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)

我使用这个过程作为DLL登录到我们开发的其他应用程序… (我们目前正在使用OpenEdge Progress)

public static string AzureLogin(string user, string password) {

    string status;

    try {
        new DirectorySearcher(new DirectoryEntry("LDAP://yourdomain.com", user, password) {
            AuthenticationType = AuthenticationTypes.Secure,
            Username = user,
            Password = password
        })  {
            Filter = "(objectclass=user)"
        }.FindOne().Properties["displayname"][0].ToString();

        status = $"SUCCESS - User {user} has logged in.";

    } catch(System.Exception e) {
        status = $"ERROR - While logging in: {e}";
    }

    return status;
}

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

这里介绍的几个解决方案缺乏区分错误用户/密码和需要更改的密码的能力。这可以通过以下方式来实现:

using System;
using System.DirectoryServices.Protocols;
using System.Net;

namespace ProtocolTest
{
    class Program
    {
        static void Main(string[] args)
        {
            try
            {
                LdapConnection connection = new LdapConnection("ldap.fabrikam.com");
                NetworkCredential credential = new NetworkCredential("user", "password");
                connection.Credential = credential;
                connection.Bind();
                Console.WriteLine("logged in");
            }
            catch (LdapException lexc)
            {
                String error = lexc.ServerErrorMessage;
                Console.WriteLine(lexc);
            }
            catch (Exception exc)
            {
                Console.WriteLine(exc);
            }
        }
    }
}

如果用户密码错误，或者用户不存在，error将包含

“8009030C: LdapErr: DSID-0C0904DC，注释:AcceptSecurityContext错误，数据52e, v1db1”，

如果用户密码需要修改，则包含

"8009030C: LdapErr: DSID-0C0904DC，注释:AcceptSecurityContext错误，数据773,v1db1"

lexc。ServerErrorMessage数据值是Win32错误码的十六进制表示。这些是通过调用Win32 LogonUser API调用返回的相同错误代码。下面的列表总结了一些常见的十六进制和十进制值:

525​ user not found ​(1317)
52e​ invalid credentials ​(1326)
530​ not permitted to logon at this time​ (1328)
531​ not permitted to logon at this workstation​ (1329)
532​ password expired ​(1330)
533​ account disabled ​(1331) 
701​ account expired ​(1793)
773​ user must reset password (1907)
775​ user account locked (1909)

如果你使用的是。net 3.5或更新版本，你可以使用System.DirectoryServices.AccountManagement命名空间轻松验证你的凭证:

// create a "principal context" - e.g. your domain (could be machine, too)
using(PrincipalContext pc = new PrincipalContext(ContextType.Domain, "YOURDOMAIN"))
{
    // validate the credentials
    bool isValid = pc.ValidateCredentials("myuser", "mypassword");
}

它简单，可靠，在你的端是100%的c#托管代码——你还能要求什么呢?: -)

在这里阅读所有内容:

在.NET Framework 3.5中管理目录安全主体 System.DirectoryServices.AccountManagement上的MSDN文档

更新:

正如在另一个SO问题(及其答案)中概述的那样，对于用户的旧密码，这个调用可能会返回True。只是要注意这种行为，如果发生这种情况不要太惊讶:-)(感谢@MikeGledhill指出这一点!)