XMPPFramework：我可以确定用户的在线/离线状态吗？

Question

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。有没有办法要求出席或某事？

谢谢。

Answer 1

您是否查看过XMPPFramework示例项目的源代码？

如果我没记错的话，这应该是相关的代码段：

// 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];

你的流代表获得的回调应该是

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

我假设您已经拥有xmmpframework源，如果没有，您可以在此克隆存储库

hg clone https://xmppframework.googlecode.com/hg/ xmppframework

示例项目位于“Xcode”文件夹中。

Answer 2

首先，获取用户状态需要在ROSTER List中的双方订阅。

此处，如何订阅和接受在线请求。

    let senderName = strReceiverEjabberedName + "@" + eJabVirtualHost
    let roomJID = XMPPJID(string: senderName)
    
    // Send subscribe request
    let detxTag:XMLElement = XMLElement(name: "presence")
    detxTag.addAttribute(withName: "to", stringValue: roomJID?.bare ?? "")
    detxTag.addAttribute(withName: "type", stringValue: "subscribe")
    xmppController.xmppStream.send(detxTag)
    
    // Send subscribed request
    let detxTaged:XMLElement = XMLElement(name: "presence")
    detxTaged.addAttribute(withName: "to", stringValue: roomJID?.bare ?? "")
    detxTaged.addAttribute(withName: "type", stringValue: "subscribed")
    xmppController.xmppStream.send(detxTaged)
    
    // Subscribe & Accept Presence Request
    xmppController.xmppRoster?.subscribePresence(toUser: roomJID!)
    xmppController.xmppRoster?.acceptPresenceSubscriptionRequest(from: roomJID!, andAddToRoster: true)

如果您的朋友向您发送请求，那么您会在名册代表中收到请求。

extension XMPPServiceController: XMPPRosterDelegate {
    func xmppRoster(_ sender: XMPPRoster, didReceivePresenceSubscriptionRequest presence: XMPPPresence) {
        sender.acceptPresenceSubscriptionRequest(from: presence.from!, andAddToRoster: true)
    }
}

完成上述步骤后，您将在下面的 XMPP 委托方法中收到您朋友的在线/离线状态。

func xmppStream(sender: XMPPStream?, didReceivePresence presence: XMPPPresence?) {
    let presenceType = presence?.type
    if presenceType == "available" { //ONLINE }
    else if presenceType == "unavailable" { //OFFLINE }
}