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

It looks like jquery is not correctly attached to the page.

Check your linking to jQuery.

Answer 2

It happened to me because of slim version of Jquery library. Full version of Jquery library includes animation.

Jquery CDN available at https://code.jquery.com/

Answer 3

Try keeping it inside

$(document).ready(function(){
// your code. and don't forget the '#' in front of item.
});

Looks like you're trying to call the function before jQuery / the DOM loads.

Answer 4

I had this error because I was using a slim version of jQuery. If you download the full version you should be okay.

Answer 5

Even if the selector didn't return any items in the collection the function call would have worked (not generated this error anyway) if jQuery was loaded correctly. Either there is a conflict in the page, or it didn't load at all. You can try

jQuery(myDiv).fadeOut("slow");

or look into why jQuery hasn't been loaded.

P.S.: don't forget the # in the selector if selecting by id.

Answer 6

Also, you probably forgot a # in the selector (unless you've got something like <item_1 /> 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.