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copied!<p>Setting the <code>dataType</code> as "json" means the response from <code>get.php</code> is parsed as JSON. If it's not valid JSON or the response is empty, the request will fail.</p> <p>If the URL is incorrect (can't be found...HTTP 404 error), then the request will fail.</p> <p>The default <code>type</code> of request is "GET", so if <code>get.php</code> doesn't allow "GET" (for some reason), it will return an HTTP error, and the request will fail.</p> <p>If there's an error on the server, it will likely return an HTTP 500 error, and the request will fail.</p> <p>Something to help debug would be to add the <code>error</code> option to the <code>$.ajax</code> call and see if that's called. Instead, I use the <code>.fail()</code> method...it does the same thing.</p> <p>Of course, the more direct way of debugging is opening your browser's console and viewing the AJAX request. It should show multiple details about it, that can help you determine any problems.</p> <p>It might seem as if the AJAX request was never executed/sent, because you don't see the <code>alert</code> in the middle. Well, just because the request wasn't successful, doesn't mean it was skipped. There are plenty of reasons (I named several above) why the request may fail. And the <code>.fail()</code> method will help you determine the cause.</p> <p>Also, the universal convention for handling deferred objects in jQuery is to use the <code>done</code> and <code>fail</code> methods, so that is an option. Of course, <code>$.ajax</code> has specific options you can specify (<code>success</code>, <code>error</code>, and <code>complete</code> - which is for something else), so that is also an option. You can also use special methods (<code>.success()</code>, <code>.error()</code>, <code>.complete()</code>) that are part of the object returned from <code>$.ajax</code>, but those are deprecated as of version 1.8 - take a look at the <code>.ajax</code> docs towards the bottom - <a href="http://api.jquery.com/jQuery.ajax/" rel="nofollow">http://api.jquery.com/jQuery.ajax/</a> . But here's how I'd set it up, which shouldn't be any different from yours, but does catch errors:</p> <pre><code>$.ajax({ url: 'get.php', dataType: 'json' }).done(data) { console.log("successful response"); }).fail(jqXHR, textStatus, errorThrown) { console.log("error: " + textStatus + ", " + errorThrown); }); </code></pre>

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