Yes, you can certainly call $() with a variable instead of a string literal.
jQuery is just JavaScript, $ is just a function name, and you can always
rewrite:
alert( 'woot' );
as:
var woot = 'wo' + 'ot';
alert( woot );
So you just need to do some debugging.
What are the values of nID and speed? Are they what you expected?
Is the jQuery object returned by $(theID) empty or does it include the DOM
element?
You can check these in Firebug with this code:
function fadeOut(inID, speed) {
console.debug( nID, speed );
var theID = "#" + inID;
console.debug( $(theID) );
$(theID).fadeOut(speed);
}
If you have a valid DOM element, then the problem must lie in the fadeOut
function.
-Mike
> From: Donovan Walker
>
> i have some code that looks like
>
> $("#myID").fadeIn("slow");
>
> and it works fine, I have some other code that looks like
>
> function fadeOut(inID, speed) {
> var theID = "#" + inID;
> $(theID).fadeOut(speed);
> }
>
> And it does NOTHING.
>
> The fadeOut function I wrote because I need to be able to
> dynamically assign which objects get faded out. I have NO
> idea what the objects will be named.
>
> Why doesn't creating the string 'theID' and using that as the
> address work? How do I make it, or something like it work?
