Reputation: 81
I want 1 list with all users there are member of 2 (both) identity.
I have used this, but it returns first all users in the first identity and then the next identity.
$groups = "SMSxxx", "Personalxxxx"
$results = foreach ($group in $groups) {
Get-ADGroupMember $group | select samaccountname, name, @{n='GroupName';e={$group}}, @{n='Description';e={(Get-ADGroup $group -Properties description).description}}
}
$results
$results | Export-csv C:\Temp\GroupMemberShip.txt -NoTypeInformation
Best regards, Peter
Upvotes: 0
Views: 329
Reputation: 25061
You can continue with your current logic and use Group-Object to find users that exist in all groups.
$groups = "SMSxxx", "Personalxxxx"
$results = foreach ($group in $groups) {
$description = (Get-ADGroup $group -Properties description).description
Get-ADGroupMember $group | select SamAccountName,Name,@{n='GroupName';e={$group}}, @{n='Description';e={$description}}
}
$results | Group-Object SamAccountName |
Where Count -eq $groups.Count | Select -Expand Group |
Export-csv C:\Temp\GroupMemberShip.csv -NoTypeInformation
Upvotes: 1