可能最简单的方法是PInvoke LogonUser Win32 api。

http://www.pinvoke.net/default.aspx/advapi32/LogonUser.html

MSDN参考在这里…

http://msdn.microsoft.com/en-us/library/aa378184.aspx

绝对要使用登录类型

LOGON32_LOGON_NETWORK (3)

这只创建了一个轻量级的令牌-完美的认证检查。(其他类型可用于构建交互式会话等)

使用DirectoryServices非常简单的解决方案:

using System.DirectoryServices;

//srvr = ldap server, e.g. LDAP://domain.com
//usr = user name
//pwd = user password
public bool IsAuthenticated(string srvr, string usr, string pwd)
{
    bool authenticated = false;

    try
    {
        DirectoryEntry entry = new DirectoryEntry(srvr, usr, pwd);
        object nativeObject = entry.NativeObject;
        authenticated = true;
    }
    catch (DirectoryServicesCOMException cex)
    {
        //not authenticated; reason why is in cex
    }
    catch (Exception ex)
    {
        //not authenticated due to some other exception [this is optional]
    }

    return authenticated;
}

需要NativeObject访问来检测坏的用户/密码

我们在内部网做这些

你必须使用System.DirectoryServices;

下面是代码的核心部分

using (DirectoryEntry adsEntry = new DirectoryEntry(path, strAccountId, strPassword))
{
    using (DirectorySearcher adsSearcher = new DirectorySearcher(adsEntry))
    {
        //adsSearcher.Filter = "(&(objectClass=user)(objectCategory=person))";
        adsSearcher.Filter = "(sAMAccountName=" + strAccountId + ")";

        try
        {
            SearchResult adsSearchResult = adsSearcher.FindOne();
            bSucceeded = true;

            strAuthenticatedBy = "Active Directory";
            strError = "User has been authenticated by Active Directory.";
        }
        catch (Exception ex)
        {
            // Failed to authenticate. Most likely it is caused by unknown user
            // id or bad strPassword.
            strError = ex.Message;
        }
        finally
        {
            adsEntry.Close();
        }
    }
}

我的简单功能

 private bool IsValidActiveDirectoryUser(string activeDirectoryServerDomain, string username, string password)
    {
        try
        {
            DirectoryEntry de = new DirectoryEntry("LDAP://" + activeDirectoryServerDomain, username + "@" + activeDirectoryServerDomain, password, AuthenticationTypes.Secure);
            DirectorySearcher ds = new DirectorySearcher(de);
            ds.FindOne();
            return true;
        }
        catch //(Exception ex)
        {
            return false;
        }
    }

可能最简单的方法是PInvoke LogonUser Win32 api。

http://www.pinvoke.net/default.aspx/advapi32/LogonUser.html

MSDN参考在这里…

http://msdn.microsoft.com/en-us/library/aa378184.aspx

绝对要使用登录类型

LOGON32_LOGON_NETWORK (3)

这只创建了一个轻量级的令牌-完美的认证检查。(其他类型可用于构建交互式会话等)

试试这段代码 (注意:报告不能在windows server 2000上工作)

#region NTLogonUser
#region Direct OS LogonUser Code
[DllImport( "advapi32.dll")]
private static extern bool LogonUser(String lpszUsername, 
    String lpszDomain, String lpszPassword, int dwLogonType, 
    int dwLogonProvider, out int phToken);

[DllImport("Kernel32.dll")]
private static extern int GetLastError();

public static bool LogOnXP(String sDomain, String sUser, String sPassword)
{
   int token1, ret;
   int attmpts = 0;

   bool LoggedOn = false;

   while (!LoggedOn && attmpts < 2)
   {
      LoggedOn= LogonUser(sUser, sDomain, sPassword, 3, 0, out token1);
      if (LoggedOn) return (true);
      else
      {
         switch (ret = GetLastError())
         {
            case (126): ; 
               if (attmpts++ > 2)
                  throw new LogonException(
                      "Specified module could not be found. error code: " + 
                      ret.ToString());
               break;

            case (1314): 
               throw new LogonException(
                  "Specified module could not be found. error code: " + 
                      ret.ToString());

            case (1326): 
               // edited out based on comment
               //  throw new LogonException(
               //   "Unknown user name or bad password.");
            return false;

            default: 
               throw new LogonException(
                  "Unexpected Logon Failure. Contact Administrator");
              }
          }
       }
   return(false);
}
#endregion Direct Logon Code
#endregion NTLogonUser

除非你需要为“LogonException”创建自己的自定义异常