Why do you have a 2nd arg in your lambda called undefined? Also, you will want to check for the existence of console.log before you call it, otherwise you will get a runtime error if firebug is switched off.
On 4 Jul 2011 20:30, Sam3k <srolda...@gmail.com> wrote: Why does the following script works within the application code but it doesn't when I type it in the firebug console. It throws the following error: TypeError: debug.log is not a function My goal with this script is to make an object from where I can log things to the console, store javascript speed performance in different areas, and turn off if no cookie is set. file: Debugger.js (function (window, undefined) { function Debugger() { this.test = 'xxxxxx'; } Debugger.prototype.log = function(data) { console.log(data); } var _debug = new Debugger(); window.debug = _debug; debug.log('DEFE"DEFEF'); // this works })(window); debug.log('OUTSIDE DKEGFYG'); // this works too Also any constructive feedback is greatly appreciate it. -- To view archived discussions from the original JSMentors Mailman list: http://www.mail-archive.com/jsmentors@jsmentors.com/ To search via a non-Google archive, visit here: http://www.mail-archive.com/jsmentors@googlegroups.com/ To unsubscribe from this group, send email to jsmentors+unsubscr...@googlegroups.com -- To view archived discussions from the original JSMentors Mailman list: http://www.mail-archive.com/jsmentors@jsmentors.com/ To search via a non-Google archive, visit here: http://www.mail-archive.com/jsmentors@googlegroups.com/ To unsubscribe from this group, send email to jsmentors+unsubscr...@googlegroups.com