Javascript jQuery图像缩放不工作
我从你那里得到这个密码。我在那个网站上尝试了最后一种方法。然而,我试图复制它,但在我这方面仍然不起作用 首先,我的HTML编码是这样的Javascript jQuery图像缩放不工作,javascript,jquery,html,Javascript,Jquery,Html,我从你那里得到这个密码。我在那个网站上尝试了最后一种方法。然而,我试图复制它,但在我这方面仍然不起作用 首先,我的HTML编码是这样的 <img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png" data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/ima
<img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"
data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"/>
<div id="gal1">
<a href="#" data-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"
data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png">
<img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png" /> </a>
<a href="#" data-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"
data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png">
<img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png" /> </a>
<a href="#" data-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"
data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png">
<img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png" /> </a>
<a href="#" data-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png"
data-zoom-image="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png">
<img id="img_01" src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/images/test.png" /> </a>
</div>
/assets/images/test.png”
数据缩放图像=”http:///assets/images/test.png"/>
我写的剧本是这样的
<script type="text/javascript">
$("#zoom_03").elevateZoom({
constrainType:"height",
constrainSize:274,
zoomType: "lens",
containLensZoom: true,
gallery:'gallery_01',
cursor: 'pointer',
galleryActiveClass: "active"});
$("#zoom_03").bind("click", function(e) {
var ez = $('#zoom_03').data('elevateZoom');
$.fancybox(ez.getGalleryList());
return false;
});
</script>
$(“#缩放03”).ElevateToom({
限制类型:“高度”,
尺寸:274,
zoomType:“镜头”,
是的,
画廊:'gallery_01',
光标:“指针”,
galleryActiveClass:“活动”});
$(“缩放03”).bind(“单击”,函数(e){
var ez=$('zoom_03')。数据('elevateToom');
$.fancybox(ez.getGalleryList());
返回false;
});
我还将调用jQuery库:
<script src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/scripts/elevateZoom/jquery-1.8.3.min.js"></script>
<script src="http://<?php echo $_SERVER['HTTP_HOST'] ?>/assets/scripts/elevateZoom/jquery.elevatezoom.js"></script>
检查控制台?是否有错误?您在id为zoom_03的元素上绑定了事件,但html中没有id为zoom_3的元素。我查看了ElevateWeb,在示例代码中似乎有很多类似的错误。不,在我将html中的id更改为#zoom_3后,它才起作用。谢谢您的提示。是的,示例是错误的。@mondjunge@conanak99它正在工作。谢谢。:)