check if jQuery UI tabs have been initialized (without checking for class)

Question

Using jQuery UI tabs - and I've written a plugin that integrates with ui tabs. I've got the plugin setup to initiate jQuery UI tabs if it .tabs() hasn't been called, but this just does a simple class check:

 if(!$globalTabs.hasClass("ui-tabs")){
    $globalTabs.tabs();
 }

But this is problematic, because often to avoid FOUC, developers add in the UI classes to the the tabs to get a better initial render before document.ready.

I could check for a different class, such as `ui-widget1, but wondering if there's another/better way?

Issuant

But why do you care if it's been called or not? You can always call .tabs() again without anything bad happening. – Specie

@Specie - it's usually good practice to avoid making unnecessary calls. – Issuant

Agreed, but you're worried about making one (possibly unnecessary) call? Look at this fiddle: jsfiddle.net/gBZbd. I'm calling tabs 1000 times, and it only takes 10 milliseconds. Seems like an awfully big premature optimization to me. @Frédéric Hamidi is absolutely correct in his answer, but AFAIK, this isn't documented anywhere. That code could break at any time. Just my two cents of course. – Specie

@aquinas, this is actually documented, see my updated answer. – Measles

Ah ha. I guess I've seen plugins do that but didn't realize it was "officially" sanctioned. Good reference. +1. (But I still think worrying about one call to tabs() is not worth thinking about. :) – Specie

@Specie - we have a dinosaur of a site - it's huge/enterprise level and I try to minimize calls as much as possible, because our codebase stretches back for years and years, and a ton of developers have had their hand in the pot. So the code is bloated, so I try to refactor and minimize impact as much as possible. Someone's got to be the shepherd of the code... – Issuant

Answer 1

You can query the attached widget with data():

if (!$globalTabs.data("tabs")) {
    $globalTabs.tabs();
}

This behavior is documented in the Widget factory page of jQuery UI's Development & Planning Wiki:

• Plugin instance accessible via $( "#something" ).data( "pluginname" )

  • A reference to a jQuery object containing the DOM element is available as a property of the instance as this.element, so it is easy to go back and forth between the object and the element.

Update: From jQuery UI 1.9 onwards, the widget key becomes the widget's fully qualified name, with dots replaced with dashes, as in:

if (!$globalTabs.data("ui-tabs")) {
    $globalTabs.tabs();
}

Using unqualified names is still supported in 1.9 but is deprecated, and support will be dropped in 1.10.

Extradition

+1 the answer. BUT with regard to my comment about premature optimization: running the if statement actually makes the code slower: See: jsfiddle.net/gBZbd versus jsfiddle.net/gBZbd/2. – Specie

@aquinas, indeed, but the question does not mention performance, so we do not know if it's about speed. The questioner might actually need to know if a widget has been instantiated on an element, for reasons we cannot guess. So, I tried to answer that, just in case :) – Measles

Answer 2

somehow it causes some error when i try to check the instance.

i need to reinitialize the tabs and for me a try catch did the trick:

try {
    $('.mytabs').tabs('destroy');
} catch (exception) {}

and after that just initialize again:

$('.mytabs').tabs();

Hubbub

May be due to newer jQuery UI versions, which changed some of the methods such as "data". – Issuant

unfortunately i am a little bit limited by having to support ie8... so updating plugins has to be done very carefully (or better don't update anything :D ) – Hubbub