谷歌分析代码404错误跟踪在WordPress PHP函数文件

我想在每个页面上显示谷歌分析，除了我想在404模板页面上显示它略有不同。

我有两个分析代码。每一页上都显示一个。另一个是只在404页面上显示。

这是我正在工作的php代码，插入到函数文件。我不能让它工作。到目前为止，它所做的只是执行else语句。请帮助。

if ( is_404() ) {
// Include the Google Analytics Tracking Code in 404
function google_analytics_tracking_code_404(){ ?>
<script>
  (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
  (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o),
  m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m)
  })(window,document,'script','//www.google-analytics.com/analytics.js','ga');
  ga('create', 'UA-XXXXXXXX-23', 'mydomainname.com');
  ga('send', 'pageview', '404/?url='+ document.location.pathname + document.location.search +'&ref=' + document.referrer);
</script>
<?php }
// include tracking code before the closing head tag
add_action('wp_head', 'google_analytics_tracking_code_404');
// OR include tracking code before the closing body tag
// add_action('wp_footer', 'google_analytics_tracking_code');
}
else {
// Include the Google Analytics Tracking Code
function google_analytics_tracking_code(){ ?>
<script>
  (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
  (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o),
  m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m)
  })(window,document,'script','//www.google-analytics.com/analytics.js','ga');
  ga('create', 'UA-XXXXXXXX-23', 'mydomainname.com');
  ga('send', 'pageview');
</script>
<?php }
// include tracking code before the closing head tag
add_action('wp_head', 'google_analytics_tracking_code');
// OR include tracking code before the closing body tag
// add_action('wp_footer', 'google_analytics_tracking_code');
}