How to get mailbox user identity list and inject it to a command?
For some unknown reason, the "Full Access Permission" of every mailbox in my Exchange 2007 has an SID stuck to it. Since it's shown as an SID instead of a name, it must be some "leftover" from Exchange 2000 or during migration to Exchange 2007. I want to remove this unknown SID from every mailbox's "Full Access Permission". I have about 60 mailboxes. I have removed the SID from one mailbox and got the command to do so. It is: Add-MailboxPermission -Identity 'CN=userABCD,OU=OU-XYZ,DC=domain,DC=com' -User 'S-1-5-21-..............' -Deny -AccessRights 'FullAccess' I am thinking of getting the list of all mailbox, but the problem is the Identity part. Not only CN is different for every user, the OU part is different. What command should I use to get the list of all identity? I have tried get-identity but such command doesn't exist. To be honest, I'm not good at all in PowerShell. TIA
October 13th, 2009 4:57pm

If you want to remove or add one user to all mailboxes you can add the get-mailbox in front of the add-mailboxpermission. get-mailbox | Add-MailboxPermission -User 'S-1-5-21-..............' -Deny -AccessRights 'FullAccess'Chris Christopher C
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October 13th, 2009 5:12pm

Yes, this works. Thanks :) ... although I don't understand how PowerShell determines how the list data is to be used for -Identity parameter but not another parameter.
October 13th, 2009 6:54pm

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