[英]How to add user to existing ROOM in XMPP iOS?
I am working in XMPP chat module. 我在XMPP聊天模块中工作。 I have created group yesterday, Now I want to add some more member in this group.
我昨天创建了组,现在我想在这个组中添加更多成员。 What will be the process to add member in existing group.
在现有组中添加成员的过程是什么。
Here is my code to create group: 这是我创建组的代码:
XMPPJID *roomJID = [XMPPJID jidWithString:@"xyz@conference.abc"];
XMPPRoomMemoryStorage *roomMemoryStorage = [[XMPPRoomMemoryStorage alloc] init];
XMPPRoom *newxmppRoom = [[XMPPRoom alloc]
initWithRoomStorage:roomMemoryStorage
jid:roomJID
dispatchQueue:dispatch_get_main_queue()];
[newxmppRoom activate:xmppStream];
[newxmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
[newxmppRoom joinRoomUsingNickname:@"MY_NICKNAME" history:nil];
Should I have to write above code every time when I want to add user in room? 每当我想在房间里添加用户时,我是否应该编写上面的代码?
Yes, all this code is required to join a room. 是的,所有这些代码是需要加入一个房间。 To invite a user, you should use the method:
要邀请用户,您应该使用以下方法:
XMPPJID * userID = [XMPPJID jidWithString:user.entityID];
[room inviteUser:userID withMessage:@""];
Then in the callback: 然后在回调中:
- (void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *)roomJID didReceiveInvitation:(XMPPMessage *)message {
// User your code here to join
}
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