I just created a simple webpage where people can browse funny photos and share it with their friends.

But I'm having problems with the Facebook like button. I have used the Facebook like button before on a static URL with iframe. And that seemed to work fine.

I'mm using the XFBML version and not the iframe version because the iframe version affects the pages layout.

This is the code I got from Facebook:

<div id="fb-root"></div><script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script><fb:like href="" send="false" layout="button_count" width="450" show_faces="false" font="tahoma"></fb:like>

Facebook says that "href - the URL to like. The XFBML version defaults to the current page. " So i left that one open.


When users click Like the counter won't update. It still shows nothing as if no one ever clicked the button.

3 Answers 3


Your problem might be with the Javascript SDK.

According to FB, it's needed for rendering XFBML plugins. It looks as if you need to register, too.


Also, according to the same page,

In order to use XFBML on your webpage, you must add an XML namespace attribute to the root element of your page. Without this declaration, XFBML tags will not render in Internet Explorer.

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


Try removing the href attribute altogethr:

<div id="fb-root"></div><script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script><fb:like send="false" layout="button_count" width="450" show_faces="false" font="tahoma"></fb:like>

href is a means to add your site or page as a link to the code. It is generally put in a format with an open and closed <a> tag as follows:

<a href="">Your Web or Facebook page</a>

This code will help link your likes, but needs the href link to know what it is linking. So, removing the href is not the answer - and this code alone is not the answer:

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

The two must be combined.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.