Why does Google's Analytics tracking code execute first?

Question

I was playing around with custom variables in Google Analytics and wasn't sure why the following actually works. Why does Google's tracking code get executed first, especially since it is at the bottom of the page?

Both scripts are in self-executing functions, so how does javascript determine which one to execute first?

 // top of the page
  <script type="text/javascript">
              $(function ()
              {
                    _gaq.push(['_setCustomVar', 1, 'Account', 'PartTime', 1]);
              });
        </script> 


// bottom of the page
        <script type="text/javascript">

            var _gaq = _gaq || [];
            _gaq.push(['_setAccount', 'UA-xxxxxxxxxx']);
            _gaq.push(['_setDomainName', '.xxxxxxxx.com']);
            _gaq.push(['_trackPageview']);

            (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/ga.js';
                var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(ga, s);
            })();

         </script>

Quentin · Accepted Answer · 2011-07-17 17:56:52Z

5

The first one does execute first, and isn't in a self executing function.

It consists of a call to the $ function and has one argument: an anonymous function.

$ is a very badly named function. The name itself is meaningless and it has been adopted by half a dozen different libraries to do half a dozen different things.

In jQuery, if you pass a function to $ it runs that function when the ready event is triggered (but the end of the HTML document being parsed). This is probably what is happening here.

answered Jul 17, 2011 at 17:56

Quentin

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2 Comments

chobo Over a year ago

I used firebug and put breakpoints in both functions and it goes to the google one first, then it goes into the one at the top. You are right about the first not being a self-executing function which is why it isn't being executed first.

Quentin Over a year ago

The first script gets executed first. The function it defines doesn't get executed first (it gets passed to $()). If you stick a breakpoint in that function, then that breakpoint one be hit until the function is called.

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