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✓ ANSWER ACCEPTED

Help with simple Custon Admin Command?

Asked 232 views1 answers

Whoever is an admin and if they say "Kick all players" then it should kick all players, it does not work. :(

Admins = {"PreyStar"} --Only person that could say remove and does the command

game.Players.PlayerAdded:connect(function(Player)
	Player.Chatted:connect(function(Msg, Person)
		if Person == (Admins[1]) then
			print 'Worked?' --Does not print!!!!
			if Msg == "Kick all players" then
				for i,v in pairs (game.Players:GetChildren()) do
					v:Kick() --I wanna kick the players if an admin says the msg.
				end
			end
		end
	end)
end)
Anonymous

1 answer

✓ Accepted answer

In line 5 of this code block, Person should be Person.Name since you're identifying a string value. Also, I would personally recommend that you connect the listening event AFTER the if statement because you don't need the event to fire every time a non-admin chats.

Admins = {PreyStar = true, Player1 = true} --Only people that can use commands.

game.Players.PlayerAdded:connect(function(Player)
	if Admins[Player.Name] then
		Player.Chatted:connect(function(Msg, Person)
			print 'Worked?'
			if string.lower(Msg) == "kick all players" then
				for i,v in pairs (game.Players:GetChildren()) do
					v:Kick() -- Kick the players if an admin says the msg, including myself.
				end
			end
		end)
	end
end)
AnonymousEdited Permalink

2 comments

1

Just a little tip, it's better to use :GetPlayers() rather than :GetChildren() in this case. If an object that wasn't a "Player" was in Players, the script would error, causing not all players to be kicked. Also, you don't need to put = true on line 1. Just putting quotes around the name will do.

Anonymous#
0

@Shawnyg, You're right on the first point, but your second point is wrong. When using brackets, you have to give a table's key, not its value. table = {"a", "b"} print(table["a"]) --> nil

Anonymous#

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