问题描述

我需要仅使用用户名而非密码来验证公司中的用户.

I need to verify users in the company using only their username - not their password.

所以我需要这样的方法

public bool UserExists(string username)
{ ... }

我知道 System.DirectoryServices 命名空间，但不知道从哪里开始.

I am aware of the System.DirectoryServices namespace but don't know where to start.

有什么想法吗?

有 80,000 多条记录，因此请记住这一点.

There are 80,000+ records so try to bear that in mind.

谢谢.

我已经完成了 - 我的代码是:

I have done it - my code is:

private bool UserExists(string userName, string domain)
{
    try
    {
        DirectoryEntry.Exists("WinNT://" + domain + ".[hidden].com/" + userName);
        return true;
    }
    catch (COMException)
    {
        return false;
    }
}

我不知道它是否正确，但它似乎到目前为止有效.

I don't know if it is correct, but it seems to work so far.

迈克尔的回答有两个相关部分:

Michael's answer has two relevant parts:

• http://www.codeproject.com/KB/system/everythingInAD.aspx#22
• http://www.codeproject.com/KB/system/everythingInAD.aspx#35

更新 #2:

我实际使用过这个:

public static bool LoggedOnUserExists()
{
    var domain = new PrincipalContext(ContextType.Domain);

    UserPrincipal foundUser = UserPrincipal.FindByIdentity(domain, IdentityType.SamAccountName, Environment.UserName);

    return foundUser != null;
}

推荐答案

在 .NET 3.5 及更高版本中，您可以非常简单地使用 System.DirectoryServices.AccountManagement 命名空间:

In .NET 3.5 and up, you can use the System.DirectoryServices.AccountManagement namespaces to do this quite simply:

public bool UserExists(string username)
{
   // create your domain context
   using (PrincipalContext domain = new PrincipalContext(ContextType.Domain))
   {
       // find the user
       UserPrincipal foundUser = UserPrincipal.FindByIdentity(domain, IdentityType.Name, username);

       return foundUser != null;
    }
}

这将适用于常规用户名 John Doe，或者您也可以使用用户的电子邮件地址 (john.doe@company.com)，或者他的专有名称 (CN=John Doe) - 看看 IdentityType 枚举必须提供什么 :-)

This will work with the regular user name John Doe, or alternatively you can use the user's e-mail address (john.doe@company.com), or his distinguished name (CN=John Doe) - see what the IdentityType enumeration has to offer :-)

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