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Home/ Questions/Q 7438023
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T10:29:26+00:00 2026-05-29T10:29:26+00:00

I’m using Chrome to debug my websites. However, sometime I come up with errors

  • 0

I’m using Chrome to debug my websites. However, sometime I come up with errors like this one:

Error

Uncaught TypeError: Cannot set property 'display' of undefined

Stack Trace

f.extend.ajax
ClassLoader.performLoadWave

So, I know that it’s something in my Ajax call, probably in my success function. But, my success function calls a lot of things.

I’m not really looking for someone to help me debug my code, but just asking how I could debug this. I use a lot of console.log(); but they’re not really helpful, how can I properly find errors like this one?

EDIT

To add to my question, in this example, the error is in a file that gets loaded by my AJAX call, since the dataType is set to ‘script’, but how can I know this without having to dig up for a few hours?


Code

For anyone that cares about my code, here it is:

$.ajax({
    url: filePath,
    async: false,
    cache: !Settings.debugging,
    dataType: 'script',
    error: function(httpRequest, message, errorThrown){
        if(errorThrown == 'Not Found')
        {
            throw 'Include cannot be found.';
        }
        else
        {
            throw errorThrown + ': ' + (message == 'parsererror' ? 'This doesn\'t look like JavaScript!' : message);
        }
    },
    success: function(){

        $.each(newNeedsLoading, function(index, element){

            if(element !== undefined)
            {

                var className = element.className;
                if(window[className] && window[className]['included'])
                {
                    window[className]['included']();
                }

                className = 'Plugin_' + element.className;
                if(window[className] && window[className]['included'])
                {
                    window[className]['included']();
                }

                className = 'Plugin_Administration_' + element.className;
                if(window[className] && window[className]['included'])
                {
                    window[className]['included']();
                }

                if(element.completed)
                {
                    element.completed();
                }

            }

        });

        ClassLoader.isLoading = false;

    }
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T10:29:27+00:00Added an answer on May 29, 2026 at 10:29 am

    After some research, this question was pretty much a duplicate of this one:

    Debugging scripts added via jQuery getScript function

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