title: How does $.ajaxError work?
body: <p>I use $.get, $.post, and $.getJSON extensively, so I wanted to see how .ajaxError worked. I tried this...</p>
  
  <pre><code>$(document).ajaxError(function (e, xhr, options, error)
  {
      console.log(e);
      console.log(xhr);
      console.log(options);
      console.log(error);
  });
  </code></pre>
  
  <p>and I also tried putting it inside my <code>$(document).ready().</code>. And then I disabled my network connection. A slew of errors flew up from the timed <code>POST</code>s in AJAX (because they were getting unexpected responses), and the <code>GET</code>s were coming back instantaneously (cached), but nothing else appeared in Firebug, as if ajaxError were never being called. What am I doing wrong? I just tried again with no console logs and just <code>alert('foo')</code>, and again nothing happened. Lots of errors, but ajaxError never firing.</p>