XMPPFramework：我可以确定用户的在线/离线状态吗？ [英] XMPPFramework : can I determine the online/offline status of a user at all?

问题描述

如何使用XMPPFramework for iPhone确定用户是否在线/离线？

我有他们的JID等等。有没有办法请求存在或

谢谢。

解决方案

XMPPFramework示例项目的源代码？

如果我没记错，应该是相关的代码片段：

  //订阅好友的存在
 // 
 //< presence to =bareJIDtype =subscribe/> 
 
 NSXMLElement * presence = [NSXMLElement elementWithName：@presence]; 
 [presence addAttributeWithName：@tostringValue：[jid bare]]; 
 [presence addAttributeWithName：@typestringValue：@subscribe]; 
 
 [xmppStream sendElement：presence]; 
  

您的流委托获得的回调应为

   - （void）xmppStream：（XMPPStream *）sender didReceivePresence：（XMPPPresence *）presence; 
  

我假设你已经有xmmpframework源代码，如果没有， p>

  hg clone https://xmppframework.googlecode.com/hg/ xmppframework 
  pre> 
 
 
示例项目位于Xcode文件夹中。
 
How would I determine if a user is online/offline using XMPPFramework for iPhone?

I have their JID etc etc. Is there a way of requesting a presence or something?

Thank you.
解决方案
Have you checked out the source code of the XMPPFramework sample project?

If I remember correctly, this should be the relevant code snippet:
// Subscribe to the buddy's presence
// 
// <presence to="bareJID" type="subscribe"/>

NSXMLElement *presence = [NSXMLElement elementWithName:@"presence"];
[presence addAttributeWithName:@"to" stringValue:[jid bare]];
[presence addAttributeWithName:@"type" stringValue:@"subscribe"];

[xmppStream sendElement:presence];
And the callback your stream delegate gets should be
- (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence;
I assume that you already have the xmmpframework source, if not, you can clone the repository here
hg clone https://xmppframework.googlecode.com/hg/ xmppframework
The sample projects are in the "Xcode" folder.

                        
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