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我如何确定用户是否使用 XMPPFramework for iPhone 在线/离线?

我有他们的 JID 等。有没有办法请求出席或其他什么?

谢谢你。

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2 回答 2

7

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片段:

// 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];

您的流委托获得的回调应该是

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

我假设您已经有了 xmmpframework 源,如果没有,您可以在此处克隆存储库

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

示例项目位于“Xcode”文件夹中。

于 2011-02-10T23:09:36.440 回答
0

首先,获取用户状态必须需要 ROSTER List 中的双方订阅。

在此处输入图像描述

在这里,如何订阅和接受出席请求。

    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出席请求,那么您在名册代表中收到了请求。

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

完成上述步骤后,您将在以下 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 }
}
于 2021-06-01T05:22:03.810 回答