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Editorial Team
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Asked: 2026-05-26T05:21:02+00:00

I just finished a small website 100% based on ajax.But i have a problem

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I just finished a small website 100% based on ajax.But i have a problem in updating the links of the facebook Like buttons.

I used simply the facebook generated code and i included the graph meta tags. when the users presses a button it loads a new youtube video, i update the meta tag

<meta property="og:url" content="http://www.mySite.com/"/>

using jQuery this way :

$("meta[property=og:url]").attr("content", nUrl);

nUrl is the new page url that contains the new randomly selected video.

 When i hit “Like” it simply grab hit “Like” it simply grab the original og:url meta tag content and now the new on set by ajax.
 How to update with javascript the like button url ? What is wrong in what i did ?
Thank you

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1 Answer

  1. Editorial Team

    So here’s what i did to solve my problem for other people who have the same:
    Add this to your site :

    <html xmlns="http://www.w3.org/1999/xhtml"
  xmlns:og="http://ogp.me/ns#"
  xmlns:fb="http://www.facebook.com/2008/fbml" >

    This will add the fbml to your site. Then use the FBML version of the like button (or comment system, or whatever), for like i used this :

    <span id="fbLike"><fb:like href="http://www.your-site.com?v=1" send="false" layout="button_count" width="80" show_faces="true"></fb:like></span>

    Now to have a new facebook like button each time the user make an Ajax action, just add this jquery code to where ever you handle this action in your client-side code:

    var nUrl = "http://www.your-site.com?v="+toPlayId.substring(1);
$( '#fbLike' ).html('<fb:like href="'+nUrl+'" send="false" layout="button_count" width="80" show_faces="true" />');
        if (typeof FB  != "undefined"){
        FB.XFBML.parse(document.getElementById('fbLike'));}

    You basically just replace the where your button is with a new one.
    nUrl is the new link you want your new button to point to.
    You don’t need a real new URL just add a parameter to the url so that your ajax knows what to display when he finds the parameter.
    That’s all.

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