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How do I make a localscript that opens a GUI when you click on it's parent?[Solved]

Asked by
JarFullOfMayonnaise 48
6 years ago
Edited 6 years ago

I'm trying to make a part that you have to insert an answer into. I added a ScreenGUI to starterGUI, and added this localscript to open it when you click on the script's parent.

local function onMC(playerwhoclicked)
    playerwhoclicked.PlayerGui.submit.TextBox.Visible = true
    playerwhoclicked.PlayerGui.submit.TextButton.Visible = true
end

script.Parent.ClickDetector.MouseClick:connect(onMC)

The problem is, the script doesn't open anything. Also, none of the errors in console link to the script when clicked on. Keep in mind, I don't know anything about localscripts other than that they are scripts that are local.

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Is the script located in a part or in a gui and is the script a local script or server script. Earthkingiv 51 — 6y
0
It's a local script and of course it's in a part, as he mentioned in his question. LeadRDRK 437 — 6y

1 answer

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Answered by
TheSkyofIndia 150
6 years ago 
--Place your gui in server storage
local name="ScreenGui" -- Place the name of your Gui here

function onclick(plr)
local x = game.ServerStorage.name:Clone()
x.parent = plr.PlayerGui --if your button is derectly in a screen gui then it whould be better if you type x.parent = script.Parent.Parent.Parent
end

script.Parent.ClickDetector.MouseButton1Click:Connect(onclick)

There can be mistakes in this script, if you find mistake please fix it, hope script help. And place it in normal script not in local script.

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