didReceiveInvitation() 未在 XMPPFramework 中调用，Swift 2

Question

我正在创建像 Whatsapp.
这样的聊天应用程序我已经成功编写了文本聊天、图像、音频、视频传输的功能。现在我正在创建多用户聊天。经过长时间的研发，我问这个问题。请告诉我我的代码做错了什么。我已经遵循了所有这些教程，但不是运气

https://github.com/robbiehanson/XMPPFramework/issues/640

MUC How-to with XMPPFramework

Accepting chatroom invitation

好的下面是我的代码

1.成功设置 STREAM 后，我在 goOnline Method

中为邀请设置了 XMPPMUC 委托

private func goOnline() {
    let presence = XMPPPresence()
    let domain = xmppStream.myJID.domain

    if domain == "gmail.com" || domain == "gtalk.com" || domain == "talk.google.com"
        //        || domain == "chat.alqatech.com"
    {
        let priority = DDXMLElement.elementWithName("priority", stringValue: "24") as! DDXMLElement
        presence.addChild(priority)
    }
    xmppMUC = XMPPMUC(dispatchQueue: dispatch_get_main_queue())
    xmppMUC!.activate(self.xmppStream)
    xmppMUC!.addDelegate(self, delegateQueue: dispatch_get_main_queue())

    xmppStream.sendElement(presence)
}

2。创建群组

func createGroupChat(members:[String],groupName:String){
        membersToInvite = members
        xmppRoomMemoryStorage = XMPPRoomMemoryStorage()
        let xmppJid = XMPPJID.jidWithString("\(groupName)@conference.chat.xxxxxx.com")
        let xmppRoom = XMPPRoom.init(roomStorage: xmppRoomMemoryStorage, jid: xmppJid)
        xmppRoom.activate(xmppStream)
        xmppRoom.addDelegate(self, delegateQueue: dispatch_get_main_queue())
        xmppRoom.joinRoomUsingNickname(xmppStream.myJID.user, history: nil)
    }

3。群组创建成功

func xmppRoomDidCreate(sender: XMPPRoom!) {
        print(sender)
    }

4。 xmppRoomDidJoin 调用成功然后在这里我邀请用户

func xmppRoomDidJoin(sender: XMPPRoom!) {
        sender.fetchConfigurationForm()
        for JID in membersToInvite! {
            sender.editRoomPrivileges([XMPPRoom.itemWithAffiliation("member", jid: XMPPJID.jidWithString(JID))])
            sender.inviteUser(XMPPJID.jidWithString(JID), withMessage: "THIS IS GROUP MESSAGE")

        }

    }

5。 didFetchConfigurationForm 调用成功

func xmppRoom(sender: XMPPRoom!, didFetchConfigurationForm configForm: DDXMLElement!) 
{

        let newConfig: DDXMLElement = configForm.copy() as! DDXMLElement
        let fields: [AnyObject] = newConfig.elementsForName("field")
        for field in fields {
            let vars: String = field.attributeStringValueForName("var")
            // Make Room Persistent
            if (vars == "muc#roomconfig_persistentroom") {
                field.removeChildAtIndex(0)
                field.addChild(DDXMLElement(name: "value", stringValue : "1"))
            }
        }
        sender.configureRoomUsingOptions(newConfig)


    }

6. didReceiveInvitation 它没有被调用。

func xmppMUC(sender: XMPPMUC!, roomJID: XMPPJID!, didReceiveInvitation message: XMPPMessage!) {
        print(roomJID)
        xmppRoomMemoryStorage = XMPPRoomMemoryStorage()
        let xmppRoom = XMPPRoom.init(roomStorage: xmppRoomMemoryStorage, jid: roomJID)
        xmppRoom.activate(xmppStream)
        xmppRoom.addDelegate(self, delegateQueue: dispatch_get_main_queue())
        xmppRoom.joinRoomUsingNickname(xmppStream.myJID.user, history: nil)

    }

Answer 1

如果你在文件XMPPMUC.m第317行设置断点

- (void)xmppStream:(XMPPStream *)sender didReceiveMessage:(XMPPMessage *)message

向您的用户发送邀请时是否调用该方法？如果是，继续调试，检查382行是否正在执行。

那一行是这样的：

[multicastDelegate xmppMUC:self roomJID:roomJID didReceiveInvitation:message];

Answer 2

问题出在这段代码中。实际上，每次我邀请任何用户时，我都会发送裸 ID。

func xmppRoomDidJoin(sender: XMPPRoom!) {
        sender.fetchConfigurationForm()
        for JID in membersToInvite! {
            sender.editRoomPrivileges([XMPPRoom.itemWithAffiliation("member", jid: XMPPJID.jidWithString(JID))])
            sender.inviteUser(XMPPJID.jidWithString(JID), withMessage: "THIS IS GROUP MESSAGE")

        }

    }

但要邀请任何用户，您必须使用完整 ID。

对于那些不知道的人

BareID = username@domain 全名 = 用户名@domain/resource

解决这个问题

我在整个应用程序的 JID 中对 resource = APPNAME 进行了硬编码。

didReceiveInvitation() 未在 XMPPFramework 中调用，Swift 2

didReceiveInvitation() is not being called in XMPPFramework and Swift 2

chat

ios

xmppframework

multiuserchat

swift