9

I am creating a chat application using XMPP Framework in iphone. i could get received messages but i am not able to send a message. can any one give me solution for this??

4

4 回答 4

11 
- (void)sendMessage:(NSString *)msgContent
{

    NSString *messageStr = textField.text;

    if([messageStr length] > 0)
    {
        NSXMLElement *body = [NSXMLElement elementWithName:@"body"];
        [body setStringValue:messageStr];

        NSXMLElement *message = [NSXMLElement elementWithName:@"message"];
        [message addAttributeWithName:@"type" stringValue:@"chat"];
        [message addAttributeWithName:@"to" stringValue:[jid full]];
        [message addChild:body];

        [xmppStream sendElement:message];



    }
}

在你的chatViewcontroller 中使用上面的代码..它对我来说工作正常。

于 2011-03-02T07:00:44.000 回答
3

尝试这个 ：

XMPPUserCoreDataStorage *user = [[self fetchedResultsController] objectAtIndexPath:indexPath];

NSXMLElement *body = [NSXMLElement elementWithName:@"body"];
[body setStringValue:strSendMsg];

NSXMLElement *message = [NSXMLElement elementWithName:@"message"];
[message addAttributeWithName:@"type" stringValue:@"chat"];
[message addAttributeWithName:@"to" stringValue:[user.jid full]];
[message addChild:body];

[[self xmppStream] sendElement:message];
于 2010-12-20T04:50:43.440 回答
2

如果您使用的是 xmpp iPhone 示例应用程序...您可以使用以下内容，它应该可以帮助您入门：

NSString *msgText = @"test reply";

XMPPMessage* msg = [[XMPPMessage alloc] initWithType:@"chat" to:[XMPPJID jidWithString:displayName]];
[msg addBody:msgText];

[_xmppStream sendElement:msg];

只需将其放在 xmppStream 委托方法中的警报下方

iPhoneXMPPAppDelegate.m:

-(void)xmppStream:(XMPPStream *)sender didReceiveMessage:(XMPPMessage *)message

这将自动将“测试回复”发送回最初向您发送消息的 jid

哈哈！

于 2013-09-18T07:02:11.817 回答
0

斯威夫特 3 答案：

let user = XMPPJID(string: "user@example.com")
let msg = XMPPMessage(type: "chat", to: user)
msg?.addBody("test message")
self.xmppStream.send(msg)
于 2017-07-19T10:12:35.513 回答