bootstrap-popover - 如何从 Bootstrap 中的弹出窗口获取弹出窗口

Question

我尝试了下面的代码，但我没有收到第二个弹出窗口，我的方法似乎有一些问题，我不知道如何获得第二个弹出窗口...

$(function() {
    $(".pendingList, .pendings").popover();
 });

.icon3{
	background: url('http://blog.flattr.net/wp-content/uploads/2011/09/stackoverflow.png') 1px 1px no-repeat;
	background-position: center;
	height:100px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet"/>

<div class="icon3 pendingList" data-toggle="popover" data-trigger="click" data-html="true" data-placement="bottom" 
													data-content="<a href='#' class='pendings' data-toggle='popover' data-trigger='click' data-html='true' data-placement='bottom' 
												data-content='test' >Pending(5)</a>"></div>

Answer 1

您的代码不起作用，因为执行 popover 方法时元素不存在。尝试下面的代码，效果很好!

$(function() {
    $(".pendingList").popover();
    $('.pendingList').on('inserted.bs.popover', function () {
      $(".pendings").popover();
    });
 });

.icon3{
	background: url('http://blog.flattr.net/wp-content/uploads/2011/09/stackoverflow.png') 1px 1px no-repeat;
	background-position: center;
	height:100px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet"/>

<div class="icon3 pendingList" data-toggle="popover" data-trigger="click" data-html="true" data-placement="bottom" data-content="<a href='#' class='pendings' data-toggle='popover' data-trigger='click' data-html='true' data-placement='bottom' data-content='test' >Pending(5)</a>"></div>