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I've got some li's, each with an image inside. The first three are visible and the rest are hidden. What I want to do is to hide one of the visible images (a random one - that bit works..), replace the image with one of the hidden ones and display it again, but I can't seem to get it visible again. Here's the codepen: 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");

}

What am I doing wrong?

4

1 回答 1

4

您没有正确设置 img 标签上的 src 属性。改变：

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

进入：

$(imageDisplayed).attr("src", srcNotDisplayed);
$(imageNotDisplayed).attr("src", srcDisplayed);
于 2013-10-04T15:28:05.907 回答