How do I make a gui button that responds to both mouseclicks and keypresses?

Help and Feedback Scripting Support
#1

I am trying to make a button that can only be activated when a tool is equipped and that the button can be activated either by clicking on it or pressing a key but I’m stumped on a couple issues.
The code(Excerpt):

local function PressOrClick(ActionName, Inputstate, Inputobject1, Inputobject2)
	if ActionName == "Click" then
		UIS.InputBegan:Connect(function(Input,gameprocessed)
			if gameprocessed then return end
			print(Inputobject1)
			if Input.UserInputType == Inputobject1 then
				print("it's working 1")
			elseif Input.KeyCode == Inputobject2 then
				print("it's working 2")
			end
		end)
	end
end
Player.Backpack:WaitForChild("tool").Equipped:Connect(function()
	ToolMenu = function()
		ContextActionService:BindAction("Click", PressOrClick, false, Enum.UserInputType.MouseButton1,Enum.KeyCode.F)
	end
	ToolMenu()
end)

Player.Backpack:WaitForChild("tool").Unequipped:Connect(function()
	print("its not working")
	ToolMenu = function()
		ContextActionService:UnbindAction("Click", PressOrClick, false, Enum.UserInputType.MouseButton1,Enum.KeyCode.F)
	end
	ToolMenu()
end)

The “working 1” and “working 2” texts are not printing. What’s the issue here?

#2

The way I typically achieve this is by simply calling the callback function that is connected to a button’s MouseButton1Click event/signal whenever a particular key/input is pressed/entered.

local Enumeration = Enum
local Game = game
local UserInputService = Game:GetService("UserInputService")
local Script = script
local TextButton = Script.Parent

local function OnTextButtonClicked()
	print("Hello world!")
end

local function OnInputBegan(InputObject, GameProcessed)
	if GameProcessed then return end
	if InputObject.KeyCode == Enumeration.KeyCode.Q then
		OnTextButtonClicked()
	end
end

TextButton.MouseButton1Click:Connect(OnTextButtonClicked)
UserInputService.InputBegan:Connect(OnInputBegan)
#3

Hmm I have thought of that but is there no other way than using two separate functions? Is it not possible to find two avenues to using the same function?

#4

Well, I would first see what the keycode actually is.

I would also remove the elseif statement.

#5

Do you mind elaborating? First seeing what the keycode actually is?

#6

Simply just print out InputObject.KeyCode and InputObject.UserInputType.

#7

Maybe something like this will work for you:
ContextActionButtonBugHelp_.rbxl (37.9 KB)