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I have some sub-domains. I use "site:mydomain.com" in Google search input to find out the number of pages Google has indexed from my domain overall.

How can I find out which sub-domain has been indexed more? Can I do it for others domain?

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2 Answers 2

up vote 2 down vote accepted

You can also use the site: search operator to return the number of results for subdomains and URLs.

For example, if you use site:voice.google.com you'll see only a few hundred results (at the time of this post). Same with URLs like site:en.wikipedia.org/wiki/Google which returns roughly ten results.

For even more details on indexed URLs, you can add each subdomain as a site in Google Webmaster Tools.

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but what if I don't know the sub-domain names ? for example if I want to check other company sub-domains indexed result ? –  hd. Sep 21 '13 at 6:51
I don't know of a way to obtain a list of indexed subdomains for someone else's domain without first knowing them, other than using a script/tool to query Google for their site:domain.com and parsing out the subdirectory string pattern in the returned URLs to create a list of subdomains. That should be doable with Perl. –  dan Sep 21 '13 at 7:36

It is simple! All You need is to add an extra dot after "site:" and before "domain.com". For Example, if you want to check index pages for "about.com" you type "site:about.com" (without quotes) in Google. , But if you want to check the indexed subdomians of "about.com" Simply type "site:.about.com" (without quotes).

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Although this will return a list of various subdomains indexed, it doesn't answer the question of how to find out which subdomain is indexed more. –  dan Jan 16 at 20:34
Well, It would be a 2 step process. First, follow the syntax I shared above then do what you normally do to check how many pages has been indexed for a site. Don't forget to add your desired subdomain. e.g. "site:subdomain.domain.com" –  Rehan Gillani Jan 17 at 21:03
then do what you normally do <- that doesn't really answer anything, and you're just repeating what I indicated in the answer above. –  dan Jan 17 at 21:41

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