iOS XMPP群聊方法的实现

来源：互联网发布：淘宝开店半年怎么算编辑：程序博客网时间：2024/06/13 12:20

首先需要创建一个房间：

xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:self jid:[XMPPJID jidWithString:[NSString stringWithFormat:@"%@@%@", jidString, @"conference.192.168.1.117"]]];    [xmppRoom activate:xmppStream];    [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];

加入房间：

[[self xmppRoom] joinRoomUsingNickname:[[[NSUserDefaults standardUserDefaults] objectForKey:@"UserInfo"] valueForKey:@"userId"] history:nil];

此时房间ID就出现在好友列表内，可以正常进入聊天。

其次邀请好友：

[[self xmppRoom] inviteUser:[XMPPJID jidWithString:[NSString stringWithFormat:@"%@@192.168.1.117", friendIDFeild.text]] withMessage:@"join room!"];

客户端要对收到邀请做出响应，方法如下：

新建一个XMPPMUC：

xmppMUC = [[XMPPMUC alloc] initWithDispatchQueue:dispatch_get_main_queue()];    [xmppMUC activate:xmppStream];    [xmppMUC addDelegate:self delegateQueue:dispatch_get_main_queue()];

类要实现XMPPMUCDelegate

-(void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *)roomJID didReceiveInvitation:(XMPPMessage *)message{    NSLog(@"%@", message);    xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:self jid:roomJID];    [xmppRoom activate:xmppStream];    [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];    [xmppRoom joinRoomUsingNickname:[[[NSUserDefaults standardUserDefaults] objectForKey:@"UserInfo"] valueForKey:@"userId"] history:nil];}

其中

[xmppRoom joinRoomUsingNickname:[[[NSUserDefaults standardUserDefaults] objectForKey:@"UserInfo"] valueForKey:@"userId"] history:nil];

为确认加入房间。

当你接受请求后，房间id会出现在联系人列表中，正常聊天就可以了，聊天信息的type为groupchat，这部分在XMPP基础中有，不做赘述。此为初步，后续补充。

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