get-acl information with special format in file

Question

Hi, I would like to write the permissions of a directory to a file in a special format. (Fields from left to right)

Example

Get-Acl -Path "\sv008\workgroup\edv" -Filter * | Select-Object -ExpandProperty Access

Get-Acl -Path "\svgeb008\workgroup\edv" -Filter * | select-object -Property IdentityReference,Filesystemrights,Accesscontroltype

If I use both

And this result i need in an file.
But if I use the following

I get this information

What am I doing wrong?

can someone help me and send me some code?

Best regards Andreas

Answer 1

Something like this?

$path = "C:\Temp"
$rpt = "c:\temp\output.txt"
"Permissions for $path" | out-file $rpt  
(Get-Acl -Path "c:\temp" -Filter *).access | select-object -Property IdentityReference,Filesystemrights,Accesscontroltype | out-file $rpt -append 
get-content $rpt

Answer 2

The "Access" property is an array of objects, each with the type "FileSystemAccessRule". Each FileSystemAccessRule has a set of named properties (AccessControlType, FileSystemRights, IdentityReference, etc).

I think this is what you set out to achieve:

Get-Acl -Path "\sv008\workgroup\edv" -Filter * |
	Select-Object -ExpandProperty Access |
		Select-Object IdentityReference,Filesystemrights,Accesscontroltype |
			Out-File c:\temp\output.txt
			

Answer 3

For my purpose, Rich's code is perfect.