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Is there a way that AD can be used to block access to certain websites (e.g. Facebook) between certain times? Ideally I'd like to only to be available at (say) lunchtimes.

(I can live with the fact that someone could work around it using a proxy... most of my users won't know how to do that)

Thanks

4 Answers 4

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In a word, no. AD has no interaction with the users browsing experience, web traffic is directed through your proxy if you have one, then your gateway and so it is at these points you would need to filter the traffic. AD is used to authenticate users, validate the users access rights for resources and can be used to prevent the user accessing the internet at all by blocking access to browsing applications, or directing to a non-existant proxy, but it cannot filter sites.

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  • I agree with your description of the traffic routing... but I would have thought that the DNS request for facebook.com (or whatever) would go through AD.... is there anything that could be done to block this between certain times?
    – cagcowboy
    Commented Apr 30, 2009 at 12:15
  • DNS request go through the DNS server, which is more than likely an AD integrated zone, so its stored in AD, but thats about it, AD does not monitor DNS requests
    – Sam Cogan
    Commented Apr 30, 2009 at 12:48
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This sort of functionality is offered by all good web content filter systems, and some bad ones.

I work for SmoothWall - our "Network Guardian" software will enable you to do just what you describe.

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I have checked out AD and there's no way you can blocked website but use Content Advisor (GPO -> User Config -> Windows Settings -> IE Maintenance). You can add the allowed/blocked URLs on the Content Advisor Approved Sites taB

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not sure where u went to school but YES AD can blcok website... it can add a password to view the page. if the password is entered correctly then they are allowed to view the website

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    Hi, James - you have no idea what you're talking about or what the OP was asking. Additionally, you answered (badly) on a 4 year old question that had already been answered.
    – Dan
    Commented Oct 16, 2013 at 15:18

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