jQuery Event Listener for JavaScript objects

Question

Is there a way to add an event listener/handler to a JavaScript object? Preferably using JQuery.

Such as:

var foo;
$(foo).bind('change', function() {
   alert("Foo has changed!");
 });

I have tried this, but nothing seems to happen. Does this only work with DOM elements?

EDIT: I need an event fired every time that the audio or video tags throw an error. Originally, I was using an interval to check whether or not the error, media.error, object was null, but this uses excess processing power and I would like to avoid it.

EDIT 2: Apparently I was going about it wrong, easiest way I found was to add the onerror property to the video/audio tag.

Nope · Accepted Answer · 2012-11-21 16:27:57Z

1

I agree with Cheeso that it's more important for you to state what you actually want to do, however one workaround for your specific question could be to store your variable within an object and only provide access through getter / setter, then you can do what you want in the setter. e.g.

function data() {
    var foo = 0;

    this.setFoo = function(newVal) {
        foo = newVal;
        alert(foo);
    };
}

var theData = new data();

theData.setFoo(5);

edited Nov 21, 2012 at 16:27

Nope

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answered May 14, 2011 at 0:02

tomfumb

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Comments

Cheeso · Accepted Answer · 2011-05-13 23:40:03Z

0

Yes, that's correct. You cannot make a "variable watcher". There is no event fired for variables when they are changed.

What are you really trying to do?

answered May 13, 2011 at 23:40

Cheeso

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Comments

Jonathan Hendler · Accepted Answer · 2011-07-28 05:35:20Z

0

You could try: http://higginsforpresident.net/js/static/jq.pubsub.js

See http://weblog.bocoup.com/publishsubscribe-with-jquery-custom-events

Or use a framework like backbone/underscore or knockout.js.

answered Jul 28, 2011 at 5:35

Jonathan Hendler

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Comments

user2009879 · Accepted Answer · 2013-01-25 05:44:46Z

0

HTML/Elements/audio

    var foo;
    $(foo).bind('error', function() {     
       //your code here
    });

answered Jan 25, 2013 at 5:44

user2009879

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