这可能吗?代码示例会很好。
答案 0 :(得分:85)
实际上,问题是如何获得.NET 3.5 (System.DirectoryServices.AccountManagement.)UserPrincipal
的两个属性 - 没有给出userPrincipalName
的对象。
以下是如何使用extension method:
执行此操作using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.DirectoryServices;
using System.DirectoryServices.AccountManagement;
namespace MyExtensions
{
public static class AccountManagementExtensions
{
public static String GetProperty(this Principal principal, String property)
{
DirectoryEntry directoryEntry = principal.GetUnderlyingObject() as DirectoryEntry;
if (directoryEntry.Properties.Contains(property))
return directoryEntry.Properties[property].Value.ToString();
else
return String.Empty;
}
public static String GetCompany(this Principal principal)
{
return principal.GetProperty("company");
}
public static String GetDepartment(this Principal principal)
{
return principal.GetProperty("department");
}
}
}
上述代码在大多数情况下都适用(即它适用于标准文本/字符串单值Active Directory属性)。您需要修改代码并为您的环境添加更多错误处理代码。
您可以通过将“扩展类”添加到项目中来使用它,然后您可以执行此操作:
PrincipalContext domain = new PrincipalContext(ContextType.Domain);
UserPrincipal userPrincipal = UserPrincipal.FindByIdentity(domain, "youruser");
Console.WriteLine(userPrincipal.GetCompany());
Console.WriteLine(userPrincipal.GetDepartment());
Console.WriteLine(userPrincipal.GetProperty("userAccountControl"));
(BTW;这对扩展属性非常有用 - too bad it won't be in C# 4 either。)
答案 1 :(得分:14)
如果用户存在部门和公司属性,则应该这样做。
DirectoryEntry de = new DirectoryEntry();
de.Path = "LDAP://dnsNameOfYourDC.my.company.com";
DirectorySearcher deSearch = new DirectorySearcher(de);
deSearch.PropertiesToLoad.Add("department");
deSearch.PropertiesToLoad.Add("company");
deSearch.SearchScope = SearchScope.Subtree;
deSearch.Filter = "(&(objectClass=User)(userPrincipalName=MyPrincipalName))";
SearchResultCollection results = deSearch.FindAll():
foreach (SearchResult result in results)
{
ResultPropertyCollection props = result.Properties;
foreach (string propName in props.PropertyNames)
{
//Loop properties and pick out company,department
string tmp = (string)props[propName][0];
}
}