Running multiple Kivy apps at same time that communicate with each other

Question

I would like my Kivy application to be able to spawn multiple apps (i.e. new windows) on a Windows machine that can communicate with each other.

ScreenManager and Popup options will not cut it because they live in the same window..I need to be able to drag new screens across multiple monitors and therefore need multiple windows.

Kivy docs explicitly state that "Kivy supports only one window per application: please don't try to create more than one."

A google search produces this simple approach of simple spawning a new app from within another app, like so:

from kivy.app import App
from kivy.uix.button import Button
from kivy.uix.label import Label


class ChildApp(App):
    def build(self):
        return Label(text='Child')


class MainApp(App):

    def build(self):
        b = Button(text='Launch Child App')
        b.bind(on_press=self.launchChild)
        return b

    def launchChild(self, button):
        ChildApp().run()

if __name__ == '__main__':
    MainApp().run()

However, when I do this, it launches the app within the same window and crashes, and my terminal spits out like crazy:

Original exception was:
Error in sys.exceptionhook:

I get the same result if instead of ChildApp().run() I do multiprocessing.Process(target=ChildApp().run()).start()

Using the subprocess library gets me closer to what I want:

# filename: test2.py

from kivy.app import App
from kivy.uix.label import Label


class ChildApp(App):
    def build(self):
        return Label(text='Child')

if __name__ == '__main__':
    ChildApp().run()

---

# filename: test.py

from kivy.app import App
from kivy.uix.button import Button

import subprocess


class MainApp(App):

    def build(self):
        b = Button(text='Launch Child App')
        b.bind(on_press=self.launchChild)
        return b

    def launchChild(self, button):
        subprocess.call('ipython test2.py', shell=True)

if __name__ == '__main__':
    MainApp().run()

This spawns the child window without error, however now the main window is locked (white canvas) and if I close the child window, it just gets reopened.

They need to be able pass data between one another. Any ideas on how to do this correctly in Windows? This post seems to suggest that this is possible but I'm not sure where to start.

Answer 1

I tried baconwichsand's code and can confirm with Python 3.6 and Windows 10 it does not work. Apparently only top level object classes can be pickled, and since both apps inherit from the App class python throws an error. However a top level definition that simply executes the ChildApp().run() command can be pickled and works. Here is my working code.

import multiprocessing
from kivy.app import App
from kivy.uix.label import Label

class MainApp(App):
    def build(self):
        return Label(text='Main App Window')

class OtherApp(App):
    def build(self):
        return Label(text='Other App Window')

def open_parent():
    MainApp().run()

def open_child():
    OtherApp().run()

if __name__ == '__main__':
    a = multiprocessing.Process(target=open_parent)
    b = multiprocessing.Process(target=open_child)
    a.start()
    b.start()

And here is the code I am using, including the Builder to use a shared .kv file for both windows.

import multiprocessing
from kivy.lang import Builder
from kivy.app import App
from kivy.uix.button import Button
from kivy.uix.label import Label
from kivy.uix.widget import Widget

class MainRoot(Widget):
    pass

class OtherRoot(Widget):
    pass

class MainApp(App):
    def build(self):
        Builder.load_file('B:\Python_Codes\Testing Grounds\shared.kv')
        main = MainRoot()
        return main

class OtherApp(App):
    def build(self):
        Builder.load_file('B:\Python_Codes\Testing Grounds\shared.kv')
        other = OtherRoot()
        return other

def open_parent():
    MainApp().run()

def open_child():
    OtherApp().run()

if __name__ == '__main__':
    a = multiprocessing.Process(target=open_parent)
    b = multiprocessing.Process(target=open_child)
    a.start()
    b.start()