You kidding me? Thats way too easy! :-)
>From now on, if I'm working on something and it takes me more than 10
minutes, I'm posting the question. :-)
Question:
I was thinking usng .apply here. But I wasn't 100% sure how that
worked.
In general, if I see something like this:
xxx.apply(x,[y,z]);
>From what I am understanding, x is "this" inside the function xxx and
x,y is the parameter to the function?
Is that correct in general when using .apply? or am I off base
completely? <g>
---
HLS
On Oct 9, 1:49 am, "Michael Geary" <[EMAIL PROTECTED]> wrote:
> > From: Pops
> > var how = (settings.show!="")?settings.show:"show";
> > eval("$box."+how+"()");
>
> foo.bar means the same thing as foo['bar'], so this code is the same as:
>
> var how = (settings.show!="")?settings.show:"show";
> $box[how]();
>
> Or a very clean and simple version:
>
> $box[ settings.show || 'show' ]();
>
> -Mike
