列出具有每个组名称的多个AD组中的用户的命令

Question

I'm trying to get a list of AD group members with each Group's members listed separately. I can list out all members of all groups, but how to separate the membership of each group?

This is what I'm running:

get-ADGroup -Server "<server_name>" -filter {Name -like  "*<common_text_in_each_group's_name*"} | get-ADGroupMember

Unfortunately I'm trying to do this in a one-liner because when I try to create a ps1 file and run it, I get an error that it cannot find the groups in the domain my local machine is logged in to. IE, the -Server argument doesn't work. All the groups are in a different domain from the machine I'm running PS on.

When I did try creating a script/ps1 file, this is what I tried:

$Groups = (Get-AdGroup -server '<server_name>' -filter * | Where {$_.name -like "*_some_text*"} | select name -expandproperty name)


$Table = @()

$Record = [ordered]@{
"Group Name" = ""
"Name" = ""
"Username" = ""
}



Foreach ($Group in $Groups)
{

$Arrayofmembers = Get-ADGroupMember -identity $Group | select name,samaccountname

foreach ($Member in $Arrayofmembers)
{
$Record."Group Name" = $Group
$Record."Name" = $Member.name
$Record."UserName" = $Member.samaccountname
$objRecord = New-Object PSObject -property $Record
$Table += $objrecord

}

}

$Table | export-csv "C:\temp\x_Groups.csv" -NoTypeInformation

Answer 1

I believe you just need to pass the -Server parameter to Get-ADGroupMember too:

$Arrayofmembers = Get-ADGroupMember -identity $Group -server '<server_name>' | select name,samaccountname

It's kind of dumb that way.