2个jquery手风琴在一页上，第二个手风琴的开放式面板来自外部链接

Question

我在一个页面上有两个手风琴。目前，我可以使用mysite.htm？panel = 0，mysite.htm？panel = 1，mysite.htm？panel = 2从外部链接打开 Accordion 1 中的特定面板，等。

但是，我无法弄清楚如何从外部页面打开 Accordion 2 中的特定面板。

   <link href="../jQueryAssets/jquery.ui.core.min.css" rel="stylesheet" type="text/css">
<link href="../jQueryAssets/jquery.ui.theme.min.css" rel="stylesheet" type="text/css">
<link href="../jQueryAssets/jquery.ui.accordion.min.css" rel="stylesheet" type="text/css">
<script src="../jQueryAssets/jquery-1.11.1.min.js" type="text/javascript"></script>
<script src="../jQueryAssets/jquery.ui-1.10.4.accordion.min.js" type="text/javascript"></script>
<div id="Accordion1">
  <h3><a href="#">Section 1</a></h3>
  <div>
    <p>Content 1</p>
  </div>
  <h3><a href="#">Section 2</a></h3>
  <div>
    <p>Content 2</p>
  </div>
  <h3><a href="#">Section 3</a></h3>
  <div>
    <p>Content 3</p>
  </div>
</div>
<p>&nbsp; </p>
<p>&nbsp; </p>
<p>&nbsp; </p>


<div id="Accordion2">
  <h3><a href="#">Section 1</a></h3>
  <div>
    <p>Content 1</p>
  </div>
  <h3><a href="#">Section 2</a></h3>
  <div>
    <p>Content 2</p>
  </div>
  <h3><a href="#">Section 3</a></h3>
  <div>
    <p>Content 3</p>
  </div>
</div>
<p>&nbsp;</p>
<script type="text/javascript">

$(function() {
      var defaultPanel = parseInt(getParam('panel'));
    $( "#Accordion1" ).accordion( 
    {active: defaultPanel}
    ); 

});

function getParam(name) {
    var query = location.search.substring(1);
    if (query.length) {
        var parts = query.split('&');
        for (var i = 0; i < parts.length; i++) {
            var pos = parts[i].indexOf('=');
            if (parts[i].substring(0,pos) == name) {
                return parts[i].substring(pos+1);
            }
        }
    }
    return 0;
};
</script>

<script type="text/javascript">

$(function() {
      var defaultPanel = parseInt(getParam('panel'));
    $( "#Accordion2" ).accordion( 
    {active: defaultPanel}
    ); 

});

function getParam(name) {
    var query = location.search.substring(1);
    if (query.length) {
        var parts = query.split('&');
        for (var i = 0; i < parts.length; i++) {
            var pos = parts[i].indexOf('=');
            if (parts[i].substring(0,pos) == name) {
                return parts[i].substring(pos+1);
            }
        }
    }
    return 0;
};
</script>