Question

目标：配置xmpp空间以限制仅限成员访问，然后添加成员。在swift4中使用xmppframework

有一些关于xmppframework github的文档，但它通常是客观的c，针对那些知道他们正在做什么的人

我正在创建一个小程序，它登录到我的xmpp server4，然后加入一个房间，并根据我的员工生产力开始发布警报。我创造了房间，它有历史，我可以很好地张贴到房间。我想限制房间，以便只有某些用户可以加入

 func xmppStreamDidAuthenticate(_ sender: XMPPStream) {
 self.xmppStream.send(XMPPPresence())
 print("Connected, and Authenticated")

 // Join Room
 let roomStorage: XMPPRoomMemoryStorage = XMPPRoomMemoryStorage()
 let roomJID = XMPPJID(string: "chatalerts@muc.im.taudy.com") ?? 
 XMPPJID(string: "chatalerts@muc.im.taudy.com")

 let xmppRoom: XMPPRoom = XMPPRoom(roomStroage: roomStorage, jid: roomJID!, 
 dispatchQueue: DispatchQueue.main)
 xmppRoom.fetchConfigurationForm()
 }

我现在需要配置房间，我相信我需要使用

xmppRoom.configureRoom(usingOptions:XMLElement )

作为初学程序员，我不知道如何继续，所以我正在寻找一个房间的示例配置，仅限于成员，我可以根据需要添加或删除成员。

Answer 1

这是XMPP-MUC的详细信息，要限制会员的空间，您只能使用此XEP-0045。

要限制房间，您可以使用以下代表。

func xmppRoom(_ sender: XMPPRoom, didFetchConfigurationForm configForm: DDXMLElement) {
    
    print("XMPPManager xmppRoom didFetch ConfigurationForm : \(configForm)")
    
    let newForm = configForm.copy() as! DDXMLElement
    
    let groupName = "Your Room Name"
    
    
    for field in newForm.elements(forName: "field")
    {
        
        if let _var = field.attributeStringValue(forName: "var")
        {
            
            switch _var
            {
            case "muc#roomconfig_persistentroom":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

            case "muc#roomconfig_roomname":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name : "value", stringValue :groupName))

            case "muc#roomconfig_publicroom":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

            case "muc#roomconfig_whois":
                field.removeChild(at: 0)
                field.insertChild(DDXMLElement(name: "value", stringValue: "anyone"), at: 0)

            case "muc#roomconfig_allow_subscription":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

           // Here goes your value. in your case you have to send 1.
            case "muc#roomconfig_membersonly":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))


            case "muc#roomconfig_getmemberlist":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "moderator"))
                field.addChild(DDXMLElement(name: "value", stringValue: "participant"))
                field.addChild(DDXMLElement(name: "value", stringValue: "visitor"))

            case "muc#roomconfig_moderatedroom":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

            case "public_list":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

            case "muc#roomconfig_allowinvites":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "1"))

            case "muc#roomconfig_changesubject":
                field.removeChild(at: 0)
                field.addChild(DDXMLElement(name: "value", stringValue: "true"))
                
            // other configures
            default:
                break
            }
            
        }
        
    }
    
    sender.configureRoom(usingOptions: newForm)
    
}

如何使用swift 4（robbiehanson xmppframework）配置xmpp会议室，以允许成员资格和密码

1 个答案: