Question

我有一个基于JQuery Mobile的HTML5移动应用程序。它从服务器获取任何内容，有时连接到Internet但也具有脱机模式。 Google Analytics是否支持离线模式？注意，我没有使用Android / iOS SDK。只是在JQueryMobile页面中的ga.js。感谢

下面是我的html5 JQueryMobile代码：

    <!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>

Answer 1

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。

Google Analytics如何合并到具有离线/在线模式的HTML5移动应用中？

1 个答案: