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

Also, you probably forgot a # in the selector (unless you've got something like in the markup).

var myDiv = '#item_' + itemID;

jQuery uses CSS selectors to search for elements, so without the #, you'd get every element with the tag item_x instead of the ID.