尝试从脚本中的子文件夹获取ACL时出错

Question

I have been doing this code lately trying to get only everyones and authenticated users from the ACL in shared folders.

The script does get the shares and iterate over them, but I'm unable to get the ACL's from parent folder nor its childs

And also, I would like to know how to export not using export-csv iterating over every share instead over every server.

Tried with -Depth Parameter, changed parameter get-childitem to get-item .

The only way it worked is to leave open the where-object comenting #-like 'Everyone' -or $_.IdentityReference -like 'NT Authority\\Authenticated Users' } |

$Server="Server"
$shares = gwmi win32_Share -ComputerName $Server |
  Where-Object {$_.type -eq '0'} | 
  Where {$_.name -notlike "*$*"} | Select-Object -ExpandProperty Name
  $Shares
  foreach ($share in $shares){
$root = "\\$Server\$share"
$csv  = "C:\temp\$Server-$share.csv"
Remove-Item $csv
New-Item $csv
Get-ChildItem -literalPath $root -Recurse -directory |
  ForEach-Object {
    $dir = $_
    ##Test-Path $dir}}
    Get-Acl $dir | Select-Object -Expand Access |
      Where-Object { $_.IdentityReference }|#-like 'Everyone' -or $_.IdentityReference -like 'NT Authority\Authenticated Users' } |
      ForEach-Object {
        New-Object PSObject -Property @{
          Folder       = $dir.FullName
          Access       = $_.FileSystemRights
          Control      = $_.AccessControlType
          User         = $_.IdentityReference
          Inheritance  = $_.IsInherited
          LastModified = $dir.LastWriteTime

        }
      } 
  } | Export-Csv $csv -Force
} 

Expected An export to a csv file but i have

Get-Acl : Cannot Find Path 'XXXX' because it does not exists + Get-Acl $Dir

Answer 1

You need to parenthesise your conditionals for a start:

Get-Acl $dir | Select-Object -Expand Access |
  Where-Object { ($_.IdentityReference -like 'Everyone') -or ($_.IdentityReference -like 'NT Authority\Authenticated Users') } |
  ForEach-Object {