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If a website a called foo.com has the following css,


and the css id LinkBuilder is actually not getting used anywhere in the html of foo.com, would Google bot crawl it as a backlink from foo.com to LoremEpsum.com and LoremEpsum.com gets the link juice from foo.com without getting displayed on the website?

Please help out. Thanks

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Explain your question more clearly. There's no such thing as "CSS tags." – Su' Jun 6 '12 at 20:40
@Su': Done. Hope it turns from -1 to +1 now :) – Fahad Uddin Jun 6 '12 at 20:43
Please read how to ask and the FAQ and then come back here and improve your question. You include too little information for us to help you. – user 99572 is fine Jun 6 '12 at 21:07
why would this effect anything about any site? Is having a unused css property going to affect rankings? And, I'm still not sure what you mean by crawled: if google loads css files, then of course it's going to see all of the properties, even if they aren't going to be used. – Hamlet Jun 6 '12 at 22:25
@Christofian: If someone has left links to another website like this, then would such links be indexed? – Fahad Uddin Jun 6 '12 at 22:42
up vote 1 down vote accepted

First of all, that's a background property, so I'm not sure why google would think it's a link.

However, assuming that you meant does it index the background image. I'm going to assume no, because if you look at their image guidelines, you will find that they make no reference to background images, and all of their advice seems to be geared towards the <img> tag.

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Linking to images on other domains in a CSS style sheet is not counted as a back link. It's not an actual anchor tag or <a> to be exact. It's not the type of link which carries weight with search engines.

While an image can be indexed in Google images, it's being indexed from your web page having it displayed.

If you run a search for jquery you do not see http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js appear in search results yet this is linked to from thousands of domains, just not linked to by <a> tags

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