bumblebee-status/bumblebee/popup_v2.py
Tobias Witek 6deb80edda [core/popup_v2] Destroy menu on leave
When mouse leaves the menu, destroy it.
2019-07-03 18:52:12 +02:00

41 lines
1,016 B
Python

"""Pop-up menus."""
import logging
try:
import Tkinter as tk
except ImportError:
# python 3
try:
import tkinter as tk
except ImportError:
logging.warning("failed to import tkinter - bumblebee popups won't work!")
import functools
class PopupMenu(object):
def __init__(self):
self._root = tk.Tk()
self._root.withdraw()
self._menu = tk.Menu(self._root)
self._menu.bind("<FocusOut>", self._on_focus_out)
self._menu.bind("<Leave>", self._on_focus_out)
def _on_focus_out(self, event=None):
self._root.destroy()
def _on_click(self, callback):
self._root.destroy()
callback()
def add_menuitem(self, menuitem, callback):
self._menu.add_command(label=menuitem, command=functools.partial(self._on_click, callback))
def show(self, event):
try:
self._menu.tk_popup(event['x'], event['y'])
finally:
self._menu.grab_release()
self._root.mainloop()