What about this:
function swapImages(imageName) {
$("#beautyImage").html('textaroo').fadeIn('slow');
}
On Thu, Oct 21, 2010 at 2:52 PM, Chad Gray <cg...@careyweb.com> wrote:
>
> If I do this it snaps into place also. No fade in.
>
> function swapImages(imageName) {
> $("#beautyImage").html('<img
> src="/output/9305B06559F4D04C37ED27103EBFB959.jpg"
> style="margin-left:5px;">').fadeIn('slow');
> }
>
>
>
> -----Original Message-----
> From: Michael Grant [mailto:mgr...@modus.bz]
> Sent: Thursday, October 21, 2010 2:43 PM
> To: cf-talk
> Subject: Re: (ot) jQuery
>
>
> I don't see why it wouldn't be. Does fadeIn() work if you do it on it's
> own?
>
> On Thu, Oct 21, 2010 at 2:40 PM, Chad Gray <cg...@careyweb.com> wrote:
>
> >
> > Thanks for the suggestions, but still now luck. Maybe the get() is not
> > compatible with fadeIn()?
> >
> >
> >
> > -----Original Message-----
> > From: Tony Bentley [mailto:cascadefreehee...@gmail.com]
> > Sent: Thursday, October 21, 2010 2:39 PM
> > To: cf-talk
> > Subject: Re: (ot) jQuery
> >
> >
> > function swapImages(imageName) {
> > $.get('/shop/act_getBeautyImage.cfm?beautyImage='+ imageName +
> > '&width=372&height=465',{},function(data) {
> > $("#beautyImage").html(data).fadeIn('slow');
> > });
> > }
> >
> >
> >
> >
> >
>
>
>
>
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