CODEWITHSUNDEEP
pythonwindows-10
Sergey Dudkin
Sergey Dudkin

Reputation: 199

After toast notification program leaves Tray Icon loop

I am trying to build simple Python application, which will add icon to Tray. I am also adding 2 menuitems to this tray icon, one to show Windows 10 toast message, and another one to stop application.

Program is launching correctly, icon appears, menu also appears. So, process, which I ran with self.icon.run() should wait, untill I will not call self.icon.stop()

The problem is, when I press first menuitem ("Test Notification") - toast notification is shown correctly, but AFTER this process goes out of self.icon.run() loop, and program finishes itself.

And I do not understand why is it so. According to documentation self.icon.run() should wait untill stop() call.

Am I doing something incorrecly? Here is my code:

from win10toast import ToastNotifier
from pystray import *
from PIL import Image

class SysTrayUI:
    def __init__(self):
        self.icon_path = "icon.png"

        self.menu_items = [
            MenuItem("Test Notification", lambda: toaster.show_toast("Test")),
            MenuItem("Exit", lambda: self.icon.stop()),
        ]
        self.menu = Menu(*self.menu_items)

        self.icon = Icon("Test Name", menu=self.menu)
        self.icon.icon = Image.open(self.icon_path)

        self.icon.run()

toaster = ToastNotifier()
app = SysTrayUI()

Upvotes: 1

Views: 919

Answers (2)

Be Champzz
Be Champzz

Reputation: 939

Threaded = True

This will help, Better you create a separate function and pass the message as shown below (This worked for me). It didn't work for me when I did the same without creating a function. So follow below format

def notify(msg):
    n = ToastNotifier()
    n.show_toast("Champzz", msg, duration= 10, threaded=True)

Upvotes: 2

Sergey Dudkin
Sergey Dudkin

Reputation: 199

Found the solution

toaster.show_toast function has threaded parameter, which allows to launch function in separate thread

So, if you write it like this:

MenuItem("Test Notification", lambda: toaster.show_toast("Test", threaded=True)),

Then this will not influence execution of self.icon.run()

Upvotes: 0

Related Questions