UPDATE [Sun 08/26/2012] This answer has become so popular that I decided to make it into a full-fledged blog/tutorial
Please visit My Blog Here to see the latest in easy access information to working with tabs in jQueryUI
Also included (in the blog too) is a jsFiddle
Update! Please note: In newer versions of jQueryUI (1.9+), ui-tabs-selected has been replaced with ui-tabs-active. !!!
I know this thread is old, but something I didn't see mentioned was how to get the "selected tab" (Currently dropped down panel) from somewhere other than the "tab events".
I do have a simply way ...
var curTab = $('.ui-tabs-panel:not(.ui-tabs-hide)');
And to easily get the index, of course there is the way listed on the site ...
var $tabs = $('#example').tabs();
var selected = $tabs.tabs('option', 'selected'); // => 0
However, you could use my first method to get the index and anything you want about that panel pretty easy ...
var curTab = $('.ui-tabs-panel:not(.ui-tabs-hide)'),
curTabIndex = curTab.index(),
curTabID = curTab.prop("id"),
curTabCls = curTab.attr("class");
// etc ....
PS. If you use an iframe variable then .find('.ui-tabs-panel:not(.ui-tabs-hide)'), you will find it easy to do this for selected tabs in frames as well.
Remember, jQuery already did all the hard work, no need to reinvent the wheel!
Question was brought up to me, "What if there are more than one tabs areas on the view?"
Again, just think simple, use my same setup but use an ID to identify which tabs you want to get hold of.
And you want the current panel of the second tab set:
var curTabPanel = $('#example-2 .ui-tabs-panel:not(.ui-tabs-hide)');
And if you want the ACTUAL tab and not the panel (really easy, which is why I ddn't mention it before but I suppose I will now, just to be thorough)
// for page with only one set of tabs
var curTab = $('.ui-tabs-selected'); // '.ui-tabs-active' in jQuery 1.9+
// for page with multiple sets of tabs
var curTab2 = $('#example-2 .ui-tabs-selected'); // '.ui-tabs-active' in jQuery 1.9+
Again, remember, jQuery did all the hard work, don't think so hard.
Excellent, thanks! If you want you could also provide this as an answer to stackoverflow.com/q/1864219/11992.