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我有一些div的图像,当我点击这些图像时,我想让另一个div打开,并点击了该图像。将图像src设置为另一个图像jquery
JS
$('.examples img').click(function() {
var loc = $(this).attr("src");
$('#image-zoom').attr("src",loc);
});
HTML
<div class="container examples" >
<div id="image-zoom">
<img class="img-thumbnail zoom" src="" alt="dental">
</div>
<div class="row">
<div class="col-sm-12">
<img id="zoom" class="img-thumbnail zoom" src="images/01.png" alt="dental">
<img class="img-thumbnail zoom" src="images/02.png" alt="dental">
<img class="img-thumbnail zoom" src="images/03.png" alt="dental">
</div>
</div>
<div class="row">
<div class="col-sm-12">
<img class="img-thumbnail zoom" src="images/04.png" alt="dental">
<img class="img-thumbnail zoom" src="images/05.png" alt="dental">
<img class="img-thumbnail zoom" src="images/06.png" alt="dental">
</div>
</div>
</div>
当我试图隐藏在div它的工作,所以我有错误的语法我觉得
什么? – ambarox
'#image-zoom''不是'img'标签,所以它不能拥有'src'属性 – JFK