Tuesday, May 29, 2012

How to debug AJAX

One of the things I do is mentor junior and young web developers. One of the most asked question is how to debug AJAX, XMLHttpRequests, also known as XHR.

Well, it is very easy. In most modern web browsers, there are web developer tools. There are native tools built in IE, Safari and Chrome. They are often called Web/Developer Inspectors. With Firefox, you can use Firebug.

Each browsers are different in how you access the XHR console. I won't go into details but it is very easy to find. Launch your web inspector or right-click and "inspect element" before you make your AJAX call. When the web developer tool (called by various names and differing across versions of the same browsers). Look for resources or Network. Then narrow down for scripts or XHR (short for XMLHttpRequest) options.

Below are screenshots of Chromium under Ubuntu 10.10 and Safari under Mac OSX.

Whenever I make an AJAX call, I look for the script/resource I am calling. In my screenshot above, I am both calling a script called ajax/_ajax_sample.php.

Here is the HTML I use to call my AJAX. It is a simple AJAX post, passing some variables in a form-like POST.

 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">  
 <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">  
 <script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js" type="text/javascript"</script>  
 <script type="text/javascript">  
 $(document).ready(function() {  
           type: "POST", url: "ajax/_ajax_sample.php", data: { action: 'update file', user_name : 'joe blow', user_id: 100 },  
           dataType: "json",  
           success: function(data){  
                if(data.status == 'success'){  
                     if (data.post_action == 'show_div') { $("#results_div").html(data.msg);     }  
                     } else {  
                          alert( "There was an error: " + data.msg);  
                } // end of data.status = success  
           }); // end of ajax  
 <div id ="results_div" style="display:block;width:400px;background-color:#CCCCCC">Nothing Yet</div>  

If you go back and look at the two browser screen grabs, I am passing the following POST data. This is indicated in the "Header" tabs. The Header's tab is what you post to your resource. You will see other behind-the-scene info such as Request Headers, the method, and the Response Headers (how the server informs the browser it will process).

For our purposes, they only thing I am looking for is the "Form Data" of what I posted. I just need to know if what I posted was correct. They are:

 action:update file  
 user_name:joe blow  

My script or resource will takes the Form Data Post  from my AJAX call in the same manner you would be sending data through a HTML form or query string.

This is how you check you are posting the right data or not.

When your CGI resource or script gets the request, it will eventually return some data. The web inspector will also you give a read-out between the time it gets the request to the time it finishes. You can see how efficient your back-end code is.  See below for example of latency and processing time.

The results  are often HTML data or JSON. To see what it returns, simply click on the "Content" tab. There may be other tabs such as preview or JSON to show the same results.

As you can see in the next screenshot, my result was in JSON. If you go back and look at the same HTML file posted, I take the JSON and use it to execute another javascript function based on what I got back. Namely, filling the data into a DIV called results_div. The status return a 'success'.

 {"status":"success","msg":"your ajax worked mr\/mrs joe blow","post_action":"show_div","record_id":40}  

Now, in this next screenshot, I introduced some errors. My HTML page tried to make 2 ajax calls and got 404 errors which means the page could not find the two file resources.

The third error, I introduced some errors on my PHP script. The web inspector returned a 500 Internal Server Error. My AJAX is passing the correct data but there was something wrong with the PHP. 

Fortunately, I have a terminal window above to tail (monitor) my apache logs for PHP errors. By running tail -f  on my apache error_log, I can see the error is in line 6 of my PHP script. It is handy to have a big enough monitor to see and debug your web application.

This is the most common problem I see when a new web developer starts to use plugins, free lightboxes, modals, and unknown scripts. They simply see an animated AJAX spinning wheel and can't figure our why the page isn't do anything. By using their browser developer's inspection tool, they can see if they are not linking correctly or if simply their back-end is broken.

That is pretty much it. Debugging AJAX isn't so difficult.

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