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Old 09-16-2021, 09:07 AM   #1
gnai29
Human being with feelings
 
Join Date: Aug 2021
Posts: 3
Default Exiting REAPER after python script in a function is called with a button

Hello all,
I have this GUI made with the Scythe module, very straight forward and useful tool for this purpose.
I got recently blocked at trying to exit Reaper by calling one function when the user clicks that Button.
Code:
local function launch_potion()
  local potion_path = "~/Documents/REAPER/extract_potion.py"
  local cmd = string.format("~/venv/bin/python3 %s", potion_path)
  --reaper.atexit(os.execute('~/venv/bin/python3 ~/Documents/REAPER/extract_potion.py'))
  os.execute(cmd)
  --Scythe.quit = true

end

local window = GUI.createWindow({
    name = "SAUCE",
    w = 50,
    h = 50,
})

layer = GUI.createLayer({name = "Layer"})

layer:addElements(GUI.createElements(

{
    name = "launch_btn",
    type = "Button",
    x = 40,
    y = 40,
    w = 250,
    caption = "Launch Potion",
    func = launch_potion
}
))

window:addLayers(layer)
window:open()


GUI.Main()
So whenever the user click that `Launch Potion` Button, an external python script GUI is being launched, and my reaper GUI or even Reaper itself stay open unless I force quit from terminal or the usual force quit of MacOS.
I even tried including a `subprocess` in the py script calling `pkill REAPER` but no luck.
Not sure if that's even supported by Reaper.
I would appreciate it if someone can show me another way to do this please.
Thank you!

Last edited by gnai29; 09-16-2021 at 02:43 PM.
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