Question

So, with var $moving we are creating javascript object, right? But what is $('<li/>')? It is not even an html element.

Also, offset values describes the values of the $this element? So, in this case of the element #slidingMenu li:first?

var $menu = $("#slidingMenu");
var $selected = $menu.find('li:first');
var $moving   = $('<li/>',{
    className : 'move',
    top       : $selected[0].offsetTop + 'px',
    width     : $selected[0].offsetWidth + 'px'
});

Answer 1

• $("#slidingMenu"); takes the reference of #slidingMenu and $(‘#slidingMenu’) returns a jQuery object.

• $menu.find('li:first'); find the first li within the #slidingMenu.

• $moving = $('<li/>', properties) means, create a jQuery-wrapped li element on the fly but not added to DOM. For more.

• $selected is reference to jQuery object and $selected[0] gives an element reference/ JavaScript object.But it doesn’t means $selected[0] is a reference to the first object and $selected[1] to the second and so on.