l.behm/bumblebee-status
1
1
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity

[util.popup] Deduplicate code, "close" button only if leave=False

Browse source
This commit is contained in:
Naya Verdier 2020-07-25 08:43:26 -07:00
parent d358790c6a
commit 4aff0499f0
No known key found for this signature in database
GPG key ID: 1A59389D46A94A4C
1 changed files with 10 additions and 14 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

18
bumblebee_status/util/popup.py
View file

@ -16,23 +16,19 @@ class menu(object):
def __init__(self, parent=None, leave=True):
self.running = True
self.parent = None
if not parent:
self._root = tk.Tk()
self.parent = parent
self._root = parent.root() if parent else tk.Tk()
self._root.withdraw()
self._menu = tk.Menu(self._root, tearoff=0)
self._menu.bind("<FocusOut>", self.__on_focus_out)
self.add_menuitem("close", self.__on_focus_out)
self.add_separator()
else:
self._root = parent.root()
self._root.withdraw()
self._menu = tk.Menu(self._root, tearoff=0)
self._menu.bind("<FocusOut>", self.__on_focus_out)
self.parent = parent
if leave:
self._menu.bind("<Leave>", self.__on_focus_out)
elif not parent:
self.add_menuitem("close", self.__on_focus_out)
self.add_separator()
self._menu.bind("<ButtonRelease-1>", self.release)