Question

Classic Like button code :

<div class="fb-like" style="position:relative;float:left;" data-href="http://www.mywebsite.com/?IDL=1" data-send="false" data-layout="button_count" data-width="450" data-show-faces="false"></div>

<div id="fb-root"></div>
<script type="text/javascript">    
    (function (d, s, id) {
        var js, fjs = d.getElementsByTagName(s)[0];
        if (d.getElementById(id)) return;
        js = d.createElement(s); js.id = id;
        js.src = "//connect.facebook.net/en_US/all.js#xfbml=1";
        fjs.parentNode.insertBefore(js, fjs);
    } (document, 'script', 'facebook-jssdk'));
</script>

it render the I Like Facebook button to the http://www.mywebsite.com/?IDL=1 page. Now, suppose I’d like, due to some action on the page, change the destination to http://www.mywebsite.com/?IDL=2 : how can I do it? I need to destroy and create again the button? I mean dynamically, without “refresh” the page (which will be easy)

Answer 1

You’ll need to destroy the already initialized button and replace it with another one then call FB.XFBML.parse() which will reparse the XFBML and re-initialize the new button.