How to stop receiving LWT retained message? - mqtt

For testing, I have created a client with clean sessin = false and setWill(WILL_TOPIC, WILL_MSG.getBytes(), 1, true); and ran the App. Later, i set setWill(WILL_TOPIC, WILL_MSG.getBytes(), 1, false); when i run the App, i receive the previously reatined LWT messages. How to stop receiving them?

You aren't receiving LWT messages as such, but a retained message. To clear a retained message, send a zero length retained message to the same topic. For example, with the Paho Python client you would do:
client.publish(WILL_TOPIC, payload=None, retain=True)

You could send a CONNECT message with “CleanSession=false” and “Will=true”, and sends an empty WILLTOPIC message
check this section 6.3

Related

How to send a message from Helix controller to participant?

I want to send a message to all Participant nodes of my Helix cluster from Controller node. I tried following piece of code to send message to all registered participants of my cluster but their registered Participant message listeners are not receiving the message notification from Helix.
Message msg = new Message(factory.getMessageTypes().get(0), msgId);
msg.setMsgId(msgId);
msg.setSrcName(hostSrc);
msg.setTgtSessionId("*");
msg.setMsgState(MessageState.NEW);
msg.getRecord().setSimpleField("TestMessage", "Message from controller");
Criteria recipientCriteria = new Criteria();
recipientCriteria.setRecipientInstanceType(InstanceType.PARTICIPANT);
recipientCriteria.setInstanceName("%"); // To all recipients
recipientCriteria.setSessionSpecific(true); // To deliver only live participants
recipientCriteria.setSessionSpecific("DEV_CLUSTER"); // To only participants of this cluster
messagingService.send(recipientCriteria,msg);
Note that, when I am sending this message there is no resource exists in the cluster.
After debugging further what I have observed is CriteriaEvaluator.evaluateCriteria(....) operation is returning an empty list which further results into 0 messages to be sent to Participants nodes.
Kindly let me know if I am missing anything here while defining my criteria for Participants.
Thanks !
Update-1: our observation on this issue is as follows:
The received message at the participant side is read by both the Participant message listener(Say L1) and the handler created through MessageHandlerFactory(Which internally creating a listener HelixtaskExecutor (Say L2)).
In case if the message is read by HelixTaskExecutor(L2) first, it then immediately deletes the Znode in Zookeeper and the additionally configured message listener(L1) doesn't receive this message.
In case if the message is first read my additional message listener i.e. L1 then in such scenarios we don't face this problem as this additionally added listener doesn't delete the Znode from ZooKeeper.
We are still not sure how can we handle this problem as we want to use both the listener and MessageHandlers but facing the same problem I stated above.
Any inputs are appreciated.

Raid doesn't receive C_ChatInfo.SendAddonMessage

I'm making this addons that have to send to the raid my interrupt cooldown.
The problem is that whenever i send a message to the raid i am the only one that receive it.
This is the code that send the message:
C_ChatInfo.SendAddonMessage("KickRotation",string.format( "%0.2f",remainingCd ), "RAID")
This is the event handler:
frame:RegisterEvent("PLAYER_ENTERING_WORLD")
frame:RegisterEvent("CHAT_MSG_ADDON")
frame:SetScript("OnEvent", function(self, event, ...)
local prefix, msg, msgType, sender = ...;
if event == "CHAT_MSG_ADDON" then
if prefix == "KickRotation" then
print("[KickRotation]" ..tostring(sender) .." potrà interrompere tra: " ..msg);
end
end
if event == "PLAYER_ENTERING_WORLD" then
print("[KickRotation] v0.1 by Galfrad")
end
end)
Basically when the message is sended it is printed only to me.
Network messages are handled and transferred to the recipient channel (in this case, Raid Group) by the server. The reason that you are seeing the message locally, but the other people do not see it is that the message will be handled on the local system (sender) to reduce the repetition of data transmit.
Server however, only accepts and sends messages that are registered to it.
Therefore, you must first register your add-on messages to the server so the other players in the requested channel be able to receive it.
First, register your add-on messages with the name you have given already (But be sure to call the registration method only once per client):
local success = C_ChatInfo.RegisterAddonMessagePrefix("KickRotation") -- Addon name.
Next, check if your message was accepted and registered to the server. In case success is set to false (failure), you may want to handle proper warning messages and notifications to the user. The case of failure means that either server has disabled add-on messages or you have reached the limit of add-on message registrations.
Finally, send your message again check if it did not fail.
if not C_ChatInfo.SendAddonMessage("KickRotation",string.format( "%0.2f",remainingCd ), "RAID") then
print("[KickRotation] Failed to send add-on message, message rejected by the server.")
end

