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javascript - 尝试将随机图像与其他随机图像交换

转载 作者:行者123 更新时间:2023-11-28 15:55:13 25 4
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我有一些李,每个里面都有一个图像。前三个是可见的,其余的是隐藏的。我想要做的是隐藏其中一张可见图像(随机一张 - 这一点有效..),用其中一张隐藏图像替换该图像并再次显示它,但我似乎无法再次使其可见。这是代码笔:http://codepen.io/anon/pen/sBvkC

HTML:

<ul>
<li class="imageIn"><img src="http://placehold.it/300x200" /></li>
<li class="imageIn"><img src="http://placehold.it/300x200" /></li>
<li class="imageIn"><img src="http://placehold.it/300x200" /></li>
<li class="imageOut"><img src="http://placehold.it/300x200" /></li>
<li class="imageOut"><img src="http://placehold.it/300x200" /></li>
</ul>

CSS:

* { margin: 0; padding: 0; }    
ul { list-style-type: none; width: 100%; }
li { width: 33%; display: inline-block; }
img { width: 100%; }
.imageOut { display: none; }
.hidden { visibility: hidden; }
.visible { visibility: visible; }

JS:

window.setInterval(function(){swapImage()},1000);

function swapImage() {

/* select random image that is currently displayed */
ins = $('.imageIn > img');
var imageDisplayed = $(ins[Math.floor(Math.random()*ins.length)]);

/* select random image that is NOT currently displayed */
outs = $('.imageOut > img');
var imageNotDisplayed = $(outs[Math.floor(Math.random()*outs.length)]);

/* switch displayed image to visibility:hidden */
$(imageDisplayed).css("visibility", "hidden");

/* SWAP src of imageDisplayed with that of imageNotDisplayed */
var srcDisplayed = $(imageDisplayed).attr("src");
var srcNotDisplayed = $(imageNotDisplayed).attr("src");
$(imageDisplayed).attr("src") = srcNotDisplayed;
$(imageNotDisplayed).attr("src") = srcDisplayed;

/* switch to visibility:visible */
$(imageDisplayed).css("visibility", "visible");

}

我做错了什么?

最佳答案

您没有正确设置 img 标签上的 src 属性。变化:

$(imageDisplayed).attr("src") = srcNotDisplayed;
$(imageNotDisplayed).attr("src") = srcDisplayed;

进入:

$(imageDisplayed).attr("src", srcNotDisplayed);
$(imageNotDisplayed).attr("src", srcDisplayed);

关于javascript - 尝试将随机图像与其他随机图像交换,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19184988/

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