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Editorial Team
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Asked: 2026-05-20T01:39:47+00:00

I’m in the process of writing a website that includes a reasonably large gallery.

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I’m in the process of writing a website that includes a reasonably large gallery. First page of the gallery the user will be displayed a bunch of thumbnail images with a url of: website.com/gallery.php

When they click a thumbnail image, if javaScript is turned off it will follow the url in the href and go to a page called gallery.php?img=67. If javaScript is turned on the href click will not execute, instead it will perform an ajax request to display the larger image and some text about it. The url changes to gallery.php#!img=67. The back button will take you back to the thumbnails, pressing f5 will keep the big image displayed with the text. If someone copies the address with the #! and sends it to someone they will get the same image displayed (assuming the receiver has javaScript turned on).

My question is, have I sorted this out correctly for google to index the individual gallery pages? Will google index them twice, once with the ?img=67 and once with the #! and if so is that a bad thing? I’m using javaScript/Ajax to preload the larger images once the thumbnail page is loaded for speed. I’ve read a lot of backlash against using hasbang ajaxy things recently and wondering if you think can justify using it here?

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  1. Editorial Team

    Google will follow your links and index the ?img=67 pages, and will not index your #! pages, because it can’t see those links. You can tell Google about those links by doing the following:

    1. Add <meta name="fragment" content="!"> to the <head> of your document, and
    2. Handle requests for /?_escaped_fragment_= by returning an “HTML Snapshot” of your page that has all your #! links in the <A> tags.

    Also, to make the most of this feature, you should also handle requests for /?_escaped_fragment_=img=67 by returning an HTML snapshot page with the big image displayed. Remember, GoogleBot doesn’t execute Javascript. Using the #! URL tells Google to retrieve an alternate version of the page (A version where #! has been replaced with ?_escaped_fragment_=) that should render without Javascript.

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