XMPPFramework - 如何创建MUC会议室并邀请用户? [英] XMPPFramework - How to create a MUC room and invite users?
问题描述
我正在使用Robbiehanson的iOS XMPPFramework。我正在尝试创建一个MUC会议室并邀请用户加入群聊室,但它无法正常工作。
I am using Robbiehanson's iOS XMPPFramework. I am trying to create a MUC room and invite a user to the group chat room but it is not working.
我使用以下代码:
XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"user101@conference.jabber.org/room" nickName:@"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:@"ABC@jabber.org"];
[room activate:[self xmppStream]];
[room inviteUser:jid1 withMessage:@"hello please join."];
[room sendMessage:@"HELLO"];
用户ABC@jabber.org应该收到邀请信息但没有任何事情发生。
The user ABC@jabber.org should receive the invite message but nothing is happening.
任何帮助将不胜感激。 :)
Any help will be appreciated. :)
推荐答案
在探索各种解决方案之后,我决定在这里编译和分享我的实现:
After exploring various solutions, I've decided to compile and share my implementation here:
-
创建XMPP会议室:
XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];
/**
* Remember to add 'conference' in your JID like this:
* e.g. uniqueRoomJID@conference.yourserverdomain
*/
XMPPJID *roomJID = [XMPPJID jidWithString:@"chat@conference.shakespeare"];
XMPPRoom *xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomStorage
jid:roomJID
dispatchQueue:dispatch_get_main_queue()];
[xmppRoom activate:[self appDelegate].xmppStream];
[xmppRoom addDelegate:self
delegateQueue:dispatch_get_main_queue()];
[xmppRoom joinRoomUsingNickname:[self appDelegate].xmppStream.myJID.user
history:nil
password:nil];
检查此代表是否成功创建了房间:
Check if room is successfully created in this delegate:
- (void)xmppRoomDidCreate:(XMPPRoom *)sender
检查您是否已加入此代表中的房间:
Check if you've joined the room in this delegate:
- (void)xmppRoomDidJoin:(XMPPRoom *)sender
创建房间后,获取房间配置表格:
After room is created, fetch room configuration form:
- (void)xmppRoomDidJoin:(XMPPRoom *)sender {
[sender fetchConfigurationForm];
}
配置房间
Configure your room
/**
* Necessary to prevent this message:
* "This room is locked from entry until configuration is confirmed."
*/
- (void)xmppRoom:(XMPPRoom *)sender didFetchConfigurationForm:(NSXMLElement *)configForm
{
NSXMLElement *newConfig = [configForm copy];
NSArray *fields = [newConfig elementsForName:@"field"];
for (NSXMLElement *field in fields)
{
NSString *var = [field attributeStringValueForName:@"var"];
// Make Room Persistent
if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) {
[field removeChildAtIndex:0];
[field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
}
}
[sender configureRoomUsingOptions:newConfig];
}
参考文献: XEP-0045:多用户聊天,实施群聊
邀请用户
- (void)xmppRoomDidJoin:(XMPPRoom *)sender
{
/**
* You can read from an array containing participants in a for-loop
* and send multiple invites in the same way here
*/
[sender inviteUser:[XMPPJID jidWithString:@"keithoys"] withMessage:@"Greetings!"];
}
在那里,你' ve创建了一个XMPP多用户/组聊天室,并邀请了一位用户。 :)
There, you've created a XMPP multi-user/group chat room, and invited a user. :)
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