我正在尝试获取所单击的特定图像的 img src。我在尝试中使用了 this
函数,所以我不确定我做错了什么。
有人看到这个问题吗?
HTML:
<div id="zoomPop" data-popup="pop2">
<div id="zoomInner">
<a class="sharePopClose" data-popup-close="pop2" href="#"><img src="/icon_close.png" alt="Close Project Image" class="xClose">
</a>
<img src="" alt="Project Enlarge" id="zoomImg">
</div>
</div>
//Project Container Zoom
$('#projectGallery').on('click', '.projectCont', function (event) {
event.stopPropagation();
$('#zoomPop').fadeIn(350);
$('body').css('overflow', 'hidden');
var currentImg = $(this).attr('src');
console.log(currentImg);
});
#zoomPop {
width: 100%;
height: 100%;
color: #FFF;
position: fixed;
z-index: 999999;
margin: 0;
padding: 0;
top: 0;
right: 0;
bottom: 0;
overflow-y: scroll;
display: none;
}
#zoomPop {
background: rgba(0,0,0,.7);
}
#zoomInner {
position: relative;
padding: 60px 0;
margin: 0 auto;
width: 90%;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="projectGallery">
<div class="projectCont">
<img src="https://images.pexels.com/photos/2422/sky-earth-galaxy-universe.jpg?auto=compress&cs=tinysrgb&dpr=1&w=500" alt="Pic">
</div>
<div class="projectCont">
<img src="https://geology.com/google-earth/google-earth.jpg" alt="Pic">
</div>
</div>
<div id="zoomPop" data-popup="pop2">
<div id="zoomInner">
<img src="" alt="Project Enlarge" id="zoomImg">
</div>
</div>
最佳答案
您的this
引用DOM节点<div class="projectCont">
使用
console.log($(this).find('img').attr('src'));
或将事件处理程序的选择器更改为
$('#projectGallery').on('click', '.projectCont img', function (event) {
关于javascript - 如何获取点击图像的img src,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57362416/