Because .each() doesn't return anything. var a = [0,1,2].each(fn); console.log(a); // undefined
On Mar 6, 3:15 am, electronbender <[email protected]> wrote: > Why use this ? > Just use imgArray = $$('#img_holder img') > and then imgArray.each.... > > On Mar 3, 3:52 am, Matt Thomson <[email protected]> wrote: > > > Hi does anyone know how I would do this: > > > html: > > > <div id="img_holder"> > > <img src="img1.gif" /> > > <img src="img2.gif" /> > > <img src="img3.gif" /> > > </div> > > > javascript: > > > this.imageArray = $('img_holder').getElements('img'); > > > this.imageArray.each(function(el,index) > > { > > el.addEvent('click', function() > > { > > alert(index + 1); > > //if image 1 is clicked I want it to alert "1" > > //if image 2 is clicked I want it to alert "2" > > //presently index is undefined > > }); > > > }); > > > Thanks, > > > Matt.
