XMPPFramework消息未发送

Question

我接收来自Google Talk帐户的消息，它们显示在Ios仿真器的表视图中，但是当我发送它时，它不显示在Google Talk客户端（在另一台计算机中）中。这是代码：

-(IBAction)sendchat:(id)sender 
{ 
General *general = [General sharedManager];//It is a singleton class used to store some values that need to be accesible in the whole application. 


NSXMLElement *body = [NSXMLElement elementWithName:@"body"]; 
text=[mensaje text]; 
NSLog(@"Texto en el body: %@", text); 
[body setStringValue:text]; 
NSArray *dest=[general.firstfrom componentsSeparatedByString:@"/"];//in firstfrom is stored the account from wich we receive the first message. This app cannot start a conversation itself, must only answer 

NSLog(@"Destination trimmed: %@", [dest objectAtIndex:0]);//Here, the destination account shows correctly (without the /xxxx stuff, just name@gmail.com) 
XMPPMessage *mens=[[XMPPMessage alloc]init]; 
[mens addAttributeWithName:@"body" stringValue:text]; 
[mens addAttributeWithName:@"sender" stringValue:general.userlogin]; 
NSLog(@"text vale: %@", text); 
NSXMLElement *messagetosend = [NSXMLElement elementWithName:@"message"]; 
[messagetosend addAttributeWithName:@"type" stringValue:@"chat"]; 
[messagetosend addAttributeWithName:@"to" stringValue:[dest objectAtIndex:0]]; 
[messagetosend addChild:body]; 
NSLog(@"We are sending to: %@", [dest objectAtIndex:0]); 
[self.xmppStream sendElement:messagetosend]; 

[self xmppStream:xmppStream didReceiveMessage:mens];//manage the sent message as it was received, to show it in the Table View 
self.mensaje.text=@""; 
} 

正如我所说，邮件接收完美，但我不能发送。我见过很多关于如何发送的例子，他们就像我的代码。如果我调试发件人，它显示ok（namesender@gmail.com），并且“to”属性也可以（namereceiver@gmail.com）。 xmppStrem设置正确（据我所知）：

xmppStream = [[XMPPStream alloc] init]; 
[xmppStream addDelegate:self delegateQueue:dispatch_get_main_queue()]; 

在ViewDidLoad方法。

任何帮助？谢谢。

---编辑---

我忘了说，这两个帐户知道对方，并在谷歌Talk客户端，存在被发送。

Answer 1

我找到了答案。我有两个类接收消息，因为A类必须收到一条消息才能触发推送类B的视图（此应用无法单独开始聊天对话）。所以，我设置了两个xmppStream，每个类都有一个。我在我的General类中放了一个xmppStream，让这两个类都接受了这个xmppStream，现在它发送消息。