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I have an HTML5 Mobile App based on JQuery Mobile. It fetches nothing from the server, is sometimes connected to the internet but also has offline mode. Does Google Analytics support offline mode? Note, I am not using the Android/iOS SDK. Just ga.js inside JQueryMobile pages. Thanks

Below is my html5 JQueryMobile code:

    <!DOCTYPE html> 
<html> 
<head> 
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Mobile @ Your Library</title>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.0/jquery.mobile-1.0.min.css" />

 <script type="text/javascript" src="http://code.jquery.com/jquery-1.6.4.min.js">  </script>
<script type="text/javascript" src="http://code.jquery.com/mobile/1.0/jquery.mobile-1.0.min.js"></script>
<script type="text/javascript">

    var _gaq = _gaq || [];
    _gaq.push(['_setAccount', 'UA-40418-saddf']);

    _gaq.push(['_gat._anonymizeIp']);

    (function() {
    var ga = document.createElement('script'); ga.type = 'text/javascript'; ga.async = true;
    ga.src = ('https:' == document.location.protocol ? 'https://ssl' : 'http://www') + '.google-analytics.com/u/ga_debug.js';
    var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(ga, s);
     })();

</script>



</script>
<body> 


<div data-role="header" data-theme="d"> 
    <h1>Mobile @ Your Library</h1>
    <a href="#home" data-icon="home" data-iconpos="notext" data-theme="d" class="ui-btn-right">home</a>
</div><!-- /header --> 

<div data-role="content">
    <ul data-role="listview" data-inset="true">
        <li><a href="#search">Search</a></li>
        <li><a href="#hours">Hours</a></li>
        <li><a href="#ask">Ask a Librarian</a></li>
        <li><a href="#about">About</a></li>
        <li><a href="#where" data-rel="dialog" data-transition="pop">Where</a></li>
    </ul>       

</div><!-- /content --> 

<div data-role="footer" data-id="myfooter" data-position="fixed" data-theme="d"> 
    <div class="controls" data-role="controlgroup" data-type="horizontal">
        <a href="#home" data-role="button" data-icon="home">Home</a>
        <a href="#search" data-role="button" data-icon="search">Search</a>
        <a href="#ask" data-role="button" data-icon="info">Ask</a>
        <a href="/index.php" rel="external" data-role="button" data-icon="plus">Full site</a>
    </div>
</div><!-- /footer -->  
</div><!-- /page -->
<!--
<script type="text/javascript">
$('[data-role=page]').live('pageshow', function (event, ui) {
    console.log('pageshow called s');
    console.log('pageshow called s2');
     try {
    _gaq.push( ['_trackPageview', event.target.id] );
    console.log('tracked '+event.target.id);
    } catch(err) {
       console.log('error '+ err);
    }

});
</script>-->
<script type="text/javascript">
$(document).delegate('[data-role=page]', 'pageshow', function (event, ui) {
console.log('pageshow event '+ui.prevPage.id);
var url = location.href;
console.log('pageshow url = '+url);
try  {

    if (location.hash) {

        url = location.hash;
        console.log('location.hash so now url = '+url);
    }
    console.log('about to push page '+url);
    //_gaq.push(['_trackPageview', url]);
    _gaq.push( ['_trackPageview', event.target.id] );
    console.log('about to push page '+url);
} 
catch(error) {
    // error catch
    console.log('error '+error);
}
});

</script>
</body> 
</html>
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1 回答 1

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GA 代码必须能够向服务器发送请求以跟踪数据。如果您处于离线模式且未发出任何请求，则不会进行跟踪。

于 2013-04-26T12:50:33.173 回答