Har*_*iss 23 html javascript css jquery
我有一些带有图像的div,当我点击这些图像时,我想要用我点击的图像打开另一个div.
JS
$('.examples img').click(function() {
var loc = $(this).attr("src");
$('#image-zoom').attr("src",loc);
});
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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>
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当我试图隐藏div工作时,我认为我的语法有错误
Cal*_*ior 56
更改src图像,而不是div:
$('#image-zoom img').attr("src",loc);
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