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Home/ Questions/Q 8539057
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T11:16:21+00:00 2026-06-11T11:16:21+00:00

On this site I used pretty Photo image gallery. The problem is – when

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On this site I used pretty Photo image gallery.
The problem is – when a user clicks on the facebook like button – on his news is displayed only the name of the site.
what I would like is – when a user clicks fb LIKE button on a particular image, that image is shown on his news feeds.
can you help me make this?

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  1. Editorial Team
    Editorial Team
    2026-06-11T11:16:22+00:00Added an answer on June 11, 2026 at 11:16 am

    When working with Facebook, always ALWAYS check your website url with the Facebook debugger.

    Looks like the problem is that Facebook couldn’t featch the image probarly, so you will need to add a meta tag so Facebook can know the desired image for the prvded URL.

    Ex: <meta property="og:image" content="YOUR_IMAGE_PATH"/>
    

    Update 1:

    In order to modify the Meta tag value when the user changes the gallery image, you can use the following code to do so:

    $("meta[property=og\\:image]").attr("content", YOUR_IMAGE_PATH);
    

    note that we needed to escape the : character as mentioned in the documentation

    Update 2:

    you will need to alter those functions in order to make it to work:

        $pp_gallery.find('.pp_arrow_next').click(function(){
                $.prettyPhoto.changeGalleryPage('next');
            // here you will need to read the current image url, then assign it to our facebook line.
        $("meta[property=og\\:image]").attr("content", YOUR_IMAGE_PATH);
                $.prettyPhoto.stopSlideshow();
                return false;
            });
    
        $pp_gallery.find('.pp_arrow_previous').click(function(){
                $.prettyPhoto.changeGalleryPage('previous');
            // here you will need to read the current image url, then assign it to our facebook line.
    $("meta[property=og\\:image]").attr("content", YOUR_IMAGE_PATH);
                $.prettyPhoto.stopSlideshow();
                return false;
            });
    
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