简体   繁体   English

使用ExchangeService从Exchange Server 2010 SP1获取全局地址列表(GAL)

[英]Get Global Address List (GAL) from Exchange Server 2010 SP1 using ExchangeService

I want to programatically (using C#) access to the Global Address List on exchange, using ExchangeService WebService. 我想使用ExchangeService WebService以编程方式(使用C#)访问交换机上的全局地址列表。 How can I do that? 我怎样才能做到这一点? Another problem: I already know how to search for a specific user. 另一个问题:我已经知道如何搜索特定用户。 However, if that user has more than 1 email, if i search for the non-default email, he can't find it! 但是,如果该用户有超过1封电子邮件,如果我搜索非默认电子邮件,他找不到它! How can I resolve this question using ExchangeService? 如何使用ExchangeService解决此问题? I saw the question How to get contact list from Exchange Server? 我看到了如何从Exchange Server获取联系人列表的问题 but, by this way, I only can access my own address list (i want all organization address list). 但是,通过这种方式,我只能访问我自己的地址列表(我想要所有的组织地址列表)。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 如何从Exchange Server 2007 SP1获取最新的SyncState - How to get latest SyncState from Exchange Server 2007 SP1 通过Exchange Web服务(EWS)erreor SSL查询全局地址列表(GAL) - Querying the Global Address List (GAL) via Exchange Web Services (EWS) erreor SSL 从Exchange Server 2010 SP3使用C#从独立MAPI提取邮件 - Fetching mail from Standalone MAPI using C# from Exchange Server 2010 SP3 如何从GAL(全局地址列表)中获取addressEntry对象的完全匹配 - How to fetch exact match of addressEntry object from GAL (Global Address List) 允许全局地址列表(GAL)集成的ASP.NET控件 - Control for ASP.NET that allows Global Address List (GAL) integration GAL从Outlook中与交换从哪个获得? - GAL from outlook with exchange from which one to get? 通过电话号码从Exchange Server查询GAL联系人 - Query GAL contacts from Exchange Server by phone number visual Studio 2010 sp1源服务器支持 - visual studio 2010 sp1 source server support 从Outlook中的多个帐户获取非默认通讯簿的GAL - To get GAL of non default address book from multiple accounts in outlook 无法以编程方式使用MS Exchange Server 2010 SP2的Web服务发送电子邮件 - Can not programmatically send email using web service of MS Exchange Server 2010 SP2
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM