Try adding these four lines immediately before the debug.log() call inside
the wrapper function (the one that fails):
console.log( _debug );
console.log( window.debug );
console.log( debug );
console.log( debug.toString() );
That may lead to this interesting idea... What if you change these lines:
var _debug = new Debugger();
window.debug = _debug;
to:
debug = new Debugger();
-Mike
On Mon, Jul 4, 2011 at 12:42 PM, Sam3k <srolda...@gmail.com> wrote:
> Environment:
> FireFox 5
> Firebug 1.7.3
>
> On Jul 4, 3:30 pm, 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.
>
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