Question

我正在尝试执行按位或（|）操作，以将多个枚举应用于.NET程序集使用的PowerShell中的变量。但是，当单管道符号执行此操作时，我收到Expressions are only allowed as the first element of a pipeline错误。如何在PowerShell中分配多个枚举？

    $everyone = New-Object System.Security.Principal.SecurityIdentifier([System.Security.Principal.WellKnownSidType]::WorldSid, $null);
    $fsr = [System.Security.AccessControl.FileSystemRights]::Read;
    $if = [System.Security.AccessControl.InheritanceFlags]::ContainerInherit | [System.Security.AccessControl.InheritanceFlags]::ObjectInherit;
    $pf = [System.Security.AccessControl.PropagationFlags]::None;
    $act = [System.Security.AccessControl.AccessControlType]::Allow;
    $accessRule = New-Object System.Security.AccessControl.FileSystemAccessRule($everyone, $fsr, $if, $pf, $act);

Answer 1

使用-bor

[System.Security.AccessControl.InheritanceFlags]::ContainerInherit -bor [System.Security.AccessControl.InheritanceFlags]::ObjectInherit

ContainerInherit, ObjectInherit

Answer 2

PowerShell将提供字符串的自动转换，因此您也可以这样做：

$everyone = New-Object System.Security.Principal.SecurityIdentifier([System.Security.Principal.WellKnownSidType]::WorldSid, $null);
$fsr = [System.Security.AccessControl.FileSystemRights]::Read;
$if = [System.Security.AccessControl.InheritanceFlags]"ContainerInherit,ObjectInherit"
$pf = [System.Security.AccessControl.PropagationFlags]::None;
$act = [System.Security.AccessControl.AccessControlType]::Allow;
$accessRule = New-Object System.Security.AccessControl.FileSystemAccessRule($everyone, $fsr, $if, $pf, $act);

事实上，如果你希望它的输入要少得多（当你不关心以后阅读的清晰度时，有利于快速命令行），你可以这样做：

$everyone = New-Object System.Security.Principal.SecurityIdentifier([System.Security.Principal.WellKnownSidType]::WorldSid, $null);
$accessRule = New-Object System.Security.AccessControl.FileSystemAccessRule $everyone, "Read", "ContainerInherit,ObjectInherit", "None", "Allow"

Answer 3

这是使用-bOr运算符完成的。请参阅TechNet documentation about ¾ down the page。

在-bOr实现后，上面的代码示例如下所示：

    $everyone = New-Object System.Security.Principal.SecurityIdentifier([System.Security.Principal.WellKnownSidType]::WorldSid, $null);
    $fsr = [System.Security.AccessControl.FileSystemRights]::Read;
    $if = [System.Security.AccessControl.InheritanceFlags]::ContainerInherit -bOr [System.Security.AccessControl.InheritanceFlags]::ObjectInherit;
    $pf = [System.Security.AccessControl.PropagationFlags]::None;
    $act = [System.Security.AccessControl.AccessControlType]::Allow;
    $accessRule = New-Object System.Security.AccessControl.FileSystemAccessRule($everyone, $fsr, $if, $pf, $act);

Answer 4

您也可以将两个继承标志对象放在数组中：

$everyone = New-Object System.Security.Principal.SecurityIdentifier([System.Security.Principal.WellKnownSidType]::WorldSid, $null);
$fsr = [System.Security.AccessControl.FileSystemRights]::Read;
$if = [System.Security.AccessControl.InheritanceFlags]::ContainerInherit
$if2 = [System.Security.AccessControl.InheritanceFlags]::ObjectInherit;
$pf = [System.Security.AccessControl.PropagationFlags]::None;
$act = [System.Security.AccessControl.AccessControlType]::Allow;
$accessRule = New-Object System.Security.AccessControl.FileSystemAccessRule($everyone, $fsr, ($if,$if2), $pf, $act);

如何在PowerShell中对枚举执行按位或（|）？

4 个答案: