That works fine TJ, thank you! I really appreciate the help.
Matt

On Jan 15, 5:17 pm, "T.J. Crowder" <t...@crowdersoftware.com> wrote:
> Hi Edd,
>
> > Just to confuse issues slightly, wouldn’t it be better to use #map
> > instead of #each?
>
> Why?  He isn't using the return value, so no need to create the return
> array.
>
> One concern in both cases, though, is whether Effect.SlideDown will be
> bothered by the extra parameter provided by both #each and #map's
> calls to the iterator.  According to the scripty docs, it expects the
> element and an optional options object, whereas #each and #map will
> pass it the element and the element's index in the array.  Oops. -
> grin-
>
> So perhaps better to replace
>
>     .each(Effect.SlideDown);
>
> ...in the various examples with something more expicit:
>
>     .each(function(elm) {
>         Effect.SlideDown(elm);
>     });
>
> ...or something like that.
> --
> T.J. Crowder
> tj / crowder software / com
> Independent Software Engineer, consulting services available
>
> On Jan 15, 3:19 pm, redheat <ecouch...@googlemail.com> wrote:
>
> > TJ,
>
> > Just to confuse issues slightly, wouldn’t it be better to use #map
> > instead of #each?
>
> > Edd
>
> > P.S., I haven’t tested, so #map may not work.
>
> > > If you know there will only be one and don't mind if the effect
> > > applies to all of them if (for whatever reason) there's more than one,
> > > Enumerable#each makes for a concise bit of code:
>
> > > document.observe("dom:loaded", function() {
> > >   $('editProducts').select('div.test').each(Effect.SlideDown);
>
> > > });
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