C＃将PC添加到AD域组

Question

尝试将主机名添加到域组，但出现异常There is a naming violation 。

据我所知我的语法是正确的，我已经看到了几个关于此确切内容的公认答案。

try
{
    DirectoryEntry de = new DirectoryEntry("LDAP://aa.bbbbb.com/CN=Group,OU=Application,OU=Groups,OU=US,DC=aa,DC=bbbbb,DC=com");
    string hostname = "CN=" + SystemInformation.ComputerName;
    DirectoryEntry add = de.Children.Add(hostname, "Computer");
    add.CommitChanges();
}
catch (Exception ex)
{
    MessageBox.Show("Group join failed" + Environment.NewLine + Environment.NewLine + ex.ToString());
}

任何帮助将不胜感激。

Answer 1

我弄清楚了问题所在-我需要传递专有名称，而不只是主机名...如果我已经阅读了MSDN文档，那应该很明显...此外， de.Children.Add()可能是实现此目的的一种有效方法（对于.Net 3.5 IIRC，这是SE上的一个可接受的答案），但我使用了de.Properties["member"].Add()来实现。

已更新的任何Google员工的源代码：

private void DoStuff(object sender, EventArgs e)
{
    using (Process addgroup = new Process())
    {
        string hostname = Environment.MachineName;
        AddMemberToGroup("LDAP://aa.bbbbb.com/CN=Group,OU=Application,OU=Group,OU=US,DC=aa,DC=bbbbb,DC=com", hostname);
    }
}

private void AddMemberToGroup(string ldapString, string host)
{
    try
    {
        DirectoryEntry de = new DirectoryEntry(ldapString);
        string distHost = GetDistinguishedName(host);
        if (!String.IsNullOrEmpty(distHost))
        {
            de.Properties["member"].Add(distHost);
            de.CommitChanges();
        }
        else
        {
            MessageBox.Show("Distinguished Host Name returned NULL");
        }
    }
    catch(Exception ex)
    {
        MessageBox.Show("Group join failed" + Environment.NewLine + Environment.NewLine + ex.ToString());
    }
}

private string GetDistinguishedName(string compName)
{
    try
    {
        PrincipalContext pContext = new PrincipalContext(ContextType.Domain);
        ComputerPrincipal compPrincipal = ComputerPrincipal.FindByIdentity(pContext, compName);
        return compPrincipal.DistinguishedName;
    }
    catch (Exception ex)
    {
        MessageBox.Show("Failed to get the Distinguished Hostname from Active Directory" + Environment.NewLine + Environment.NewLine + ex.ToString());
        return null;
    }
}