如何在下面的示例中为每个图像simiar添加共享链接到第一个图像:
https://jsfiddle.net/w1h6mhp8/4/
$(document).ready(function () {
$(".copy-url").click(function() {
$(this).select();
document.execCommand("copy");
})
if(true) { //will be replaced by navigator.share, value will be url
$("div.copy-url").each(function() {
var img =$(this);
var imgval = $('.copy-url').val;
var sharer = "<a class="share">" + imgval + "</span>";
$(this).append(sharer)
});
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div><input type="text" class="copy-url" value="222222" readonly="yes">
<span class="share">copy</span> <a class="share" href="222222">share</a></div>
<div><input type="text" class="copy-url" value="333333" readonly="yes">
<span class="share">copy</span></div>
<div><input type="text" class="copy-url" value="444444" readonly="yes">
<span class="share">copy</span></div>
<div><input type="text" class="copy-url" value="555555" readonly="yes">
<span class="share">copy</span></div>
<div><input type="text" class="copy-url" value="666666" readonly="yes">
<span class="share">copy</span></div>
答案 0 :(得分:2)
这应该按照你的意愿行事:
$("div .copy-url").each(function() {
var imgval = $(this).find('.copy-url').val();
var sharer = "<a class='share' href='"+imgval+"'>share</a>";
$(this).append(sharer)
});
您遇到以下错误:
您"<a class="share">"
中的引号不匹配应为"<a class='share'>"
在{。{1}}行中,您的.val后缺少$('.copy-url').val
和$(this).find
。应为()
您的代码中没有$(this).find('.copy-url').val()
个类div
。
<强>演示强>
copy-url
$(document).ready(function() {
$(".copy-url").click(function() {
$(this).select();
document.execCommand("copy");
})
if (true) { //will be replaced by navigator.share, value will be url
$("div").each(function() {
var imgval = $(this).find('.copy-url').val();
var sharer = "<a class='share' href='"+imgval+"'>share</a>";
$(this).append(sharer)
});
}
});