: Should I use rel="me" or itemprop="sameAs" when linking to social networks like Google+ I'm currently using this: <a href="http://twitter.com/username" rel="me">Twitter</a> <a href="https://www.facebook.com/pages/userf

@Smith883

Posted in: #Google #Rel #SocialNetworks

I'm currently using this:

<a href="http://twitter.com/username" rel="me">Twitter</a>

<a href="https://www.facebook.com/pages/userfacebookpage" rel="me">Facebook</a>

<a href="http://www.pinterest.com/username/" rel="me">Pinterest</a>

<a href="https://www.google.com/websiteprofile" rel="me">Google+</a>

<a href="https://www.youtube.com/c/userchannel" rel="me">Youtube</a>


I see in this answer that:


rel=”me” Defines The Person

While the rel=”author” attribute is new markup, rel=”me” is not. In
fact, for a number of years now Google has encouraged its use. It
should be used in links that point from a social networking profile
back to your website and vice-versa. This tells Google that the same
name and bio information on separate websites is actually the same
person.


and Google Developer's Specify your social profiles to Google help doc says this:


Use markup on your official website to add your social profile
information to the Google Knowledge panel in some searches. Knowledge
panels can prominently display your social profile information. [...]

Note: As long as you use the same schema.org types as the example, you
can also use microdata or RDFa markup formats instead of JSON-LD. For
example, in microdata, visible links can be marked up by in a
structured data SPAN, like this:


<span itemscope itemtype="http://schema.org/Organization">
<link itemprop="url" href="http://www.your-company-site.com">
<a itemprop="sameAs" href="http://www.facebook.com/your-company">FB</a>
<a itemprop="sameAs" href="http://www.twitter.com/YourCompany">Twitter</a>
</span>


Do I need the rel="me" and the itemprop="sameAs" or will one do the 'job' of both?

Google doesn't support itemprop="sameAs" for Pinterest and Twitter already uses rel="me" in your website field.

This may be an obvious question but I'm lost.

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