Is this possible ???

[PRED_UK]

Is it possible to make a bot  register in the hub so that the hublist picks up on it and displays it`s presence so that users know that bot is there ???

awaiting replies with anticipation.......

PRED_UK

LUA LANGAUGE BOARD FAN.....


Thanks in advance

[pHaTTy]

Originally posted by PRED_UK
Is it possible to make a bot  register in the hub so that the hublist picks up on it and displays it`s presence so that users know that bot is there ???

awaiting replies with anticipation.......

PRED_UK

LUA LANGAUGE BOARD FAN.....


Thanks in advance

yes there is, but im not gonna tell ya, its called faking users, u get banned from hublist for it

[PRED_UK]

ah i see, that would be pretty stupid in that case so i don`t think I`ll bother as I like my hub......

Thanks anyway ......

All I wanted to do was let people know that i have hub protection at work seeing as they keep trying to take my hub out.....

no problem

thanks m8

[VERMiN]

Most bot's have a name (or you may even choose one yourself for it) and will display themself's as a user.
So people will see that you have a protecting bot active, if that were what you ment?

[PRED_UK]

That`s sort of what i meant but when you scan down the users that are in the hub on hublist it shows there too.......

But as it`s already been explaned that would be classed as faking a user.....

so i think it`s best if this thread was deleted as i have no intentions of causing a problem or offense.....

PRED.......

[Stravides]

Rubbish, its not faking as the "user / bot" has a legitimate usage in the hub.....

If you want the name of your bot in the list you name it, after all its a bot, the description can say its a bot and I see no problem listing it... its when you are using 50 bots or creating false accounts thats where the crap starts.

top of the script name your bot ie

BOTName="Freddybot"

then you put in your func main

function Main()
   frmHub:UnregBot(BOTName)   -- to unlist the bot
   frmHub:RegBot(BOTName)  -- and/or to list the bot
end