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    <p>You won't be able to use the <code>$ajax</code> helper here, since it just outputs a static script which can't be changed/influenced at "run-time" in the browser. It just wasn't made for something more complex than it is.</p> <p>So, you'll have to roll your own JS, which shouldn't be that hard though. All you need is:</p> <ol> <li>a Cake action that outputs a list of items based on the data it receives (shouldn't be hard)</li> <li>a bit of JS that figures out which brands and categories are selected (which you already have)</li> <li>another bit of JS that packages that data and sends it to the Cake action</li> <li>another bit of JS that updates the site with the list of items you received back</li> </ol> <p>I'd take a look at <a href="http://api.jquery.com/jQuery.post/" rel="nofollow noreferrer">jQuery's AJAX functions</a> to accomplish #3. If you POST the data in a format like this, it's very easily accessible in <code>$this-&gt;data</code> in Cake:</p> <pre><code>{ 'data[ModelName][categories]' : categories, 'data[ModelName][brands]' : brands } </code></pre> <p>Regarding your question:</p> <blockquote> <p>"How do I pass the js vars into the php method?"</p> </blockquote> <p>You don't. PHP runs on the server and is already finished by the time the Javascript runs in the browser. The only "communication" between JS and PHP is via standard HTTP GET and POST requests, and there it doesn't matter whether the request comes from a standard browser or JS or Flash or whatnot.</p> <p>The <code>$ajax</code> helper just has a bunch of pre-fabricated Javascript snippets it can put into your page, but your JS will not be able to "talk to" the <code>$ajax</code> helper in any way.</p>
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    1. POHow to pass javascript var into a ajax remoteFunction?
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    1. COHere is what I have so far: http://bin.cakephp.org/saved/55267. How can I exactly see how is the $this->data record? tried print_r and debug but got nothing :S. And second the php response im geting to my post is the full page, isnt that going to be bad to just update div #3? Btw thanks for your answer, this (JS -> PHP) was something I never had explained so well!
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    2. COLog stuff on the Cake site with `CakeLog::write('debug', $something)` (goes into `app/tmp/logs/`) and/or `console.log(something)` in Javascript, which you can see in Firefox using Firebug or Safari's or Chrome's Web Inspector's Javascript Console. To avoid the whole layout being rendered, you need to set `$this->layout = 'ajax'` in Cake.
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    3. COIf I use $this->layout ='ajax' then it won't render full page at first load, so I tried to create a function (menRefresh) only for the refresh of this page... but i get 404 @ firebug console with: jQuery.post("../items/menRefresh"). I think its because I don't have a view for that action (I'm really noob lol).
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