给定用户的SID,我如何获得AD DirectoryEntry?

时间:2011-08-17 08:57:24

标签: directoryentry sid

我在windowsPrincipal.getIdentity()。getSid()中将用户的SID作为byte []。如何从SID获取活动目录条目(DirectoryEntry)?

4 个答案:

答案 0 :(得分:9)

使用SecurityIdentifier类将sid从byte []格式转换为字符串,然后直接绑定到对象:

DirectoryEntry OpenEntry(byte[] sidAsBytes)
{
    var sid = new SecurityIdentifier(sidAsBytes, 0);

    return new DirectoryEntry(string.Format("LDAP://<SID={0}>", sid.ToString()));
}

答案 1 :(得分:4)

我在c#

中找到了这个例子
    // SID must be in Security Descriptor Description Language (SDDL) format
    // The PrincipalSearcher can help you here too (result.Sid.ToString())
    public void FindByIdentitySid()
    {
        UserPrincipal user = UserPrincipal.FindByIdentity(
            adPrincipalContext,
            IdentityType.Sid,
            "S-1-5-21-2422933499-3002364838-2613214872-12917");
        Console.WriteLine(user.DistinguishedName);
    }

转换为VB.NET:

    ' SID must be in Security Descriptor Description Language (SDDL) format
    ' The PrincipalSearcher can help you here too (result.Sid.ToString())
    Public Sub FindByIdentitySid()
        Dim user As UserPrincipal = UserPrincipal.FindByIdentity(adPrincipalContext,     IdentityType.Sid, "S-1-5-21-2422933499-3002364838-2613214872-12917")
        Console.WriteLine(user.DistinguishedName)
    End Sub

显然你可以:

    dim de as new DirectoryEntry("LDAP://" & user.DistinguishedName)

获得SID = S-1-5-21- * (对不起VB.NET)

    ' Convert ObjectSID to a String

    ' http://social.msdn.microsoft.com/forums/en-US/netfxbcl/thread/57452aab-4b68-4444-aefa-136b387dd06e

    Dim ADpropSid As Byte()
    ADpropSid = de.Properties("objectSid").Item(0)    
    ' in my test the byte field looks like this : 01 02 00 00 00 00.......37 02 00 00
    Dim SID As New System.Security.Principal.SecurityIdentifier(ADpropSid, 0)

我还没有测试过C#或者我自己使用过转换版本,但是已经使用上面的方法以SDDL格式返回SID。

答案 2 :(得分:0)

这也可以在PowerShell中完成,只要您有.Net 3.5或4.0可用(如果您没有默认情况,请参阅https://gist.github.com/882528

add-type -assemblyname system.directoryservices.accountmanagement
$adPrincipalContext = 
    New-Object System.DirectoryServices.AccountManagement.PrincipalContext( 
    [System.DirectoryServices.AccountManagement.ContextType]::Domain)
$user = [system.directoryservices.accountmanagement.userprincipal]::findbyidentity(
    $adPrincipalContext
    , [System.DirectoryServices.AccountManagement.IdentityType]::Sid
    , "S-1-5-21-2422933499-3002364838-2613214872-12917")
$user.DisplayName
$user.DistinguishedName

答案 3 :(得分:0)

我发现的最简单方法是使用LDAP绑定。与Nick Giles所说的相似。有关详情,请访问MSDN

''' <summary>
''' Gets the DirectoryEntry identified by this SecurityIdentifier.
''' </summary>
''' <param name="id">The SecurityIdentifier (SID).</param>
<System.Runtime.CompilerServices.Extension()> _
Public Function GetDirectoryEntry(ByVal id As SecurityIdentifier) As DirectoryEntry
    Const sidBindingFormat As String = "LDAP://AOT/<SID={0}>"

    Return New DirectoryEntry(String.Format(sidBindingFormat, id.Value))
End Function