Fire Chosen plugin after creating Select box via Xajax

Question

I want to bind the Jquery plugin Chosen on a select box that I create via Xajax, the Select does not figure in the DOM when the page is loaded.

So I start by returning the select box and then I execute .chosen() on the box ID:

<? $objResponse = new xajaxResponse();
$objResponse->assign($selectID, 'innerHTML', $codeThatCreatesSelectBox);
$objResponse->script('$(function(){ $("#selectID").chosen(); });');
return $objResponse; ?>

But I'm obviously doing something wrong... How could I do this ?

EDIT

I tried waiting for ajax returns before firing the plugin by using ajaxComplete() :

$objResponse->script('$("#selectID").ajaxComplete(function() {$("#selectID").chosen(); });');

But it doesn't change anything..

Ricardo Souza · Accepted Answer · 2012-12-05 14:22:27Z

1

You can create a callback to do this. From the documentation: http://www.xajax-project.org/En/docs-tutorials/upgrading-from-xajax-0-2-x-to-0-5/

myCallback = xajax.callback.create(100, 10000);
myCallback.onRequest = function()
{
    xajax. $('loadingMsg').style.display = 'block';
};
myCallback.onComplete = function()
{
    xajax. $('loadingMsg').style.display = 'none';
};

// then, on the PHP side, specify the callback option when registering your function:
$xajax->register(XAJAX_FUNCTION, 'myFunction', array(
    'callback' => 'myCallback'
));

So you can call your jQuery plugin code inside the desired callback.

answered Dec 5, 2012 at 14:22

Ricardo Souza

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