Mosquitto should not store and send offline messages

I am trying not to get the offline messages my scenario is if client1 is offline and client2 is sending messages so client1 should not receive any old messages when he reconnects and he should receive messages sent after reconnection. I am using mqtt library(npm) on client side and mosquitto server. I have tried {clean:true} and publish and subscribe using {qos:0} and its not working.This is my code
client2:
this.client = mqtt.connect(url, {
clean: true
}
this.client.publish("mqtt/location", JSON.stringify(data1) ,{qos: 0});
Client1:
this.client = mqtt.connect(url, {
clean: true
}
this.client.subscribe("mqtt/location", {qos: 0});
this.client.on("message", function(topic, payload) {
console.log(payload);
})
Thanks
If you are setting clean session to true, then the only other explanation is that the messages you are receiving were published with the retained bit set. There is nothing you can do to stop your client receiving those messages, but you can detect them. Any published message that you receive from the broker that has the retained bit set is "old".

LuaSocket - TCP 2nd message not sending

I've been searching Google for awhile and there seems to be no offers on fixing this problem I have here.
I am using LuaSocket as a simple way to connect to a external server I created, and I am able to connect to it successfully and send a signal.
However, when I try to send a second message later on, the external server does not seem to be receiving the message, even though I am still connected to the socket.
socket = require("socket")
host, port = ip, port
tcp = assert(socket.tcp())
tcp:settimeout( 0 )
tcp:connect(host, port);
msg = {
["status"]="connect",
["usrName"]=username
}
msg = Json.Encode(msg)
tcp:send(msg); -- This message, the server received this message.
-- Later in my code, I attempt to send another message.
msg = {
["status"]="anotherMessage",
["usrName"]=username
};
msg = Json.Encode(msg)
tcp:send(msg); -- This message is not sending, even though i'm still connected.
You need to show what happens on the other side as it may be simply not reading even though the connection may be open. You also don't say what exactly happens when "message is not sending"; do you get an error? the script finishes but the message is not sent?
There are several things you can try:
Switch to the (default) synchronous send until you get it working; remove tcp:settimeout(0), as your send may simply fail with "timeout" message if the other side is not ready to read the message.
Check the error message from :send call to see if it's timing out or not.
local ok, err = tcp:send(msg)
Use socket.select to check if the other side it ready to accept the message you are sending.
Try adding "\r\n" at the end of your serialized JSON.

How can I apply timeout function to LMAX Disruptor Queue?

To developers/users of LMAX Disruptor http://code.google.com/p/disruptor/ :
My question:
Can anyone suggest an approach to how apply a timeout function to Disruptor e.g. using EventHandler?
Here is one scenario that came up in my line of work:
Outbox - messages sent to the Server over a network
Inbox - ACK messages received from the Server
ACK Handler - marks outbox messages as ACKed
Timeout Handler - marks outbox message as NACKed (much needed, but where can it fit into the Disruptor design?)
Is there anyone who share the same opinion?
Or can anyone point out why it is unnecessary.
I hope the ensuing debate would be brief.
Thank you.
To clarify the timeout-handler would "fire" after a certain period of time when a message could not be delivered?
The way it works with disruptor is you have a ringbuffer for inbound and a ringbuffer for outbound messges... so email comes in, place it into the inbound ring buffer using an appropriate event. then process the message (i.e. decode, analye, log, store) and send it along to another sytem by placing it into the outbound ringbuffer... another handler takes the message and stores it into a database or sends it to another server using smtp... if a error / timeout etc. occurs, your create an event in the inbound ringbuffer signaling the error (NACK) and process this message. does that make sense?!?

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