Why does jQuery throw the error `fadeOut is not a function'?

Question

I am using jQuery and put this code in my javascript:

function HideMe(itemID) {
    var myDiv = \'item_\' + itemID;
    $(myDiv).fadeOut(\"slow\");
}
         


        

Answer 1

Do you have another javascript library on that page? It seems you have the hide function, and the $ defined (prototype, for example, also has an hide function).
If that is the case, try:

jQuery("#item_0").fadeOut("slow");