Get All The Members Of The Distributions Lists That A User Is A Member Of

This is kind of a weird script tip but I bumped into a need for this kind of script so I thought I’d share it. In this post, I have a user and I want to get all the members of all the distribution lists that the user is a member of. That is to say, if the user is a member of DL1, DL2 and DL3 distribution lists, I want to get all the other members of all those distribution lists. You’re going to need a remote Exchange shell for this.

Here’s the code I came up with.

$DN = 'CN=ThmsRynr,OU=BestUsers,DC=lab,DC=workingsysadmin,DC=com'
Get-DistributionGroup -filter "members -eq '$DN'" | Select-Object Name,@{l='Members';e={(Get-DistributionGroupMember $_.SamAccountName -ResultSize Unlimited | % { $_.Name }) -join '; ' }}

Line 1 is just declaring a variable to hold the DistinguishedName attribute for the user I am interested in. Line 2 is the work line. The first thing I’m doing is getting all the distribution groups which have a member equal to the DN of the user I’m interested in. Now, the weirdness happens…

When you do a Get-DistributionGroup, you do not get the members of that group back with it. Here are the properties that come back that contain the string “mem” in the name.

PS C:\> Get-DistributionGroup -filter "members -eq '$DN'" | Select-Object -First 1 | Get-Member | Where-Object Name -match 'mem' | Select-Object Name


Nothing in there contains the members. So back to the command I wrote to accomplish my goal.

I’m piping the Distribution Groups returned into a Select-Object cmdlet to return the Name property and then a custom column. The label is Members and the content is going to just be a string of all the Distribution Group members’ names separated by semicolons. The expression for my custom column is a Get-DistributionGroupMember command for the Distribution Group piped into a Foreach-Object (alias is “%”) which returns an array of all the names of the members in the Distribution Group. I use the -join command to take the array and convert it into a string separated by semicolons.

It’s just that easy!

Written on December 16, 2015