Android 如何接受MUC的邀请?
我正在用xmpp开发聊天应用程序。我已经使用MUC创建了组,并向其他用户发送了邀请。但我不知道如何接受和拒绝邀请 以下是我发送邀请的代码:Android 如何接受MUC的邀请?,android,xmpp,smack,multiuserchat,Android,Xmpp,Smack,Multiuserchat,我正在用xmpp开发聊天应用程序。我已经使用MUC创建了组,并向其他用户发送了邀请。但我不知道如何接受和拒绝邀请 以下是我发送邀请的代码: EntityBareJid userInviteJID = JidCreate.entityBareFrom("user2@servicename"); muc2.invite(userInviteJID, "Meet me in this excellent room"); 我尝试了多用户聊天。拒绝(conn,room,inviter.asBareJ
EntityBareJid userInviteJID = JidCreate.entityBareFrom("user2@servicename");
muc2.invite(userInviteJID, "Meet me in this excellent room");
我尝试了多用户聊天。拒绝(conn,room,inviter.asBareJid()s,“我现在很忙”)代码>invitationReceived()方法中的方法。但问题是MultiUserChat.Decept()方法给出了错误:
无法解析方法拒绝()
有人能帮我吗?我找到了拒绝邀请的答案
此函数被移动到MultiUserChatManager,它与MultiUserChat的特定实例没有关系,因此它是静态的,现在是管理器的一个函数
MultiUserChatManager.getInstanceFor(connection).decline(roomJID,inviter.asEntityBareJid(),"reason");
但如何接受邀请呢?任何人都可以回答我,请?您需要在收到邀请时自动加入,以下是连接完成时的代码
MultiUserChatManager.getInstanceFor(MyApplication.connection).addInvitationListener(new InvitationListener() {
@Override
public void invitationReceived(XMPPConnection conn, MultiUserChat room, EntityJid inviter, String reason, String password, Message message, MUCUser.Invite invitation) {
// Log.e(TAG, "invitationReceived() called with: conn = [" + conn + "], room = [" + room + "], inviter = [" + inviter + "], reason = [" + reason + "], password = [" + password + "], message = [" + message + "], invitation = [" + invitation + "]");
LogM.e("invitationReceived() called with: conn = [" + conn + "], room = [" + room + "], inviter = [" + inviter + "], reason = [" + reason + "], password = [" + password + "], message = [" + message + "], invitation = [" + invitation + "]");
try {
Resourcepart nickname = null;
try {
nickname = Resourcepart.from("MY_JID_HERE");
} catch (XmppStringprepException e) {
e.printStackTrace();
}
try {
room.join(nickname); //while get invitation you need to join that room
room.getRoom().getLocalpart();
} catch (SmackException.NoResponseException e) {
e.printStackTrace();
} catch (SmackException.NotConnectedException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (MultiUserChatException.NotAMucServiceException e) {
e.printStackTrace();
}
Log.e(TAG, "join room successfully");
} catch (XMPPException e) {
e.printStackTrace();
Log.e(TAG, "join room failed!");
}
}
});
你需要编辑你的问题,而不是给出答案。