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Anonymous
June 19, 2019
Question

Loading an Image from URL String Variable?

  • June 19, 2019
  • 1 reply
  • 356 views

Hi everybody,

I have a situation where I have a gallery that we show internally with our work peeps. We want to be able to see this gallery with a particular tag showing before we post to the client. However, we don't want the client to see it.

So rather than adding and image tag for 7+ images then removing it everytime, I thought it'd be great if I could somehow access CSS to show and image from the URL string by adding a variable reference. Like...

www.mygallery.com/gallery.html&showIMGtagClass=true

Is that possible somehow if I have a CSS class referring to a <div><img></img></div> ??

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    1 reply

    BenPleysier
    Community Expert
    BenPleysierCommunity Expert
    Community Expert
    June 20, 2019

    Using a URL variable it would be possible to show/hide parts of the HTML, however, this could be added by those that are not supposed to see the tag thus nullifying the intended outcome.

    There are many other ways of achieving what you want, but this is purely dependent on the application. Is it a PHP page?  Do you have a secured section?

    To answer your question, we would need a lot more detail.

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      C
      CCM19
      Participating Frequently
      June 20, 2019

      No, it's not PHP. It's entirely client-side. I'm using something called fancybox http://fancybox.net/ ​

      There's not a whole lot of security to be concerned with. It's a shared link showcasing ads amongst probably a dozen people and it's only up for a couple months. I post this same showcase page in a template for each one. So I'd have the URL for the template. If I could then add a URL variable for my team to add at the end to see an additional overlay (AdChoices) that would be great.

      We don't want the client to see the AdChoices badge because they would get confused, but we like to see it so we can know where to position stuff. So if I could enter a URL like flipping a switch, that would be great.

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