XMPPFramework – 如何获得在线和离线的朋友列表？

如何在iOS中使用XMPPFramework获得在线和离线的朋友列表？

我想发送好友请求给用户。 那么我怎么能做到这一点呢？ 有人可以为我分享一些示例代码吗？

谢谢。

Answer 1

通过使用这些方法，您将得到通知整个用户去在线/离线，添加，删除，更新等通过使用这些方法，您可以更新您的联系人列表

- (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didAddUser:(XMPPUserMemoryStorageObject *)user - (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didUpdateUser:(XMPPUserMemoryStorageObject *)user - (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didRemoveUser:(XMPPUserMemoryStorageObject *)user - (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didAddResource:(XMPPResourceMemoryStorageObject *)resource withUser:(XMPPUserMemoryStorageObject *)user - (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didUpdateResource:(XMPPResourceMemoryStorageObject *)resource withUser:(XMPPUserMemoryStorageObject *)user - (void)xmppRoster:(XMPPRosterMemoryStorage *)sender didRemoveResource:(XMPPResourceMemoryStorageObject *)resource withUser:(XMPPUserMemoryStorageObject *)user 

//添加好友

 - (void)addContactWithUserName:(NSString *)userName andNickName:(NSString *)nickName; { if(userName) { XMPPJID *jid = [XMPPJID jidWithString:[NSString stringWithFormat:@"%@@%@",userName,self.hostName]]; if(nickName) { [_xmppRoster addUser:jid withNickname:nickName]; } else { [_xmppRoster addUser:jid withNickname:nil]; } } else { NSLog(@"missing userName"); } }