need alert if more than 1 monster in same SQM

[Etee]

Hi

I need alert if somewhere on my screen is more than 1 monster in same SQM

Can someone help me?

[Etee]

Someone knows how to do it?

[Etee]

still waiting for that, Can someone help me?

[Imba]

https://tibiawindbot.com/docs.html#iterators-creature
https://tibiawindbot.com/docs.html#types-posx
https://tibiawindbot.com/docs.html#types-posy
https://tibiawindbot.com/docs.html#types-posz
https://tibiawindbot.com/docs.html#functions-playsound

Write if you won't manage to do it alone.

[Etee]

https://tibiawindbot.com/docs.html#iterators-creature
https://tibiawindbot.com/docs.html#types-posx
https://tibiawindbot.com/docs.html#types-posy
https://tibiawindbot.com/docs.html#types-posz
https://tibiawindbot.com/docs.html#functions-playsound

Write if you won't manage to do it alone.

@Imba

I really try to do that persistent and only lost my time.
I know how to do it if more than 1 monster is in exact position xyz. Here you are:

function floorCheck(x,y,z, dist, ...)

	local Monsters = {...}

	local amount = 0

	foreach creature m "m" do

		if table.find(Monsters, m.name) and m.posz == z and math.abs(m.posx - x) <= dist and math.abs(m.posy - y) <= dist then

			amount = amount + 1

		end

	end

	return amount

end



--



auto(1000)



if floorCheck(33608, 31950, 7, 0, "Rat") > 1 then

	playsound('playeronscreen.wav')

end

but I dont know how to make if SOMEWHERE (random xyz) on my screen is more than 1 monster is in same sqm.
Can you help me?

[terea]

is that possible to have more than 1 monster in same sqm?

[glowingstick]

is that possible to have more than 1 monster in same sqm?

If there's a door nearby, mwall the other side, force monster to enter the door sqm and it will stack on the opposite side when the door is closed

[mistgun]

@Etee, this could work, although i didn't test it.

init start

    local function findStackedCreatures()

        for x, y in screentiles() do

            local creatures = {}



            foreach creature m 'm' do

                if m.posx == x and m.posy == y and m.posz == $posz then

                    table.insert(creatures, m)

                end

            end



            if #creatures > 1 then

                return creatures

            end

        end



        return nil

    end

init end



auto(500)

local creatures = findStackedCreatures()

if creatures then

    print('Found more than one creature on the same sqm.')

end

[Etee]

@mistgun your action doesnt work.

When I put monster name

 local creatures = {"Rat"} 

and chage

 print('Found more than one creature on the same sqm.') 

to

 playsound('playeronscreen.wav') 

then your action play alert all the time if see ANY monster on my screen and that monster dont need to be in list (something like if I have rat on the list and I see only spider it play sound also)

[Donatello]

@mistgun your action doesnt work.

When I put monster name

 local creatures = {"Rat"} 

and chage

 print('Found more than one creature on the same sqm.') 

to

 playsound('playeronscreen.wav') 

then your action play alert all the time if see ANY monster on my screen and that monster dont need to be in list (something like if I have rat on the list and I see only spider it play sound also)

init start

    local function findStackedCreatures(...)

	local monsters = {...}

        for x, y in screentiles() do

            local creatures = {}



            foreach creature m 'm' do

                if m.posx == x and m.posy == y and m.posz == $posz and table.find(monsters, m.name) then

                    table.insert(creatures, m)

                end

            end



            if #creatures > 1 then

                return true

            end

        end

        return false

    end

init end



auto(500)

if findStackedCreatures("rat", "cave rat") then

	beep()

end