javascript - 弹出框在关闭后第一次点击时不显示

Question

我正在研究 this demo .我试图发现为什么弹出窗口在通过 .close 按钮关闭后第一次点击时不起作用。

 $("#pop-One").popover({
        placement: 'right',
        html: 'true',
        title : '<span class="text-info" style=""><strong>Model Type</strong></span>'+
                '<button type="button"  class="btn btn-default close" onclick="$("#pop-One").popover("hide");">×</button>',
        content : '<p class="popup" style="color:#323232;"> \Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry standard dummy text ever since the 1500s,</p>'
    });

@import 'https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css';
@import 'https://maxcdn.bootstrapcdn.com/font-awesome/4.4.0/css/font-awesome.min.css';
body{padding:20px;}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/js/bootstrap.min.js"></script>
<div class="container">
<button type="button" class="btn btn-default" id="pop-One" data-toggle="popover"><i class="fa fa-question"></i></button>
</div>

无论出于何种原因，关闭按钮在这里不起作用。我已经看到了this post但这是一种不同的方法。

Answer 1

不知道这是不是最好的解决方案，但它的工作

  $("#pop-One").popover({
        placement: 'right',
        html: 'true',
        title : '<span class="text-info" style=""><strong>Model Type</strong></span>'+
                '<button type="button"  class="btn btn-default close"\
     onclick="$(&quot;#pop-One&quot;).popover(&quot;hide&quot;);">×</button>',
        content : '<p class="popup" style="color:#323232;"> \Lorem Ipsum is simply dummy\ text of the printing and typesetting industry. Lorem Ipsum has been the industry standard\ dummy text ever since the 1500s,</p>'
  }).on('shown.bs.popover', function() {
    var popup = $(this);
    $(this).parent().find("div.popover .close").click(function() {
      popup.click();
    });
  });

here the result of my efforts