div>
<div class="imgs slider mushroom" style="display:none">
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/mushroom-1.png');?>" alt="Mushroom" />
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/mushroom-2.png');?>" alt="Mushroom" />
<img src="<?php echo $this->getSkinUrl('images/recliners/indiana/mushroom-3.png');?>" alt="Mushroom" />
</div>我的抄本是:
jQuery(document).ready(function() {
jQuery(".fabrics a").click(function() {
var e = jQuery(this).attr("class");
jQuery(".imgs").hide(), jQuery(".imgs." + e).toggle();
})
}),
jQuery(document).ready(function() {
jQuery(".slider").owlCarousel({
loop: !0,
autoplay: !0,
autoplayTimeout: 3e3,
autoplayHoverPause: !0,
items: 1,
animateIn: "fadeIn",
animateOut: "fadeOut",
navigation: !0
})
});当单击第二项时,它确实加载了第二个旋转木马,但是width似乎很大。1260px在我的案子里。我知道这可能是因为浏览器不知道第二个旋转木马设置为无显示的width?有什么方法可以在我的图标的click事件上触发刷新吗?
发布于 2016-02-11 12:10:24
将旋转木马放入高度为0的包装元素中,然后隐藏溢出;将其切换到可见的设置高度: auto;
https://stackoverflow.com/questions/35337176
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