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